Math in Focus

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 11 Lesson 11.2 Area of a Circle detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 11 Lesson 11.2 Answer Key Area of a Circle

Math in Focus Grade 6 Chapter 11 Lesson 11.2 Guided Practice Answer Key

Complete. Use 3.14 as an approximation for π.

Question 1.
Find the area of a circle that has a radius of 18 centimeters.
Area = πr²
≈ •
= •
= cm²
The area of the circle is approximately square centimeters
Answer:
The area of the circle is approximately 1,017 square centimeters,

Explanation:
Given radius as 18 centimeters as the area of the circle is πr², So area = 3.14 X 18 cms X 18 cms = 1,017.36 cm² approximately 1,017 square centimeters.

Question 2.
Find the area of a circle that has a radius of 15 inches.
Area = πr²
≈ •
= •
= in.²
The area of the circle is approximately square inches.
Answer:
The area of the circle is approximately 707 square inches,

Explanation:
Given radius as 15 inches as the area of the circle is πr², So area = 3.14 X 15 in X 15 in =  706.5 in² approximately 707 square inches.

Question 3.
Find the area of a circle that has a diameter of 26 centimeters.
Radius = diameter ÷ 2
= ÷
= cm
Area = πr²
≈ • •
= •
= cm²
The area of the circle is approximately square centimeters.
Answer:
The area of the circle is approximately 531 square centimeters,

Explanation:
Given diameter as 26 cms so radius = diameter ÷ 2 = 26 cms ÷ 2 = 13 cms as the area of the circle is πr², So area = 3.14 X 13 cms X 13 cms = 530.66 cm² approximately 531 square centimeters.

Complete. Use \(\frac{22}{7}\) as an approximation for π.

Question 4.
Find the area of a quadrant.

The area of a quadrant is approximately square feet.
Answer:
The area of the quadrant is approximately 154 square feet,

Explanation:
Given radius is 14 ft as area of quardant is \(\frac{1}{4}\) X area of circle = \(\frac{1}{4}\) X πr² = \(\frac{1}{4}\) X 3.14 X 14 feet X 14 feet = 153.86 square feet approximately 154 square feet.

Question 5.
The diameter of a circle is 42 inches. Find the area of a quadrant.

Radius = diameter ÷ 2
= ÷ 2
= in.
Area of quadrant = • area of circle
= • πr²
≈ • •
= • • •
= in.²
The area of a quadrant is approximately square inches.
Answer:
The area of the quadrant is approximately 346 square inches,

Explanation:
Given diameter as 42 inches so radius = 42 inches/2 = 21 inches as area of quardant is \(\frac{1}{4}\) X area of circle = \(\frac{1}{4}\) X πr² = \(\frac{1}{4}\) X 3.14 X 21 inches X 21 inches = 346.185 square inches approximately 346 square inches.

Math in Focus Course 1B Practice 11.2 Answer Key

Find the area of each circle. Use 3.14 as an approximation for π.

Question 1.

Answer:
The area of the circle is 314 square centimeters,
Explanation:

Explanation:
Given radius as 10 cm as the area of the circle is πr², So area = 3.14 X 10 cm X 10 cm = 314 cm² .

Question 2.

Answer:
The area of the circle is approximately 1,963 square inches,

Explanation:
Given diameter as 50 inches so radius = diameter ÷ 2 = 50 in ÷ 2 = 25 in as the area of the circle is πr², So area = 3.14 X 25 in X 25 in =1962.5 inches² approximately 1,963 square inches.

Find the area of each semicircle. Use \(\frac{22}{7}\) as an approximation for π.

Question 3.

Answer:
The area of semicircle is 307.72 square feet,

Explanation:
Given diameter as 28 ft  so radius = diameter ÷ 2 = 28 ft ÷ 2 = 14 ft as the area of the circle is πr²,  So area = 3.14 X 14 ft X 14 ft = 615.44 square feet, The area of semicircle is half the area of the circle so it is \(\frac{1}{2}\) X
615.44 square feet = 307.72 square feet.

Question 4.

Answer:
The area of semicircle is 76.93 square meter,

Explanation:
Given radius as 7 m the area of the circle is πr²,  So area = 3.14 X 7 m X 7 m = 153.86 square meter, The area of semicircle is half the area of the circle so it is \(\frac{1}{2}\) X
153.86 square meter = 76.93 square meter.

Find the area of each quadrant to the nearest tenth. Use 3.14 as an approximation for π.

Question 5.

Answer:
The area of quardant is 113.04 square inches,

Explanation:
Given radius as 12 inches the area of the circle is πr²,  So area = 3.14 X 12 in. X 12 in. = 452.16 square inches, The area of quardant is 1/4 of the area of the circle so it is \(\frac{1}{4}\) X 452.16 square inches = 113.04 square inches.

Question 6.

Answer:
The area of quardant is 283.385 square meters,

Explanation:
Given radius as 19 meters the area of the circle is πr²,  So area = 3.14 X 19 m X 19 m = 1,133.54 square meters, The area of quardant is 1/4 of the area of the circle so it is \(\frac{1}{4}\) X 1,133.54 square meters = 283.385 square meters.

Solve. Show your work.

Question 7.
A circular pendant has a diameter of 7 centimeters. Find its area. Use \(\frac{22}{7}\) as an approximation for π.
Answer:
Area = 38.465 square centimeters,

Explanation:
Given a circular pendant has a diameter of 7 centimeters. So radius = diameter ÷ 2 = 7 cms ÷ 2 = 3.5 cms, As area of circle = πr², So area = 3.14 X 3.5 cms X 3.5 cms = 38.465 square centimeters.

Question 8.
The shape of the stage of a lecture theater is a semicircle. Find the area of the stage. Use 3.14 as an approximation for π.

Answer:
The area of the stage is 226.08 square meter,

Explanation:
Given shape of the stage of a lecture theater is a semicircle with radius 12 m, First area of circle is πr², So area = 3.14 X 12 m X 12 m = 452.16 square meter, Now area of semicircle is \(\frac{1}{2}\) X 452.16 sq mt = 226.08 square meter.

Question 9.
The shape of a balcony floor is a quadrant. Find the area of the balcony floor. Use 3.14 as an approximation for π.

Answer:
The area of the balcony floor is 113.04 square feet,

Explanation:
Given shape of a balcony floor is a quadrant with radius 12 ft, First area of circle is πr², So area = 3.14 X 12 ft X 12 ft = 452.16 square feet, Now area of quadrant is \(\frac{1}{4}\) X 452.16 square feet = 113.04 square feet.

Question 10.
The cost of an 8-inch pizza is $4. The cost of a 16-inch pizza is $13. Use 3.14 as an approximation for π.
a) How much greater is the area of the 16-inch pizza than the area of the 8-inch pizza?
Answer:
Greater is the area of the 16-inch pizza than the area of the 8-inch pizza is 150.72 square inches,

Explanation:
Area of 8 inch pizza radius is 8 inch ÷ 2 = 4 inch, Area of 8 inch pizza = 3.14 X 4 inch X 4 inch = 50.24 square inch,
Area of 16 inch pizza is 16 inch ÷ 2 = 8 inch, Area of 16 inch pizza = 3.14 X 8 inch X 8 inch = 200.96 square inch,
Now the area of the 16-inch pizza than the area of the 8-inch pizza is 200.96 square inch – 50.24 square inch = 150.72 square inches.

b) Which is a better deal? Explain your reasoning.
Answer:
8 inch pizza,

Explanation:
Cost of 16 inch pizza is $13, So cost of 1 inch pizza is $13/16 = $0.8125,
Cost of 8 inch pizza is $4, So cost of 1 inch pizza is $4/8 = $0.5 therefore it is better to buy 8 inch pizza as in 1 inch it cost $0.5 than 16 inch pizza in which 1 inch is $0.8125.

Question 11.
Four identical drinking glasses each have a radius of 5 centimeters. The glasses are arranged so that they touch each other as shown in the figure below. Find the area of the green portion. Use 3.14 as an approximation for π.

Answer:
The area of the green portion is 21.5 square centimeters,

Explanation:
Given to find the area of green portion so it is the area of square minus area of 4 quadrants so ((the area of square – 4(\(\frac{1}{4}\) X πr²)) = 10 cms x 10 cms – 4(\(\frac{1}{4}\) X 3.14 X 5 cms X 5 cms) = 100 square cms – 3.14 X 25 square cms = 100 square cms – 78.5 square cms =  21.5 square centimeters.

Question 12.
The figure is made up of trapezoid ABCD and a semicircle. The height of trapezoid ABCD is \(\frac{5}{6}\) the length of \(\overline{B C}\). Find the area of the figure. Use \(\frac{22}{7}\) as an approximation for π.

Answer:
Total area of trapezoid ABCD and a semicircle is 18,235.7925 square meters,

Explanation:
The area of trapezoid is 1/2 X (base1 + base2) X height, where base1 and base2 are lengths of parallel sides, we have height as \(\frac{5}{6}\) X 124.8 m = 104 m, Now area of trapezoid is 1/2 X (124.8 m + 119 m) X 104 m = 1/2 X 243.8 m X 104 m =12,677.6 square meters. First we calculate area of circle as diameter is 119 m so radius = 119 m/2 = 59.5 m, Area = 3.14 X 59.5 m x 59.5 m = 11,116.385 square meter, Now area of semicircle is half the area of circle = 11,116.385 square mt/2 = 5,558.1925 square meter, So total area of trapezoid ABCD and a semicircle is
area trapezoid + area of semicircle = 12,677.6 square meters + 5,558.1925 square meters = 18,235.7925 square meters.

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 11 Lesson 11.1 Radius, Diameter, and Circumference of a Circle to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 11 Lesson 11.1 Answer Key Radius, Diameter, and Circumference of a Circle

Math in Focus Grade 6 Chapter 11 Lesson 11.1 Guided Practice Answer Key

Question 1.
In the figure, 0 ¡s the center of the circle with \(\overline{A B}\), \(\overline{C D}\), and \(\overline{E D}\) as shown.

a) Name all the diameters that are drawn in the circle.
Answer:
AB, CD,

Explanation:
A diameter of a circle is any straight line segment that passes through the center of the circle and whose endpoints lie on the circle, all the diameters that are drawn in the circle are AB, CD.

b) Which line segment that joins two points on the circle ¡s not a diameter? Explain why it ¡s not a diameter.
Answer:
DE,

Explanation:
A line segment that crosses the circle by passing through its center is called a diameter. As DE doesnot cross through the circle.

Question 2.
The radius of a circle is 6 centimeters. What is the length of its diameter?
Diameter = 2 • radius
=
= cm
The diameter of the circle is centimeters.
Answer:
12 centimeters,

Explanation:
Given the radius of a circle is 6 centimeters, So the length of its diameter as diameter = 2 X radius,
Diameter = 2 X 6 centimeters = 12 centimeters.

Question 3.
The diameter of a circle is 15 inches. What is the length of its radius?
Radius = diameter ÷ 2
= ²
= in.
The radius of the circle is inches.
Answer:
7.5 inches,

Explanation:
Given the diameter of a circle is 15 inches, So the length of it radius is as radius = diameter ÷ 2 = 15 inches ÷ 2 = 7.5 inches.

Hands-On Activity

Materials

• compass

DRAWING CIRCLES USING A COMPASS

Step 1: Measure 5 centimeters on a ruler with a compass.
Example

Step 2: Then draw a circle. Draw and label the center O and four radii: \(\overline{O P}\), \(\overline{O Q}\), \(\overline{O R}\) and \(\overline{O S}\).
Example

Step 3: Measure the radii \(\overline{O P}\), \(\overline{O Q}\), \(\overline{O R}\), and \(\overline{O S}\). What can you say about the lengths of \(\overline{O P}\), \(\overline{O Q}\), \(\overline{O R}\), and \(\overline{O S}\)?
Answer:
\(\overline{O P}\) = 5 centimeters, \(\overline{O Q}\) = 5 centimeters, \(\overline{O R}\), = 5 centimeters and \(\overline{O S}\) = 5 centimeters,

Explanation:
The lengths of \(\overline{O P}\), \(\overline{O Q}\), \(\overline{O R}\), and \(\overline{O S}\) are 5 centimeters as we measured 5 centimeters on a ruler with a compass and then drawa a circle and labelled the center O and four radii \(\overline{O P}\), \(\overline{O Q}\), \(\overline{O R}\) and \(\overline{O S}\) so the lengtha are 5 cm each.

Hands-On Activity

INVESTIGATING THE RELATIONSHIP BETWEEN THE CIRCUMFERENCE AND DIAMETER OF A CIRCLE

Step 1: Lisa uses a string to measure the circumference of each circle to the nearest tenth of a centimeter and records it in a table.

Copy the table. Divide the circumference of each circle by its diameter. Round your answers to the nearest tenth. Record your results.

What do you notice about the quotients in the last column?
The circumference of any circle divided by its diameter always gives the same value.

The Greek letter π is used to represent this value.

Step 2: To see the value of π up to 9 decimal places, press π Enter on your calculator. Round the value of π to
a) the nearest tenth.
b) the nearest hundredth.
c) the nearest thousandth.

Step 3: In Step 1, you learned that the circumference of any circle divided by its diameter is equal to n. You can use this fact to write a formula for the circumference of a circle.
Complete the following statement.
Since circumference ÷ diameter = π,
Circumference = π •
You also know that the diameter of a circle is 2 times its radius. You can use this fact to write a related formula for the circumference of a circle. Complete the following statement.
Circumference = π • diameter
= π • 2 •
= 2 • π •
Using C for circumference, d for diameter, and r for radius, you can write these formulas as
C = πd
C = 2πr
Math Note
πd means π • d and 2πr means 2 • π • r.
Answer:

Explanation:
The quotients in the last column is the circumference of any circle divided by its diameter always gives the same value 3.14.

Copy and complete the table. Use \(\frac{22}{7}\) as an approximation for π.

Question 4.

Answer:

Explanation:
Given A has diameterr 14 cm so as circumfrence = πd so diameter is circumfrence/ π, 14 = circumfrence/ π,
so circumfrence = 14 X π = 14 X 3.14 = 43.96 cms, Now radius = circumfrence/ 2π = 43.96/ 2 X 3.14 = 43.96/6.28 = 7 cm. Circle B has radius 21 cm so diameter =  2r = 2 X 21 cm = 42 cms, circumfrence = 2πr = 2 X 3.14 X 21 cm = 131.88 cm and Circle C has radius 10.5 cm so diameter =  2r = 2 X 10.5 cm =  21 cms, circumfrence = 2πr = 2 X 3.14 X 10.5 cm = 65.94 cm. Completed the table as shown above.

Copy and complete the table. Use 3.14 as an approximation for π.

Question 5.

Answer:

Explanation:
Given D has diameterr 25 cm so as circumfrence = πd so diameter is circumfrence/ π, 25 = circumfrence/ π,
so circumfrence = 25 X π = 25 X 3.14 = 78.5 cms, Now radius = circumfrence/ 2π = 78.5/ 2 X 3.14 =78.5/6.28 = 12.5 cm. Circle E has radius 16 cm so diameter =  2r = 2 X 16 cm = 32 cms, circumfrence = 2πr = 2 X 3.14 X 16 cm = 100.48 cm and Circle F has radius 8.25 cm so diameter =  2r = 2 X 8.25 cm = 16.5 cms, circumfrence = 2πr = 2 X 3.14 X 8.25 cm = 51.81 cm. Completed the table as shown above.

Complete.

Question 6.
A circular hoop is cut into two equal parts. Its diameter is 35 inches. Find the length of each semicircular arc. Use \(\frac{22}{7}\) as an approximation for π.
Circumference of hoop = πd
≈
= in.
Length of each semicircular arc = \(\frac{1}{2}\) • circumference of hoop
=
= in.
The length of each semicircular arc is approximately inches.
Answer:
The length of each semicircular arc is approximately 55 inches,

Explanation:
Given a circular hoop is cut into two equal parts. Its diameter is 35 inches. So the length of each semicircular arc. Using \(\frac{22}{7}\) as an approximation for π. Circumference of hoop = πd = 3.14 X 35 inches = 109.9 inches, Now Length of each semicircular arc = \(\frac{1}{2}\) • circumference of hoop = \(\frac{1}{2}\) X 109.9 inches = 54.95 approximately 55 inches.

Question 7.
A quadrant is cut from a square. The side of the square is 10 centimeters. Find the length of the arc of the quadrant. Use 3.14 as an approximation for π.

Circumference of circle = 2πr
≈ 2 • •
= cm
Length of the arc of the quadrant = circumference ÷ 4
= ÷
= cm
The length of the arc of the quadrant is approximately centimeters.
Answer:
The length of the arc of the quadrant is approximately 15.7 centimeters.,

Explanation:
Given quadrant is cut from a square. The side of the square is 10 centimeters. So the length of the arc of the quadrant by using 3.14 as an approximation for π, Circumference of circle = 2πr,  2 X 3.14 X 10 cm =62.8 cms,
Length of the arc of the quadrant = circumference ÷ 4 = 62.8 ÷ 4 = 15.7 cm, therefore the length of the arc of the quadrant is approximately 15.7 centimeters.

Hands-On Activity

DRAWING SEMICIRCLES AND QUADRANTS

Materials

• compass
• drawing triangles
• ruler
• protractor

Step 1: Use a compass to draw a circle of radius 2 inches. Label the center O.

Step 2: Draw and label a diameter of the circle \(\overline{P Q}\). What do you notice?
A diameter of a circle divides it into semicircles.
This figure is one of the semicircles.

Find the distance around the semicircle. Use 3.14 as an approximation for π.

Step 3: In your circle, draw a second diameter perpendicular to \(\overline{P Q}\) using a ruler and protractor, or a drawing triangle. Label it \(\overline{R S}\). What do you notice?
a) Two perpendicular diameters of a circle divide it into quadrants.
This figure is one of the quadrants.

Find the distance around the quadrant. Use 3.14 as an approximation for π.
Answer:
A diameter of a circle divides it into 2 semicircles,
The distance around the quadrant is 3.14 inches,

b) This figure is made up of the semicircular arc, the arc of a quadrant, and the radii \(\overline{O R}\) and \(\overline{O Q}\). Find the distance around the figure. Use 3.14 as an approximation for π.

Answer:
The distance around the fiqure is 9.42 inches,

Explanation:
Step 1: Using a compass to draw a circle of radius 2 inches. Label the center O.
Step 2: Drawn and labelled a diameter of the circle \(\overline{P Q}\).
A diameter of a circle divides it into 2 semicircles.
Step 3: In my circle, drawn a second diameter perpendicular to \(\overline{P Q}\) using a ruler and protractor, or a drawing triangle. Labeledit \(\overline{R S}\).
Length of the arc of the quadrant = circumference ÷ 4, As circumfrence = 2πr,  2 X 3.14 X 2 in =12.56 inches,
= 12.56 inches/4 = 3.14 inches, The distance around the fiqure is 3 X 3.14 inches =  9.42 inches.

Math in Focus Course 1B Practice 11.1 Answer Key

Use the figure to complete. In the figure, O is the center of the circle and \(\overline{X Y}\) is a straight line.

Question 1.
\(\overline{O X}\), \(\overline{O Y}\), and \(\overline{O Z}\) are of the circle.
Answer:
\(\overline{O X}\), \(\overline{O Y}\), and \(\overline{O Z}\) are radius of the circle,

Explanation:
Radius of a circle is the distance from the center of the circle to any point on it’s circumference. So \(\overline{O X}\), \(\overline{O Y}\), and \(\overline{O Z}\) are radius of the circle.

Question 2.
XY is a of the circle.
Answer:
Diameter,

Explanation:
The diameter of a circle is any straight line segment that passes through the center of the circle and whose endpoints lie on the circumference of the circle. So XY ia a diameter of the circle.

Question 3.
OX = =
Answer:
OX = OY = OZ,

Explanation:
As OX is radius of the circle and the distance of OY and OZ is same so OX = OY = OZ.

Question 4.
XY = • OZ
Answer:
2 X OZ,

Explanation:
As XY id diameter so XY = 2 X OZ.

Question 5.
Circumference of the circle = π •
Answer:
Circumference of the circle = 2 X π X OX or 2 X π X OY or 2 X π X OZ,

Explanation:
As radius of circle is OX, OY, OZ so Circumference of the circle = 2 X π X OX or 2 X π X OY or 2 X π X OZ.

Find the circumference of each circle. Use \(\frac{22}{7}\) as an approximation for π.

Question 6.

Answer:
The circumfrence is 43.96 cms,

Explanation:
As we know circumference of circle is C = 2πr given radius r = 7 cm so circumfrence = 2 X 3.14 X 7 cm = 43.96 cms.

Question 7.

Answer:
The circumfrence is 65.94 inches,

Explanation:
As we know circumference of circle is C = πd given diameter d = 21 inches so  circumfrence = 3.14 X 21 inches = 65.94 inches.

Question 8.

Answer:
The circumfrence is 153.86 m,

Explanation:
As we know circumference of circle is C = πd given diameter d = 49 m so circumfrence = 3.14 X 49 m = 153.86 m.

Question 9.

Answer:
The circumfrence is 17.584 cms,

Explanation:
As we know circumference of circle is C = 2πr given radius r = 2.8 ft so circumfrence = 2 X 3.14 X 2.8 ft = 17.584 cms.

Find the length of each arc. Use \(\frac{22}{7}\) as an approximation for π.

Question 10.

Answer:
Length of the arc = 12.089 cm,

Explanation:
Length of arc = \(\frac{1}{2}\) X circumference, Given diameter d = 7.7 cm as we know C = πd so length of
arc is \(\frac{1}{2}\) X 3.14 X 7.7 cm = 12.089 cm.

Question 11.

Answer:
Length of the arc = 15.386 ft,

Explanation:
Length of arc = \(\frac{1}{2}\) X circumference, Given diameter d = 9.8 ft as we know C = πd so length of
arc is \(\frac{1}{2}\) X 3.14 X 9.8 ft = 15.386 ft.

Question 12.

Answer:
Length of the arc = 3.297 m,

Explanation:
Length of the arc of the quadrant = circumference ÷ 4 and C = πd or 2πr as given radius = 2.1 m so circumfrence = 2 X 3.14 X 2.1 m = 13.188 m, So length of the arc =  13.188 m ÷ 4 = 3.297 m.

Question 13.

Answer:
Length of the arc = 17.584 cm,

Explanation:
Length of the arc of the quadrant = circumference ÷ 4 and C = πd or 2πr as given radius = 11.2 cm so circumfrence = 2 X 3.14 X 11.2 cm = 70.366 cm, So length of the arc =  70.366 cm ÷ 4 = 17.584 cm.

Find the distance around each semicircle. Use 3.14 as an approximation for π.

Question 14.

Answer:
Length of the semicircle = 31.4 in,

Explanation:
Length of semicircle = \(\frac{1}{2}\) X circumference, Given diameter d = 20 in as we know C = πd so length of semicircle is \(\frac{1}{2}\) X 3.14 X 20 in = 31.4 in.

Question 15.

Answer:
Length of the semicircle = 31.4 in,

Explanation:
Length of semicircle = \(\frac{1}{2}\) X circumference, Given diameter d = 10 cm as we know C = πd so length of semicircle is \(\frac{1}{2}\) X 3.14 X 10 cm = 31.4 in.

Question 16.

Answer:
Length of semicircle = 78.5 m,

Explanation:
Length of the semicircle = \(\frac{1}{2}\) X circumference and C = 2πr as given radius = 25 m so length of the semicircle = \(\frac{1}{2}\) X 2 X 3.14 X 25 m = 78.5 m.

Question 17.

Answer:
Length of semicircle = 23.55 ft,

Explanation:
Length of the semicircle = \(\frac{1}{2}\) X circumference and C = 2πr as given radius = 7.5 ft so length of the semicircle = \(\frac{1}{2}\) X 2 X 3.14 X 7.5 ft = 23.55 ft.

Find the distance around each quadrant. Use \(\frac{22}{7}\) as an approximation for π.

Question 18.

Answer:
Distance around quadrant = 5.495 in,

Explanation:
Distance of the quadrant = circumference ÷ 4 and C = πd or 2πr as given radius = 3.5 in. So circumfrence = 2 X 3.14 X 3.5 in = 21.98 in , So length of the quadrant =  21.98 in ÷ 4 = 5.495 in.

Question 19.

Answer:
Distance around quadrant = 16.485 cm,

Explanation:
Distance of the quadrant = circumference ÷ 4 and C = πd or 2πr as given radius = 10.5 cm so circumfrence = 2 X 3.14 X 10.5 cm = 65.94 cm, So length of the quadrant =  65.94 cm ÷ 4 = 16.485 cm.

Question 20.

Answer:
Distance around quadrant = 27.475 m,

Explanation:
Distance of the quadrant = circumference ÷ 4 and C = πd or 2πr as given radius = 17.5 m so circumfrence = 2 X 3.14 X 17.5 m = 109.9 m , So length of the quadrant =  109.9 m ÷ 4 = 27.475 m.

Question 21.

Answer:
Distance around quadrant = 43.96 ft,

Explanation:
Distance of the quadrant = circumference ÷ 4 and C = πd or 2πr as given radius = 28 ft so circumfrence = 2 X 3.14 X 28 ft = 175.84 ft, So length of the quadrant =  175.84 ft ÷ 4 = 43.96 ft.

Solve. Show your work. Use 3.14 as an approximation for π.

Question 22.
A circular garden has a diameter of 120 feet. Find its circumference.
Answer:
The circumfrence is 376.8 feet,

Explanation:
Given a circular garden has a diameter of 120 feet as we know circumference of circle is C = πd  given diameter = 120 feet so circumfrence = 3.14 X 120 ft = 376.8 feet.

Question 23.
A circular coaster has a radius of 4 centimeters. Find its circumference.
Answer:
The circumfrence is 25.12 cms,

Explanation:
Given a circular coaster has a radius of 4 centimeters as we know circumference of circle is C = 2πr given radius r = 4 cm so circumfrence = 2 X 3.14 X 4 cms =  25.12 cms.

Question 24.
The diameter of a roll of tape is 5\(\frac{1}{2}\) centimeters. Find its circumference.
Answer:
The circumfrence is 17.27 centimeters,

Explanation:
Given a diameter of a roll of tape 5\(\frac{1}{2}\) centimeters as we know circumference of circle is C = πd  given diameter = 5\(\frac{1}{2}\) centimeters so circumfrence = 3.14 X  5\(\frac{1}{2}\) centimeters = 3.14 X \(\frac{11}{2}\) centimeters = 17.27 centimeters.

Question 25.
The shape of a floor mat is a semicircle. Find the distance around the mat.

Answer:
Diatance around the mat = 119.32 cm,

Explanation:
Distance of semicircle = \(\frac{1}{2}\) X circumference, Given diameter d = 38 cm + 38 cm = 76 cm as we know C = πd so length of semicircle is \(\frac{1}{2}\) X 3.14 X 76 cm = 119.32 cm.

Question 26.
A small playground is shaped like a quadrant, as shown. Find the distance around the playground.

Answer:
The distance around the playground is 37.68 ft,

Explanation:
Given a small playground is shaped like a quadrant so distance of the quadrant = circumference ÷ 4 and C = πd or 2πr as given radius = 24 ft so circumfrence = 2 X 3.14 X 24 ft =150.72 ft, So length of the arc = 150.72 ft ÷ 4 = 37.68 ft.

Question 27.
If the radius of a wheel is 14 inches, what is the distance traveled when the wheel turns around 100 times?

Answer:
The distance traveled when the wheel turns around 100 times is 8,792 inches,

Explanation:
Given the radius of a wheel is 14 inches, the distance traveled is circumference so it is C = πd or 2πr as given radius = 14 in so circumfrence = 2 X 3.14 X 14 inchs = 87.92 inches so the distance traveled when the wheel turns around 100 times is 100 X 87.92 inches = 8,792 inches.

Find the distance around each figure. Use \(\frac{22}{7}\) as an approximation for π.

Question 28.
The figure is made up of a rectangle and a semicircle.

Answer:
Length of a rectangle is 80 cm,
Length of the semicircle = 43.96 cm,

Explanation:
Length or Perimeter of rectangle = 2(length + width) = 2(28 cm + 12 cm) = 2(40 cm) = 80 cm,
Length of semicircle = \(\frac{1}{2}\) X circumference, Given diameter d = 28 cm as we know C = πd so length of semicircle is \(\frac{1}{2}\) X 3.14 X 28 cm = 43.96 cm.

Question 29.
The figure is made up of an equilateral triangle and a semicircle.

Answer:
Length of equilateral triangle =
Length of the semicircle = 65.94 cm,

Explanation:
Length of equilateral triangle = 42 cm + 42 cm + 42 cm = 126 cm,
Length of semicircle = \(\frac{1}{2}\) X circumference, Given diameter d = 42 cm as we know C = πd so length of semicircle is \(\frac{1}{2}\) X 3.14 X 42 cm = 65.94 cm.

Find the distance around each figure. Use 3.14 as an approximation for π.

Question 30.
The figure is made up of a rectangle and two identical semicircles.

Answer:
Length of a rectangle is 172 ft,
The length of each identical semicircles is 50.24 ft,

Explanation:
Length or Perimeter of rectangle = 2(length + width) = 2(54 ft + 32 ft) = 2 X 86 ft = 172 ft,
Length of semicircle = \(\frac{1}{2}\) X circumference, Given diameter d = 32 ft as we know C = πd so length of semicircle is \(\frac{1}{2}\) X 3.14 X 32 ft = 50.24 ft.

Question 31.
The figure is made up of two identical quadrants, a semicircle, and an equilateral triangle.

Answer:
The length of each quardrants is 36.11 ft,
The length of a semicircle is 72.22 ft,
Length of equilateral triangle is 69 ft,

Explanation:
Distance of the quadrant = circumference ÷ 4 and C = πd or 2πr as given radius = 23 ft so circumfrence = 2 X 3.14 X 23 ft = 144.44 ft, So length of the each quadrant = 144.44 ft ÷ 4 = 36.11 ft.
Length of semicircle = \(\frac{1}{2}\) X circumference, Given radius = 23 ft as we know C = 2πr so length of semicircle is \(\frac{1}{2}\) X 3.14 X 2 X 23 ft = 72.22 ft. Length of equilateral triangle is 23 ft + 23 ft + 23 ft = 69 ft.

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 11 Circumference and Area of a Circle to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 11 Answer Key Circumference and Area of a Circle

Math in Focus Grade 6 Chapter 11 Ouick Check Answer Key

Add.

Question 1.
0.451 + 3.12
Answer:
3.571,

Explanation:
Adding 3.12 to 0.451 we will get
0.451
+3.120
3.571

Question 2.
0.861 + 6.95
Answer:
7.811,

Explanation:
Adding 6.95 to 0.861 we will get
0.861
+6.950
7.811

Question 3.
13.74 + 3.791
Answer:
17.531,

Explanation:
Adding 3.791 to 13.74 we will get
13.740
+3.791
17.531

Subtract.

Question 4.
5.45 – 1.78
Answer:
3.67,

Explanation:
Subtacting 1.78 from 5.45 we will get
5.45
-1.78
3.67

Question 5.
12.795 – 0.816
Answer:
11.979,

Explanation:
Subtracting 0.816 from 12.795 we will get
12.795
-0.816
11.979

Question 6.
42.781 – 36.19
Answer:
6.591

Explanation:
Subtracting 36.19 from 42.781 we will get
42.781
-36.190
6.591

Multiply

Question 7.
1.34 × 9
Answer:
12.06,

Explanation:
Given to multiply 1.34 with 9 we will get
3,3
1.34
X 9
12.06

Question 8.
4.246 × 2
Answer:
8.492,

Explanation:
Given to multiply 4.246 by 2 we will get
1
4.246
X   2
8.492

Question 9.
7.487 × 8
Answer:
59.896,

Explanation:
Given to multiply 7.487 by 8 we will get
3,6,5
7.487
X  8
59.896

Divide

Question 10.
2.56 ÷ 5
Answer:
0.512,

Explanation:
5)2.56(0.512
   2.5
0.06
0.05
0.01
0.01
0

Question 11.
2.429 ÷ 7
Answer:
0.347,

Explanation:
7)2.429(0.347
   2.1
    0.32
0.28
0.049
0.049
0

Question 12.
1.143 ÷ 9
Answer:
0.127,

Explanation:
9)1.143(0.127
   0.9
   0.24
    0.18
   0.063
    0.063
      0

Question 13.
4.671 ÷ 9
Answer:
0.519,

Explanation:
9)4.671(0.519
   4.5
   0.17
   0.09
  0.081
   0.081
  0

Question 14.
0.656 ÷ 8
Answer:
0.082,

Explanation:
8)0.656(0.082
    0.64
    0.016
     0.016
0

Question 15.
0.867 ÷ 3
Answer:
0.289,

Explanation:
3)0.867(0.289
    0.6
0.26
0.24
0.027
0.027
0

Round to the nearest whole number.

Question 16.
4.56
Answer:
5,

Explanation:
Asked to round 4.56 to the nearest whole number so if number is less than 5 then round the number down by removing the decimal part of the number, If it is 5 or more then round the number up by adding one on to the ones digit and removing the decimal part of the number. So 4.56 is 5.

Question 17.
12.05
Answer:
12,

Explanation:
Asked to round 12.05 to the nearest whole number so if number is less than 5 then round the number down by removing the decimal part of the number, If it is 5 or more then round the number up by adding one on to the ones digit and removing the decimal part of the number. So 12.05 is 12.

Question 18.
6.48
Answer:
6,

Explanation:
Asked to round 6.48 to the nearest whole number so if number is less than 5 then round the number down by removing the decimal part of the number, If it is 5 or more then round the number up by adding one on to the ones digit and removing the decimal part of the number. So 6.48 is 6.

Question 19.
6.50
Answer:
7,

Explanation:
Asked to round 6.50 to the nearest whole number so if number is less than 5 then round the number down by removing the decimal part of the number, If it is 5 or more then round the number up by adding one on to the ones digit and removing the decimal part of the number. So 6.50 is 7.

Question 20.
14.15
Answer:
14,

Explanation:
Asked to round 14.15 to the nearest whole number so if number is less than 5 then round the number down by removing the decimal part of the number, If it is 5 or more then round the number up by adding one on to the ones digit and removing the decimal part of the number. So 14.15 is 14.

Question 21.
46.59
Answer:
47,

Explanation:
Asked to round 46.59 to the nearest whole number so if number is less than 5 then round the number down by removing the decimal part of the number, If it is 5 or more then round the number up by adding one on to the ones digit and removing the decimal part of the number. So 46.59 is 47.

Round to the nearest tenth.

Question 22.
7.68
Answer:
7.7,

Explanation:
To round a number to the nearest tenth, we need to look to the right of the tenths place, which is the hundredths place. If the digit in the hundredths place is 5 or greater than 5, we need to add one to the digit in the tenths place. If it is 4 or less than 4, the tenths place digit remains the same. All the digits to the right of the tenths place should be changed to 0. That means, while rounding off a number to the nearest tenth, we will stop the rounding at the tenths place. So 7.68 is 7.7.

Question 23.
3.05
Answer:
3.00,

Explanation:
To round a number to the nearest tenth, we need to look to the right of the tenths place, which is the hundredths place. If the digit in the hundredths place is 5 or greater than 5, we need to add one to the digit in the tenths place. If it is 4 or less than 4, the tenths place digit remains the same. All the digits to the right of the tenths place should be changed to 0. That means, while rounding off a number to the nearest tenth, we will stop the rounding at the tenths place. So 3.05 is 3.00.

Question 24.
19.92
Answer:
20.00,

Explanation:
To round a number to the nearest tenth, we need to look to the right of the tenths place, which is the hundredths place. If the digit in the hundredths place is 5 or greater than 5, we need to add one to the digit in the tenths place. If it is 4 or less than 4, the tenths place digit remains the same. All the digits to the right of the tenths place should be changed to 0. That means, while rounding off a number to the nearest tenth, we will stop the rounding at the tenths place. So 19.92 is 20.00.

Question 25.
5.55
Answer:
5.60,

Explanation:
To round a number to the nearest tenth, we need to look to the right of the tenths place, which is the hundredths place. If the digit in the hundredths place is 5 or greater than 5, we need to add one to the digit in the tenths place. If it is 4 or less than 4, the tenths place digit remains the same. All the digits to the right of the tenths place should be changed to 0. That means, while rounding off a number to the nearest tenth, we will stop the rounding at the tenths place. So 5.55 is 5.60.

Question 26.
8.17
Answer:
8.0,

Explanation:
To round a number to the nearest tenth, we need to look to the right of the tenths place, which is the hundredths place. If the digit in the hundredths place is 5 or greater than 5, we need to add one to the digit in the tenths place. If it is 4 or less than 4, the tenths place digit remains the same. All the digits to the right of the tenths place should be changed to 0. That means, while rounding off a number to the nearest tenth, we will stop the rounding at the tenths place. So 8.17 is 8.0.

Question 27.
2.44
Answer:
2.4,

Explanation:
To round a number to the nearest tenth, we need to look to the right of the tenths place, which is the hundredths place. If the digit in the hundredths place is 5 or greater than 5, we need to add one to the digit in the tenths place. If it is 4 or less than 4, the tenths place digit remains the same. All the digits to the right of the tenths place should be changed to 0. That means, while rounding off a number to the nearest tenth, we will stop the rounding at the tenths place. So 2.44 is 2.4.

Question 28.
43.65
Answer:
43.7,

Explanation:
To round a number to the nearest tenth, we need to look to the right of the tenths place, which is the hundredths place. If the digit in the hundredths place is 5 or greater than 5, we need to add one to the digit in the tenths place. If it is 4 or less than 4, the tenths place digit remains the same. All the digits to the right of the tenths place should be changed to 0. That means, while rounding off a number to the nearest tenth, we will stop the rounding at the tenths place. So 43.65 is 43.7.

Question 29.
23.73
Answer:
23.8,

Explanation:
To round a number to the nearest tenth, we need to look to the right of the tenths place, which is the hundredths place. If the digit in the hundredths place is 5 or greater than 5, we need to add one to the digit in the tenths place. If it is 4 or less than 4, the tenths place digit remains the same. All the digits to the right of the tenths place should be changed to 0. That means, while rounding off a number to the nearest tenth, we will stop the rounding at the tenths place. So 23.73 is 23.8.

Question 30.
17.51
Answer:
17.5,

Explanation:
To round a number to the nearest tenth, we need to look to the right of the tenths place, which is the hundredths place. If the digit in the hundredths place is 5 or greater than 5, we need to add one to the digit in the tenths place. If it is 4 or less than 4, the tenths place digit remains the same. All the digits to the right of the tenths place should be changed to 0. That means, while rounding off a number to the nearest tenth, we will stop the rounding at the tenths place. So 17.51 is 17.5.

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 9 The Coordinate Plane to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 9 Answer Key The Coordinate Plane

Math in Focus Grade 6 Chapter 9 Quick Check Answer Key

Use the coordinate plane below.

Question 1.
Give the coordinates of points B, C, and D.
Answer:
Point B:
The first coordinate represents distance along x axis which is 3 and the second coordinate represents distance along y axis which is 3.
So, the point P will be (3,3).
Point C:
The first coordinate represents distance along x axis which is 1 and the second coordinate represents distance along y axis which is 4.
So, the point C will be (1,4).
Point D:
The first coordinate represents distance along x axis which is 5 and the second coordinate represents distance along y axis which is 2.
So, the point D will be (5,2).

Use graph paper. Plot the points on a coordinate plane.

Question 2.
P (3, 2), Q (2, 3), and R (0, 4)
Answer:
To plot point P (3, 2):
Here, the x-coordinate is 3 and the y-coordinate is 2. Start at the Origin. As the x coordinate is positive, move 3 units along the positive x-axis and 2 units along positive y -axis.
Thus, the required point P (3, 2) is marked.
To plot point Q (2, 3):
Here, the x-coordinate is 2 and the y-coordinate is 3. Start at the Origin. As the x coordinate is positive, move 2 units along the positive x-axis and 3 units along positive y -axis.
Thus, the required point P (2, 3) is marked.
To plot point R (0, 4):
Here, the x-coordinate is 0 and the y-coordinate is 4.  As the y coordinate is positive, move 5 units along the positive y-axis and mark it.
Thus, the required point(0, 4) is marked

Identify the number that each indicated point represents.

Question 3.

Answer:
To the left of the origin, the x coordinate values will be negative.
Start counting from the origin and note down the values.
The first value will be -1, next will be -4, -6 and -8.

Draw a horizontal number line to represent each set of numbers.

Question 4.
-3, 0, 1, 5, 8
Answer:
-3 is the negative value; therefore mark it to the left of the origin.
0 is the origin. 1,5,8 are the positive values, which are to be marked to the right of the origin.

Question 5.
-15, -11, -9, -7, -2
Answer:
Here, all are negative values. Hence all the values are to be marked to the left of the origin.

Use the symbol || to write the absolute values of the following numbers.

Question 6.
11
Answer:
The distance of an integer from ‘

$0$

‘ on the number line irrespective of its direction is called the absolute value of that integer.
Two vertical bars ‘| |’ are used to denote the absolute value.
The absolute value of any positive number is the number itself.
Therefore, the absolute value of 11 will be 11.

Question 7.
-16
Answer:
The absolute value of negative number is the positive of it.
Therefore, the absolute value of -16 or |-16| will be 16.

Question 8.
-21
Answer:
The absolute value of negative number is the positive of it.
Therefore, the absolute value of -21 or |-21| will be 21.

Find the perimeter of each polygon.

Question 9.
Figure ABC is an isosceles triangle.

Answer:
Given triangle ABC has two equal sides, thus it will form an isosceles triangle.
The triangle will measure 8 in, 8 in and 5 in.
Perimeter of the triangle will be the sum of all sides, 8+8+5 = 21 in

Question 10.
Figure DEFis an equilateral triangle.

Answer:
Given DEF is an equilateral triangle. Equilateral trinagle will have all equal sides.
All sides of the triangle will measure 4 in.
Perimeter of the triangle will be the sum of all sides, 4+4+4 = 12 in

Question 11.
Figure PQRS is a trapezoid.

Answer:
Given figure PQRS is a trapezoid. One of the pair of opposite sides are equal in length.
The trapezoid measures 16cm,10cm and 9cm.
Perimeter of the trapezoid will be the sum of all sides, 16+10+9+9 = 44cm

Question 12.
Figure WXYZ is a parallelogram.

Answer:
Given figure WXYZ is a parallelogram. The opposite sides are equal in length.
The parallelogram measures 8cm and 7cm.
Perimeter of the parallelogram will be the sum of all sides, 8+7+8+7 = 30cm

Question 13.
Figure JKLM is a rhombus.

Answer:
Given figure JKLM is a rhombus. All the sides are equal in length.
Each side will measure 6m in length.
Perimeter of the rhombus will be the sum of all sides, 6+6+6+6 = 24cm

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 9 Lesson 9.1 Points on the Coordinate Plane to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 9 Lesson 9.1 Answer Key Points on the Coordinate Plane

Math in Focus Grade 6 Chapter 9 Lesson 9.1 Guided Practice Answer Key

Use the coordinate plane below.

Question 1.
Give the coordinates of points P, Q, R, S, T, U, and V. In which quadrant is each point located?

Answer:
Point P:
The first coordinate represents distance along x axis which is 3 and the second coordinate represents distance along y axis which is 0.
So, the point P will be (3,0).
The point having y-coordinate as 0 lie on the X-axis and thus is not located in any quadrant.

Point Q:
The first coordinate represents distance along x axis which is -3 and the second coordinate represents distance along y axis which is 3.
So, the point Q will be (-3,3).
If the x axis coordinate is negative and y axis coordinate is positive, the point will lie in quadrant II.

Point R:
The first coordinate represents distance along x axis which is -5 and the second coordinate represents distance along y axis which is 1.
So, the point R will be (-5,1).
If the x axis coordinate is negative and y axis coordinate is positive, the point will lie in quadrant II.

Point S:
The first coordinate represents distance along x axis which is -6 and the second coordinate represents distance along y axis which is -4.
So, the point S will be (-6,-4).
If the both the coordinates are negative then, the point will lie in quadrant III.

Point T:
The first coordinate represents distance along x axis which is -4 and the second coordinate represents distance along y axis which is -7.
So, the point T will be (-4,-7).
If the both the coordinates are negative then, the point will lie in quadrant III.

Point U:
The first coordinate represents distance along x axis which is 4 and the second coordinate represents distance along y axis which is -5.
So, the point U will be (4,-5).
If the x axis coordinate is positive and y axis coordinate is negative , the point will lie in quadrant IV.

Point V:
The first coordinate represents distance along x axis which is 3 and the second coordinate represents distance along y axis which is -2.
So, the point V will be (3,-2).
If the x axis coordinate is positive and y axis coordinate is negative , the point will lie in quadrant IV.

Use graph paper.

Question 2.
Plot points A (-4, 3), 6(3, -4), C (5, 0), D (0, -5), E(-2, -1), and F(2, -1)on a coordinate plane.
Answer:
To plot point A(-4,3):
Here, the x-coordinate of A is -4 and the y-coordinate is 3. Start at the Origin. As the x coordinate is negative, move 4 units to the left along the x-axis and 3 units to the top along y -axis.
Thus, the required point A(-4,3) is marked.

To plot point B(3,-4):
Here, the x-coordinate of B is 3 and the y-coordinate is -4. Start at the Origin. As the x coordinate is positive, move 3 units to the right along the x-axis. As y cooridnate is negative, move 4 units to the down along y -axis.
Thus, the required point B(-3,4) is marked.

To plot point C(5,0):
Here, the x-coordinate of C is 5 and the y-coordinate is 0. As the x coordinate is positive, move 5 units to the right along the x-axis and mark it. Thus, the required point(5,0) is marked.

To plot point D(0,-5):
Here, the x-coordinate of D is 0 and the y-coordinate is -5.  As the y coordinate is negative, move 5 units to the down along the y-axis and mark it. Thus, the required point(0,-5) is marked.

To plot point E(-2, -1):
Here, the x-coordinate of E is -2 and the y-coordinate is -1.  As the x coordinate is negative, move 2 units to the left along the x-axis. As the y coordinate is negative, move 1 units to the down along the y-axis and mark it.
Thus, the required point (-2,-1) is marked.

To plot point F(2, -1):
Here, the x-coordinate of E is 2 and the y-coordinate is -1. As the x coordinate is positive, move 2 units to the right along the x-axis. As the y coordinate is negative, move 1 units to the down along the y-axis and mark it.
Thus, the required point (2,-1) is marked.

Question 3.
Points P and Q are reflections of each other about the x-axis. Give the coordinates of point Q if the coordinates of point P are the following:
a) (-6, 2)
Answer:
Given that the points P and Q are reflections of each other about the x-axis.
Given that Q is reflection of point P in the x-axis and point P is (-6,2).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x cordinate will be -6 and y coordinate will be -2.
Therefore, point Q will be (-6,-2).

b) (-2, -4)
Answer:
Given that Q is reflection of point P in the x-axis and point P is (-2, -4).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x cordinate will be -2 and y coordinate will be 4.
Therefore, point Q will be (-2,4).

c) (4, 5)
Answer:
Given that Q is reflection of point P in the x-axis and point P is (4, 5).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x cordinate will be 4 and y coordinate will be -5.
Therefore, point Q will be (4,-5).

d) (7, -3)
Answer:
Given that Q is reflection of point P in the x-axis and point P is (7, -3).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x cordinate will be 7 and y coordinate will be 3.
Therefore, point Q will be (7,3).

Question 4.
Points R and S are reflections of each other about the y-axis. Give the coordinates of point S if the coordinates of point R are the following:

a) (-6,2)
Answer:
Given that the points R and S are reflections of each other about the y-axis.
Given that R is reflection of point S in the y-axis and point R is (-6,2).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be 6 and y coordinate will be 2.
Therefore, point S will be (6,2).

b) (-2, -4)
Answer:
Given that R is reflection of point S in the y-axis and point R is (-2, -4).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be 2 and y coordinate will be -4.
Therefore, point S will be (2,-4).

c) (4, 5)
Answer:
Given that R is reflection of point S in the y-axis and point R is (4, 5).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be -4 and y coordinate will be 5.
Therefore, point S will be (-4,5).

d) (7, -3)
Answer:
Given that R is reflection of point S in the y-axis and point R is (7, -3).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be -7 and y coordinate will be -3.
Therefore, point S will be (-7,-3).

Use graph paper. For each exercise, plot the given points on a coordinate plane. Then join the points in order with line segments to form a closed figure. Name each figure formed.

Question 5.
A (3, 4), B (-6, -3), and C (2, -4)

Answer:

Given points are, A (3, 4), B (-6, -3), and C (2, -4).
To get a closed figure, plot the above points and join them in order.
It is forming a three-sided polygon, which is a triangle.
Thus, ABC will form a triangle as shown in the graph above.

Question 6.
D(1, 1), E(0, 0), and F (-4, 4)
Answer:

Given points are, D(1, 1), E(0, 0), and F (-4, 4).
To get a closed figure, plot the above points and join them in order.
It is forming a straight line.
Thus, DEF is a straight line as shown in the graph above.

Question 7.
J (-3, 0), K (0, 5), and L (3, 0)
Answer:

Given points are, J (-3, 0), K (0, 5), and L (3, 0).
To get a closed figure, plot the above points and join them in order.
It is forming a three-sided polygon, which is a triangle.
Thus, JKL will form a triangle as shown in the graph above.

Question 8.
P (3, 2), Q (-1, 2), R (-1, -2), and S (3, -2)
Answer:

Given points are, P (3, 2), Q (-1, 2), R (-1, -2), and S (3, -2).
To get a closed figure, plot the above points and join them in order.
It is forming a four-sided polygon. All the sides are equal in length. A square have all equal sides.
Thus, PQRS will form a square as shown in the graph above.

Question 9.
W(-3, 2), X (1, -2), Y(3, 0), and Z(-1, 4)
Answer:

Given points are, W(-3, 2), X (1, -2), Y(3, 0), and Z(-1, 4)
To get a closed figure, plot the above points and join them in order.
It is forming a four-sided polygon. The opposite sides are equal in length. A rectangle will have opposite sides equal.
Thus, WXYZ will form a rectangle as shown in the graph above.

Question 10.
A (-5, 2), B (-5, -1), C(-1, -1), and D(1, 2)
Answer:

Given points are, A (-5, 2), B (-5, -1), C(-1, -1), and D(1, 2)
To get a closed figure, plot the above points and join them in order.
It is forming a four-sided polygon. All the four sides are of different length, and a pair of opposite sides seems to be parallel which can be a trapezuim.
Thus, ABCD will form a trapezium as shown in the graph above.

Question 11.
E (-2, -2), F (-5, -5), G (-2, -5), and H (1, -2)
Answer:

Given points are E (-2, -2), F (-5, -5), G (-2, -5), and H (1, -2)
To get a closed figure, plot the above points and join them in order.
It is forming a four-sided polygon. The opposite sides are parallel to each other. A parallelogram will have opposite sides in parallel.
Thus, EFGH will form a parallelogram as shown in the graph above.

Question 12.
J (-4, 1), K (-3, -1), L (0, -1), and M (2, 1)
Answer:

Given points are, J (-4, 1), K (-3, -1), L (0, -1), and M (2, 1)
To get a closed figure, plot the above points and join them in order.
It is forming a four-sided polygon. All the four sides are of different length, and a pair of opposite sides seems to be parallel which can be a trapezuim.
Thus, JKLM will form a trapezium as shown in the graph above.

Question 13.
P (4, 0), Q (0, 4), P (-4, 0), and S (0, -4)
Answer:

Given points are, P (4, 0), Q (0, 4), R (-4, 0), and S (0, -4)
To get a closed figure, plot the above points and join them in order.
It is forming a four-sided polygon. All the four sides are equal in length and diagonals are bisecting each other.
Thus, PQRS will form a rhombus.

Question 14.
W (-2, 0), X (-3, -3), Y(1, 1), and Z(2, 4)
Answer:

Given points are W (-2, 0), X (-3, -3), Y(1, 1), and Z(2, 4)
To get a closed figure, plot the above points and join them in order.
It is forming a four-sided polygon. The opposite sides are parallel to each other. A parallelogram will have opposite sides in parallel.
Thus, WXYZ will form a parallelogram as shown in the graph above.

Hands-On Activity

IDENTIFYING QUADRILATERALS DRAWN ON A COORDINATE PLANE

Work in pairs.

Step 1.
Plot four points on a coordinate plane and connect them to form a special quadrilateral such as a parallelogram, a rectangle, or a rhombus. Do not let your partner see your quadrilateral.

Step 2.
Tell your partner the coordinates of three out of the four coordinates of the points you plotted in Step 1. Also tell your partner the type of quadrilateral you plotted, and in which quadrant the fourth point is located. Have your partner guess the coordinates of the fourth point.
Example

Points A (1, 5), B (-2, 1), C(1, -3), and D can be joined to form a rhombus. If point D is in Quadrant I, what are the coordinates of point D?

Step 3.
Switch roles with your partner and repeat the activity with other quadrilaterals.

Math in Focus Course 1B Practice 9.1 Answer Key

Use the coordinate plane below.

Question 1.
Give the coordinates of each point. In which quadrant is each point located?

Answer:

Point A:
The first coordinate represents distance along x axis which is 2 and the second coordinate represents distance along y axis which is 6.
So, the point A will be (2,6).
If the x axis coordinate and y axis coordinate are positive, the point will lie in quadrant I.

Point B:
The first coordinate represents distance along x axis which is 6 and the second coordinate represents distance along y axis which is 2.
So, the point B will be (6,2).
If the x axis coordinate and y axis coordinate are positive, the point will lie in quadrant I.

Point C:
The first coordinate represents distance along x axis which is 3 and the second coordinate represents distance along y axis which is 0.
So, the point C will be (3,0).
The point having y-coordinate as 0 lie on the X-axis and thus is not located in any quadrant.

Point D:
The first coordinate represents distance along x axis which is 4 and the second coordinate represents distance along y axis which is 9.
So, the point D will be (4,-9).
If the x axis coordinate is positive and y axis coordinate is negative, then the point will lie in quadrant IV.

Point E:
The first coordinate represents distance along x axis which is 0 and the second coordinate represents distance along y axis which is 2.
So, the point E will be (0,-2).
The point having x-coordinate as 0 lie on the y-axis and thus is not located in any quadrant.

Point F:
The first coordinate represents distance along x axis which is 2 and the second coordinate represents distance along y axis which is 4.
So, the point F will be (-2,-4).
If the x axis coordinate and y axis coordinate are positive, the point will lie in quadrant III.

Point G:
The first coordinate represents distance along x axis which is 3 and the second coordinate represents distance along y axis which is 0.
So, the point G will be (-3,0).
The point having y-coordinate as 0 lie on the x-axis and thus is not located in any quadrant.

Point H:
The first coordinate represents distance along x axis which is 1 and the second coordinate represents distance along y axis which is 9.
So, the point F will be (-1,9).
If the x axis coordinate is negative and y axis coordinate is positive, than the point will lie in quadrant II.

Use graph paper. Plot the points on a coordinate plane. In which quadrant is each point located?

Question 2.
A (3, 7), B (2, 0), C (8, -1), D (0, -6), E(-3, -5), and F(-6, 7)
Answer:

Point A (3, 7) lies in quadrant I, as x-coordinate and y-coordinate are both positive.
Point B (2, 0) is having y-coordinate as 0 lie on the x-axis does not lie in any quadrant.
Point C (8, -1) lies in quadrant IV, as x axis is positive and y-axis is negative.
Point D (0, -6) is having x-coordinate as 0 lie on the y-axis does not lie in any quadrant.
Point E (-3, -5) lies in quadrant III, as x-coordinate and y-coordinate are both negative.
Point F (-6, 7) lies in quadrant II, as x-coordinate is negative and y-coordinate is positive.

Use graph paper. Points A and B are reflections of each other about the x-axis. Give the coordinates of point B if the coordinates of point A are the following:

Question 3.
(4, 1)
Answer:
Given that the points A and B are reflections of each other about the x-axis.
Given that B is reflection of point A in the x-axis and point A is (4,1).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x cordinate will be 4 and y coordinate will be -1.
Therefore, point B will be (4,-1).

Question 4.
(-2, 3)
Answer:
Given that B is reflection of point A in the x-axis and point A is (-2,3).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x cordinate will be -2 and y coordinate will be -3.
Therefore, point B will be (-2,-3).

Question 5.
(2, -2)
Answer:
Given that B is reflection of point A in the x-axis and point A is (2,-2).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x cordinate will be 2 and y coordinate will be 2.
Therefore, point B will be (2,2).

Question 6.
(-1, -3)
Answer:
Given that B is reflection of point A in the x-axis and point A is (-1,-3).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x cordinate will be -1 and y coordinate will be 3.
Therefore, point B will be (-1,3).

Use graph paper. Points C and D are reflections of each other about the y-axis. Give the coordinates of point D if the coordinates of point C are the following:

Question 7.
(4, 1)
Answer:
Given that the points A and B are reflections of each other about the y-axis.
Given that B is reflection of point A in the x-axis and point A is (4,1).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be -4 and y coordinate will be 1.
Therefore, point S will be (-4,1).

Question 8.
(-2, 3)
Answer:
Given that the points A and B are reflections of each other about the y-axis.
Given that B is reflection of point A in the x-axis and point A is (2,-3).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be 2 and y coordinate will be -3.
Therefore, point S will be (2,-3).

Question 9.
(2, -2)
Answer:
Given that the points A and B are reflections of each other about the y-axis.
Given that B is reflection of point A in the x-axis and point A is (2,-2).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be -2 and y coordinate will be -2.
Therefore, point S will be (-2,-2).

Question 10.
(-1, -3)
Answer:
Given that the points A and B are reflections of each other about the y-axis.
Given that B is reflection of point A in the x-axis and point A is (-1,-3).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be 1 and y coordinate will be -3.
Therefore, point S will be (1,-3).

Use graph paper. For each exercise, plot the given points on a coordinate plane. Then join the points in order with line segments to form a closed figure. Name each figure formed.

Question 11.
H(-5, 1), J(-3, -1), K (-1, 1), and L(-3, 3)
Answer:

Given points are H(-5, 1), J(-3, -1), K (-1, 1), and L(-3, 3)
To get a closed figure, plot the above points and join them in order.
It is forming a four-sided polygon. All the four sides are equal in length and diagonals are bisecting each other.
Thus, HJKL will form a rhombus as shown in the graph above.

Question 12.
R (2, 1), S (-1, -3), T(4, -3), and U(7, 1)
Answer:

Given points are R (2, 1), S (-1, -3), T(4, -3), and U(7, 1)
To get a closed figure, plot the above points and join them in order.
It is forming a four-sided polygon. The opposite sides are parallel to each other. A parallelogram will have opposite sides in parallel.
Thus, RSTU will form a parallelogram as shown in the graph above.

Question 13.
W (-5, -2), X (-6, -5), Y (-1, -5), and Z (-3, -2)
Answer:

Given points are, W (-5, -2), X (-6, -5), Y (-1, -5), and Z (-3, -2)
To get a closed figure, plot the above points and join them in order.
It is forming a four-sided polygon. All the four sides are of different length, and a pair of opposite sides seems to be parallel which can be a trapezuim.
Thus, WXYZ will form a trapezium as shown in the graph above.

Use graph paper. Plot the points on a coordinate plane and answer each question.

Question 14.
a) Plot points A (-6, 5), C (5, 1), and D (5, 5) on a coordinate plane.
Answer:

Given points are, A (-6, 5), C (5, 1), and D (5, 5)
To get a closed figure, plot the above points and join them in order.
It is forming a three-sided polygon, which is a triangle. Angle CDA is forming a right angle, so it is a right angle triangle.
Thus, ADC will form a right angle triangle as shown in the graph above.

b) Figure ABCD is a rectangle. Plot point B and give its coordinates.
Answer:

Given points are, A (-6, 5), C (5, 1), and D (5, 5)
A rectangle is combination of two right angle triangle. The base length of given trinagle is 11 units. In a rectangle, opposite sides are equal in length. Therefore, if AD is of 11 units, BC will also be of 11 units and the height will also be same which is 4 units.
Move 11 units towards left from point C and 4 units towards down from point A to mark point B.
Thus, point B will be (-6,1)
Plot point B to form a rectangle.

c) Figure ACDE is a parallelogram. Plot point E above \(\overline{\mathrm{AD}}\) and give its coordinates.
Answer:
Given that figure ACDE is a parallelogram. Point E is above \(\overline{\mathrm{AD}}\).
In parallelogram, opposite sides are parallelo to each other. Move 4 units towards upwards from point A nd mark as E.
Join the point EC to form the required parallelogram ACDE.
Thus, Point E will be (-6,9)

Question 15.
a) Plot points A (-3, 2) and B (-3, -2) on a coordinate plane.
Answer:

To plot point A(-3,2):
Here, the x-coordinate of A is -3 and the y-coordinate is 2. Start at the Origin. As the x coordinate is negative, move 3 units to the left along the x-axis and 2 units to the top along y -axis.
Thus, the required point A(-3,2) is marked.
To plot point B(-3,-2):
Here, the x-coordinate of A is -3 and the y-coordinate is 2. Start at the Origin. As the x coordinate is negative, move 3 units to the left along the x-axis and 2 units to the down along y -axis as it also is negative.
Thus, the required point A(-3,-2) is marked.

b) Join points A and B with a line segment.
Answer:

Join A and B points to form a straight line.

c) \(\overline{A B}\) is a side of square ABCD. Name two possible sets of coordinates that could be the coordinates of points C and D.
Answer:

Given that AB is a side of square ABCD. In square all sides are equal. Length of AB is 4 units.
Set 1:
So, first we will start moving towards right from B along with the x-axis and mark the point as C.
Move 4 units towards up from C or towards right from A, mark the point D.
Join ABCD to form a square.
The first possible set of coordinates for C and D will be (1,-2) and (1,2) respectively.
Set 2:
Next, start moving 4 units towards the left from point B and mark it as F.
Move 4 uniits towards up from F or towards left from A, mark the point as E.
Join ABEF to form a square.
The second possible set of coordinates for C and D will be (-7,-2) and (-7,2) respectively.

Question 16.
Plot points A (2, 5) and 8 (2, -3) on a coordinate plane. Figure ABC is a right isosceles triangle. If point C is in Quadrant III, give the coordinates of point C.
Answer:
To plot point A(2,5):
Here, the x-coordinate of A is 2 and the y-coordinate is 5. Start at the Origin. As the x coordinate is positive, move 2 units to the right along the x-axis and 5 units to the top along y -axis.
Thus, the required point A(2,5) is marked.
To plot point B(2,-3):
Here, the x-coordinate of A is 2 and the y-coordinate is -3. Start at the Origin. As the x coordinate is positive, move 2 units to the right along the x-axis and 3 units to the top along y -axis.
Thus, the required point B(2,-3) is marked.
Point C is in Quadrant III and it should form a right angle triangle.
The side AB is of length 8 units, so BC will also remain the same.
Figure ABC is a right isosceles triangle.

Question 17.
Plot points A (0, 4), B (-4, 0), and C (0, -4) on a coordinate plane.

Given points A (0, 4), B (-4, 0), and C (0, -4) are to be plotted on a coordinate plane.
Joining them will form a triangle.

a) What kind of triangle is triangle ABC?
Answer:
Since the point B is at equidistant from A and C.
The lengths of BC and AB will be same.
Thus, ABC will form an isosceles triangle.

b) Figure ABCD is a square. Plot point D on the coordinate plane and give its coordinates.
Answer:

In square, all sides are equal. The length of AB and BC equals 4 units.
D is reflection of point B in the x-axis and point B is (-4,0).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Therefore, point D will be (4,0)
Thus, the required square ABCD is formed.

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 9 Lesson 9.2 Length of Line Segments to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 9 Lesson 9.2 Answer Key Length of Line Segments

Math in Focus Grade 6 Chapter 9 Lesson 9.2 Guided Practice Answer Key

Use graph paper. Plot each pair of points on a coordinate plane. Connect the points to form a line segment and find its length.

Question 1.
E(-6, 0) and D(8, 0)
Answer:
Point E(-6, 0)
Here, the x-coordinate of E is -6 and the y-coordinate is 0. As the y-coordinate is zero and x coordinate is negative, move 6 units along the negative x-axis and mark it. Thus, the required point(-6,0) is marked.
Point D(8,0)
Here, the x-coordinate of D is 8 and the y-coordinate is 0. As the y-coordinate is zero and x coordinate is positive, move 8 units along the positive x-axis and mark it. Thus, the required point(8,0) is marked.

After joining the E and D, we will get a line segment of 18 units.

Question 2.
E(-6, 0) and F(-2, 0)
Answer:
Point E(-6, 0)
Here, the x-coordinate of E is -6 and the y-coordinate is 0. As the y-coordinate is zero and x coordinate is negative, move 6 units along the negative x-axis and mark it. Thus, the required point(-6,0) is marked.
Point F(-2,0)
Here, the x-coordinate of D is -2 and the y-coordinate is 0. As the y-coordinate is zero and x coordinate is negative, move 2 units along the negative x-axis and mark it. Thus, the required point(-2,0) is marked.

After joining the E and F, we will get a line segment of 4 units.

Question 3.
G(-7, 0) and H(1, 0)
Answer:
Point E(-7, 0)
Here, the x-coordinate of E is -7 and the y-coordinate is 0. As the y-coordinate is zero and x coordinate is negative, move 7 units along the negative x-axis and mark it. Thus, the required point(-7,0) is marked.
Point H(1, 0)
Here, the x-coordinate of H is 1 and the y-coordinate is 0. As the y-coordinate is zero and x coordinate is positive, move 1 units along the positive x-axis and mark it. Thus, the required point(1,0) is marked.

After joining the G and H, we will get a line segment of 8 units.

Question 4.
J (0, 5) and K (0, 2)
Answer:
To plot J(0,5):
Here, the x-coordinate is 0 and the y-coordinate is 5. As the x-coordinate is zero and y coordinate is positive, move 5 units along positive y-axis and mark it. Thus, the required point(0,5) is marked.
To point K (0, 2):
Here, the x-coordinate is 0 and the y-coordinate is 2. As the x-coordinate is zero and y coordinate is positive, move 2 units along positive y-axis and mark it. Thus, the required point(0,2) is marked.

After joining the J and K, we will get a line segment of 3 units.

Question 5.
M(0, -6)and N(0, -3)
Answer:
To plot M(0, -6):
Here, the x-coordinate is 0 and the y-coordinate is -6. As the x-coordinate is zero and y coordinate is negative, move 6 units along negative y-axis and mark it. Thus, the required point (0,-6) is marked.
To point N(0, -3):
Here, the x-coordinate is 0 and the y-coordinate is -3. As the x-coordinate is zero and y coordinate is negative, move 3 units along negative y-axis and mark it. Thus, the required point(0,-3) is marked.

After joining the M and N, we will get a line segment of 3 units.

Question 6.
P(0, -3) and Q(0, 5)
Answer:
To plot P(0, -3):
Here, the x-coordinate is 0 and the y-coordinate is -3. As the x-coordinate is zero and y coordinate is negative, move 3 units along negative y-axis and mark it. Thus, the required point (0,-3) is marked.
To point Q(0, 5):
Here, the x-coordinate is 0 and the y-coordinate is 5. As the x-coordinate is zero and y coordinate is positive, move 5 units along positive y-axis and mark it. Thus, the required point(0,5) is marked.

After joining the P and Q, we will get a line segment of 8 units.

Use graph paper. Plot each pair of points on a coordinate plane. Connect the points to form a line segment and find its length.

Question 7.
A(1, -2) and B(6, -2)
Answer:
To plot A(1, -2):
Here, the x-coordinate is 1 and the y-coordinate is -2. Start at the Origin. As the x coordinate is positive, move 1 units to the right along the x-axis and 2 units to the bottom along y -axis.
Thus, the required point (1,-2) is marked.
To point B(6, -2):
Here, the x-coordinate is 6 and the y-coordinate is -2. Start at the Origin. As the x coordinate is positive, move 6 units to the right along the x-axis and 2 units to the bottom along y -axis.
Thus, the required point (6, -2) is marked.

After joining the A and B, we will get a line segment of 5 units.

Question 8.
C(-1, 3) and D(5, 3)
Answer:
Plot C(-1, 3):
Here, the x-coordinate is -1 and the y-coordinate is 3. Start at the Origin. As the x coordinate is negative, move 1 units to the left along the x-axis and 3 units to the above along positive y -axis and mark it.
Thus, the required point (-1, 3) is marked.
Point D(5, 3):
Here, the x-coordinate is 5 and the y-coordinate is 3. Start at the Origin. As the x-coordinate and y coordinate is positive, move 5 unit along positive x-axis and 3 units to the above along positive y -axis and mark it.
Thus, the required point (5, 3) is marked.

After joining the C and D, we will get a line segment of 6 units.

Question 9.
E (-3, 4) and F (1, 4)
Answer:
Plot E (-3, 4):
Here, the x-coordinate is -3 and the y-coordinate is 4. Start at the Origin. As the x coordinate is negative, move 3 units to the left along the x-axis and 4 units to the above along positive y -axis and mark it.
Thus, the required point (-3, 4) is marked.
Point F (1, 4):
Here, the x-coordinate is 5 and the y-coordinate is 3. Start at the Origin. As the x-coordinate and y coordinate is positive, move 1 unit along positive x-axis and 4 units to the above along positive y -axis and mark it.
Thus, the required point (1, 4) is marked.

After joining the E and F, we will get a line segment of 4 units.

Question 10.
G (-3, 2) and H (-3, 6)
Answer:
Plot G (-3, 2):
Here, the x-coordinate is -3 and the y-coordinate is 2. Start at the Origin. As the x coordinate is negative, move 3 units to the left along the x-axis and 2 units to the above along positive y -axis and then mark it.
Thus, the required point (-3, 2) is marked.
Point H (-3, 6):
Here, the x-coordinate is -3 and the y-coordinate is 6. Start at the Origin. As the x-coordinate is negative, move 3 units to the left along the x-axis and 6 units along positive y-axis and then mark it.
Thus, the required point (-3, 6) is marked.

After joining the G and H, we will get a line segment of 4 units.

Question 11.
J(-1, -6) and K(-1, 4)
Answer:
Plot J(-1, -6):
Here, the x-coordinate is -1 and the y-coordinate is -6. Start at the Origin. As the x coordinate is negative, move 1 unit to the left along the x-axis and 6 units along negative y -axis and then mark it.
Thus, the required point (-1, -6) is marked.
Point K(-1, 4):
Here, the x-coordinate is -1 and the y-coordinate is 4. Start at the Origin. As the x-coordinate is negative, move 1 units to the left along the x-axis and 4 units along positive y-axis and then mark it.
Thus, the required point (-1, 4) is marked.

After joining the J and K, we will get a line segment of 10 units.

Question 12.
L(5, 6) and M(5, 1)
Answer:
Point L(5, 6):
Here, the x-coordinate is 5 and the y-coordinate is 6. Start at the Origin. As the x-coordinate and y coordinate is positive, move 5 units along positive x-axis and 6 units to the above along positive y -axis and mark it.
Thus, the required point (5, 6) is marked.
Point M(5, 1):
Here, the x-coordinate is 5 and the y-coordinate is 1. Start at the Origin. As the x-coordinate and y coordinate is positive, move 5 units along positive x-axis and 1 unit to the above along positive y -axis and mark it.
Thus, the required point (5, 1) is marked.

After joining the L and M, we will get a line segment of 5 units.

In the diagram, triangle ABC represents a plot of land. The side length of each grid square ¡s 5 meters. Use the diagram to answer questions 13 to 15.

Question 13.
Give the coordinates of points A, B, and C.
Answer:
Point A:
The first coordinate represents distance along x axis which is 10 units and the second coordinate represents distance along y axis which is 20 units.
So, the point A will be (10,20).
Point B:
The first coordinate represents distance along x axis which is 10 units and the second coordinate represents distance along y axis which is 5 units.
So, the point B will be (10,5).
Point C:
The first coordinate represents distance along x axis which is 30 units and the second coordinate represents distance along y axis which is 20 units.
So, the point C will be (30,20).

Question 14.
Mr. Manning wants to build a fence around the plot of land. If BC is 25 meters, how many meters of fencing does he need?
AB = –
= m
AC = –
Perimeter of triangle ABC = AB + BC + AC
= + +
= m
Mr. Manning needs meters of fencing.
Answer:
Here, 1 grid square represents 5 meters. From the figure,
AB equals 3 grid square so it will measure 3×5=15m
AC equals 4 grid square so it will measure 4×5=20m
Given BC equals 25m.
Perimeter of triangle ABC = AB + BC + AC
= 15+25+20
= 60 m

Question 15.
A pole is located at point D on the plot of land at a distance of 10 meters from \(\overline{A B}\) and 5 meters from \(\overline{A C}\). Give the coordinates of point D.
1 grid square represents 5 meters.
10 m = ÷ 5
= grid squares
Answer:
Given that point D is located a distance of 10 meters from \(\overline{A B}\) and 5 meters from \(\overline{A C}\).
10 m = 10 ÷ 5
= 2 grid squares.

For point D to be on the plot of land, the x-coordinate has to be grid squares to the right of \(\overline{\mathrm{AB}}\). So, point D is + = grid squares to the right of the y-axis. The x-coordinate of point D is × = .
Answer:
For point D to be on the plot of land, the x-coordinate has to be 2 grid squares to the right of
\(\overline{\mathrm{AB}}\). So, point D is 2+2 = 4 grid squares to the right of the y-axis.
The x-coordinate of point D is 4×5  = 20.

For point D to be on the plot of land, the y-coordinate has to be grid square below \(\overline{\mathrm{AC}}\). So, point D is – = grid squares above the x-axis. The y-coordinate of point D is × = .
The coordinates of D are (, )
Answer:
For point D to be on the plot of land, the y-coordinate has to be 1 grid square below \(\overline{\mathrm{AC}}\). So, point D is 4-1 = 3 grid squares above the x-axis. The y-coordinate of point D is 3×5  = 15.
The coordinates of D are (20,15)

In the diagram, rectangle PQRS represents a parking lot of a supermarket. The side length of each grid square is 4 meters. Use the diagram to answer questions 16 to 18.

Question 16.
Give the coordinates of points P, Q, R, and S.
Answer:
Point P:
The first coordinate represents distance along x axis which is 16 units and the second coordinate represents distance along y axis which is 44 units.
So, the point P will be (16,44).
Point Q:
The first coordinate represents distance along x axis which is 16 units and the second coordinate represents distance along y axis which is 4 units.
So, the point Q will be (16,4).
Point R:
The first coordinate represents distance along x axis which is 36 units and the second coordinate represents distance along y axis which is 4 units.
So, the point QRwill be (36,4).
Point S:
The first coordinate represents distance along x axis which is 36 units and the second coordinate represents distance along y axis which is 44 units.
So, the point S will be (36,44).

Question 17.
The manager of the supermarket wants to build a concrete wall around the parking lot. What is the perimeter of the wall?
Answer:
Here, 1 grid square represents 4 meters.
Length of PQ = 44-4 = 40 m
Length of QR = 36-16 = 20m
Since PQRS is a rectangle, opposite lengths will be equal.
Perimeter of wall will be = 40+20+40+20 = 120m

Question 18.
The entrance of the supermarket is at point T. It lies on \(\overline{P Q}\), and point T is 8 meters from point P. Give the coordinates of point T.
Answer:
Point T lies on \(\overline{P Q}\), and it is 8 meters from point P.
Thus, the entrance will be on PQ.
8 m means 8÷4=2 square grid
Move 2 square grid below point P on \(\overline{P Q}\)
Thus, point T will be (16,36)

Math in Focus Course 1B Practice 9.2 Answer Key

Use graph paper. Plot each pair of points on a coordinate plane. Connect the points to form a line segment and find its length.

Question 1.
A(5, 0) and B(8, 0)
Answer:
Point A(5, 0)
Here, the x-coordinate is 5 and the y-coordinate is 0. As the y-coordinate is zero and x coordinate is positive, move 5 units along the positive x-axis and mark it. Thus, the required point(5,0) is marked.
Point B(8,0)
Here, the x-coordinate is 8 and the y-coordinate is 0. As the y-coordinate is zero and x coordinate is positive, move 8 units along the positive x-axis and mark it. Thus, the required point(8,0) is marked.

After joining the A and B, a line segment of 3 units is formed.

Question 2.
C(-3, 4)and D(3, 4)
Answer:
Plot C(-3, 4):
Here, the x-coordinate is -3 and the y-coordinate is 4. Start at the Origin. As the x coordinate is negative, move 3 units along the negative x-axis and 4 units to the above along positive y -axis and mark it.
Thus, the required point (-3, 4) is marked.
Plot D(3, 4):
Here, the x-coordinate is 3 and the y-coordinate is 4. Start at the Origin. As the x coordinate is positive, move 3 units along the positive x-axis and 4 units to the above along positive y -axis and mark it.
Thus, the required point (3, 4) is marked.

After joining the C and D, a line segment of 6 units is formed.

Question 3.
E (-5, -2) and F (8, -2)
Answer:
Plot point E(-5,-2)
Here, the x-coordinate is -5 and the y-coordinate is -2. Start at the Origin. As the x coordinate is negative, move 5 units along the negative x-axis and 2 units along negative y -axis and mark it.
Thus, the required point (-5, -2) is marked.
Plot point F(8, -2)
Here, the x-coordinate is 8 and the y-coordinate is -2. Start at the Origin. As the x coordinate is positive, move 8 units along the positive x-axis and 2 units along negative y -axis and mark it.
Thus, the required point (8, -2) is marked.

After joining the E and F, a line segment of 13 units is formed.

Question 4.
G(0, -5) and H(0, 2)
Answer:
Plot point G(0, -5)
Here, the x-coordinate is 0 and the y-coordinate is -5. As the x-coordinate is zero and y coordinate is negative, move 5 units along negative y-axis and mark it. Thus, the required point(0,-5) is marked.
Plot point H(0, 2)
Here, the x-coordinate is 0 and the y-coordinate is 2. As the x-coordinate is zero and y coordinate is positive, move 2 units along positive y-axis and mark it. Thus, the required point(0,2) is marked.

After joining the G and H, a line segment of 7 units is formed.

Question 5.
J(-5, -3) and K(-5, -8)
Answer:
Plot point J(-5, -3)
Here, the x-coordinate is -5 and the y-coordinate is -3. Start at the Origin. As the x coordinate is negative, move 5 units along the negative x-axis and 3 units along negative y -axis and mark it.
Thus, the required point (-5, -3) is marked.
Plot point K(-5,-8):
Here, the x-coordinate is -5 and the y-coordinate is -8. Start at the Origin. As the x coordinate is negative, move 5 units along the negative x-axis and 8 units along negative y -axis and mark it.
Thus, the required point (-5, -8) is marked.

After joining the J and K, a line segment of 5 units is formed.

Question 6.
M(1, 7) and N(1, -8)
Answer:
Plot point M(1, 7)
Here, the x-coordinate is 1 and the y-coordinate is 7. Start at the Origin. As the x coordinate is positive, move 1 unit along the positive x-axis and 7 units along positive y -axis and mark it.
Thus, the required point (1, 7) is marked.
Plot point N(1, -8)
Here, the x-coordinate is 1 and the y-coordinate is -8. Start at the Origin. As the x coordinate is positive, move 1 unit along the positive x-axis and 8 units along negative y -axis and mark it.
Thus, the required point (1, -8) is marked.

After joining the M and N, a line segment of 15 units is formed.

Use graph paper. Find the coordinates.

Question 7.
Rectangle PQRS is plotted on a coordinate plane. The coordinates of P are (-1, -3) and the coordinates of Q are (-1, 2). Each unit on the coordinate plane represents 1 centimeter, and the perimeter of rectangle PQRS is 20 centimeters. Find the coordinates of points R and S given these conditions:
a) Points R and S are to the left of points P and Q.
Answer:
Given that PQRS is a rectangle. The coordinates of P are (-1, -3) and the coordinates of Q are (-1, 2).
The perimeter of rectangle PQRS is 20 centimeters.
Length of PQ is 5 units or 5 cm which is equal to length of RS. Length of RQ is equal to length of SP.
Perimeter of rectangle = Sum of four sides
20 = PQ+QR+RS+SP
20 = 5+QR+5+SP
20 = 5+SP+5+SP
20 = 10+2×SP
20-10 = 10+2×SP-10
10 = 2×SP
10÷2 = (2×SP)÷2
5 = SP
Length of SP equals to 5 units or 5 cm.
Start from point P and move 5 square grids towards left and point R. Similarly, point R from Q and then join the points to form the rectangle.

b) Points R and S are to the right of points P and Q. The coordinates of P are (-1, -3) and the coordinates of Q are (-1, 2).
Answer:
We know the measurements of the PQRS, where length of all sides are equal.
Start from point P and move 5 square grids towards right and point R. Similarly, point R1 from Q and then join the points to form the rectangle.

Question 8.
Rectangle ABCD is plotted on a coordinate plane. The coordinates of A are (2, 3) and the coordinates of B are (-2, 3). Each unit on the coordinate plane represents 3 centimeters, and the perimeter of rectangle ABCD is 48 centimeters. Find the coordinates of points C and D given these conditions:
a) Points C and D are below points A and B.
Answer:
Given that ABCD is a rectangle. The coordinates of A are (2, 3) and the coordinates of B are (-2, 3).
The perimeter of rectangle ABCD is 48 centimeters.
Length of AB is 4 units or 4×3 = 12cm which is equal to length of CD. Length of AC is equal to length of BD.
Perimeter of rectangle = Sum of four sides
48 = AB+BD+DC+CA
48 = 12+3BD+12+3BD (BD will be equal to 3BD as length of each unit on the coordinate plane represents 3 cm)
48 = 24+2×6×BD
48-24 = 24+12×BD-24
24 = 12×BD
24÷12 = (12×BD)÷12
2= BD
Length of BD equals to 2 units or 2×3=6 cm.
Start from point A and move 2 square grids below and point C. Similarly, point D from B and then join the points to form the rectangle.

b) Points C and D are above points A and B.
Answer:
We know the measurements of the ABCD, where length of opposite sides are equal.
Start from point A and move 2 square grids upwards and point C1. Similarly, point D1 from B and then join the points to form the rectangle.

Question 9.
Rectangle PQRS is plotted on a coordinate plane. The coordinates of P are (-1, 4) and the coordinates of Q are (-1, -4). Each unit on the coordinate plane represents 1 centimeter, and the area of rectangle PQRS is 64 square centimeters. Find the coordinates of points R and S given these conditions:

a) Points R and S are to the left of points P and Q.
Answer:
Given that PQRS is a rectangle. The coordinates of P are (-1, 4) and the coordinates of Q are (-1, -4).
The area of rectangle PQRS is 64 sq.cm
Length of PQ is 8 units or 8 cm which is equal to length of RS. Length of QS is equal to length of RP.
Area of rectangle = length×width
64 = PQ×QR
64 = 8×QR
64÷8 = (8×QR)÷8
8 = QR
Length of QR equals to 8 units or 8 cm.
Start from point P and move 8 square grids to the left and mark point R. Similarly, mark point S from Q and then join the points to form the rectangle.

b) Points R and S are to the right of points P and Q.
Answer:
We know the measurements of the PQRS, where length of all sides are equal.
Start from point P and move 8 square grids towards right and point it as R. Similarly, point S from Q and then join the points to form the rectangle.

In the diagram, rectangle ABCD represents a shopping plaza. The side length of each grid square Is 10 meters. Use the diagram to answer questions 10 to 14.

Question 10.
Give the coordinates of points A, B, C, and D.
Answer:
Point A:
The first coordinate represents distance along x axis which is 60 units and the second coordinate represents distance along y axis which is 90 units.
So, the point A will be (-60,90).
Point B:
The first coordinate represents distance along x axis which is 60 units and the second coordinate represents distance along y axis which is 50 units.
So, the point A will be (-60,50).
Point C:
The first coordinate represents distance along x axis which is 50 units and the second coordinate represents distance along y axis which is 50 units.
So, the point A will be (50,-50).
Point D:
The first coordinate represents distance along x axis which is 50 units and the second coordinate represents distance along y axis which is 90 units.
So, the point A will be (50,90).

Question 11.
Write down the shortest distance of points A, B, C, and D from the y-axis.
Answer:
The shortest distance of points A from the y-axis is 60 units.
The shortest distance of points B from the y-axis is 60 units.
The shortest distance of points C from the y-axis is 50 units.
The shortest distance of points D from the y-axis is 50 units.

Question 12.
Write down the shortest distance of points A, B, C, and D from the x-axis.
Answer:
The shortest distance of points A from the x-axis is 90 units.
The shortest distance of points B from the x-axis is 50 units.
The shortest distance of points C from the x-axis is 50 units.
The shortest distance of points D from the x-axis is 90 units.

Question 13.
Find the area and perimeter of the shopping plaza.
Answer:
The side length of each grid square Is 10 meters.
Length of AB is 140 square grid, 140×10=1400 m
Length of BC is 110 square grid, 110×10=1100 m
Thus, the length of the shopping plaza is 1400m and width is 1100m
Area of the shopping plaza = length × width
= 1400×1100
= 1540000 sq.m
Perimeter of the shopping plaza = 2×(length+width)
= 2×(1400+1100)
= 2×2500
= 5000 m

Question 14.
A man at the shopping plaza is standing 50 meters from \(\overline{\mathrm{AD}}\), and 40 meters from \(\overline{\mathrm{DC}}\).
a) Find the coordinates of the point representing the man’s location.
Answer:
Given that:
A man at the shopping plaza is standing 50 meters from \(\overline{\mathrm{AD}}\)
AD – 50 = 90-50 = 40
and 40 meters from \(\overline{\mathrm{DC}}\).
DC – 40 = 50-40 = 10

b) Find the shortest distance in meters from the man’s location to the side \(\overline{\mathrm{BC}}\).
Answer:
Count the square grids from the man’s location to the side BC.
There are 9 square grids, 9×10 = 90m
Thus, 90m is the shortest distance in meters from the man’s location to the side BC

In the diagram, triangle PQR represents a triangular garden. The side length of each grid square Is 5 meters. Use the diagram to answer questions 15 to 19.

Question 15.
Use graph paper. A rectangular region ABCR in the garden is to be fenced in. Point A lies on \(\overline{P R}\), and is 35 meters away from point P. Point C lies below \(\overline{P R}\), and is 20 meters away from point R. Plot and label points A, B, and C on the coordinate plane. Write the coordinates of points A, B, and C.
Answer:
A rectangular region ABCR in the garden is to be fenced in. Point A lies on \(\overline{P R}\), and is 35 meters away from point P.
The side length of each grid square Is 5 meters.
Since point A is 35m away from P, 35÷5=7 square grids.
Move 7 units away from point P and mark it as A.
Thus, the point A will be (5,15)

Point C lies below \(\overline{P R}\), and is 20 meters away from point R.
Since point C is 20m away from point R, 20÷5=4 square grids.
Move 4 units to the down from point R and mark it as C.
Thus, the point C will be (30, -5)

Since ABCR is a rectangular region, the opposite sides are equal.
Start from point C and move 5 units towards left, and mark the point as B.
Thus, the point B will be (5,-5)

Quadrant 16.
If PQ is 75 meters, what is the perimeter of the triangular garden in meters?
Answer:
Given that PQ is 75 m.
Length of PR = 12 square grids.
Here, one square grid equals 5m. So, 12 square grids = 12×5 = 60m
Length of RQ = 9 square grids.
So, 9 square grids = 9×5 = 45m
Perimeter of the triangular garden = 75+60+45 = 180m

Question 17.
Find the area of the enclosed region ABCR in square meters.
Answer:
Length of AR = 5 square grids, 5×5=25m
Width of AB = 4 square grids, 4×5=20m
Area of rectangle = length×width
=25×20
=500
Area of the enclosed region ABCR will be 500 sq.m

Question 18.
Find the perimeter of the enclosed region ABCR in meters.
Answer:
Length of AR = 25m
Width of AB = 20m
Perimeter = AR+RC+CB+BA
= 25+20+25+20
= 90 m

Question 19.
If PQ is 75 meters, what is the perimeter of the garden that is not enclosed?
Answer:
Given PQ is 75m.
Length of AP = 7 square grids,7×5=35 m
Length of AB = 4 square grids,4×5=20 m
Length of BC = 5 square grids,5×5=25 m
Length of CQ = 5 square grids,5×5=25 m
Perimeter of the garden that is not enclosed = 75+35+20+25+25 = 180 m

The diagram shows the outline of a park. The side length of each grid square is 10 meters. Use the diagram to answer questions 20 to 22.

Question 20.
Find the area of the park in square meters.
Answer:
Let us assume AHGF as a rectangle, AH be the length and HG be the width.
Length of AH is 80m and length of HG is 180m.
Area of rectangle = length×width
= 80×180
= 14400 sq.m
Area of AHGFA will be 14400 sq.m
Area of rectangle BCDE:
Length of BC = 50m
Length of CD = 120m
Area of BCDE will be 50×120 = 6000 sq.m
Area of the park can be calculated by subtracting the area of rectangle BCDE from the rectangle AHGF
= 14400-6000
= 8400 sq.m

Question 21.
Brandon starts at point A and walks all the way around the perimeter of the park. If he walks at 1.5 meters per second, about how many seconds pass before he returns to point A? Round your answer to the nearest second.
Answer:
The side length of each grid square Is 10 m.
Length of HA: 8 square grids = 8×10 = 80 m
Length of AB: 3 square grids = 3×10 = 30 m
Length of BC: 5 square grids = 5×10 = 50 m
Length of CD: 12 square grids = 12×10 = 120 m
Length of DE: 5 square grids = 5×10 = 50 m
Length of EF: 3 square grids = 3×10 = 30 m
Length of FG: 8 square grids = 8×10 = 80 m
Length of GH: 8 square grids = 18×10 = 180 m
Perimeter of the park will be 80+30+50+120+50+30+80+180 = 620m
He walks at 1.5 meters per second.
620÷1.5 = 413.3
Round off for 413.3 equals to 413 sec

Question 22.
A picnic table in the park is 20 meters from \(\overline{B C}\), and is closer to point B than it is to point C. Write down two possible pairs of coordinates for the location of the picnic table.
Answer:
Given that a picnic table in the park is 20 m from \(\overline{B C}\), and is closer to point B than it is to point C.
20m away from B will be 2 square grids towards up which can be (-30,80) and (-40,80)

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 9 Lesson 9.3 Real-World Problems: Graphing to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 9 Lesson 9.3 Answer Key Real-World Problems: Graphing

Math in Focus Grade 6 Chapter 9 Lesson 9.3 Guided Practice Answer Key

Graph an equation on a coordinate plane.

Angela is driving to the Raccoon River. The distance traveled, d miles, after t hours, is given by d = 40t. Graph the relationship between d and t. Use 2 units on the horizontal axis to represent 1 hour and 2 units on the vertical axis to represent 20 miles.

a) What type of graph is it?
It is a straight line graph.This is also called a linear graph.

b) How far did Angela drive in 3.5 hours?
From the graph, Angela drove 140 miles.

c) What is the speed at which Angela is driving?

= 40 mi/h
Angela is driving at 40 miles per hour.

d) Angela has driven for 4 hours. If she drives for another hour at this • constant speed, how far will she drive in all?
Distance = speed × time
= 40 × 5
= 200 mi
She will drive 200 miles.

e) If Angela wants to drive at least 120 miles, how many hours will she need to drive? Express your answer in the form of an inequality in terms of t, where t stands for the number of hours.
t ≥ 3

f) Name the dependent and independent variables.
d is the dependent variable, and t is the independent variable.

Use graph paper. Solve.

Question 1.
A car uses 1 gallon of gas for every 20 miles traveled. The amount of gas left in the gas tank, x gallons, after traveling y miles is given by y = 240 – 20x. Copy and complete the table. Graph the relationship between x and y.
Use 1 unit on the horizontal axis to represent 1 gallon and 1 unit on the vertical axis to represent 20 miles.
a)

Answer:
Given that, A car uses 1 gallon of gas for every 20 miles traveled. The amount of gas left in the gas tank, x gallons, after traveling y miles is given by y = 240 – 20x.
If y equals to 80 miles then x will be equal to
y = 240 – 20x
80 = 240-20x
80+20x = 240-20x+20x
80+20x = 240
80+20x-80 = 240-80
20x = 160
20x÷20 = 160÷20
x = 8
If y equals to 160 miles then x will be equal to
y = 240 – 20x
160 = 240-20x
160+20x = 240-20x+20x
160+20x = 240
160+20x-160 = 240-160
20x = 80
20x÷20 = 80÷20
x = 4

b) What type of graph is it? It is a graph.
Answer:
It is a straight line graph.This is also called a linear graph.

c) How many gallons of gas will be left in the tank after the car has traveled 60 miles?
From the graph, there will be gallons of gas left,
Answer:
From the graph, there will be 8 gallons of gas left in the tank after the car has traveled 60 miles.

d) How many gallons of gas will be left in the tank after the car has traveled 100 miles?
From the graph, there will be gallons of gas left.
Answer:
From the graph, there will be 4 gallons of gas left in the tank after the car has traveled 100 miles.

e) After the car has traveled 160 miles, how much farther can the car travel before it runs out of gas?
After 160 miles, gallons of gas were left.
The car uses 1 gallon for every miles traveled.
Distance = number of gallons × mileage
= ×
= mi
The car can travel another miles.
Answer:
After 160 miles, 4 gallons of gas were left.
The car uses 1 gallon for every 20 miles traveled.
Distance = 4 × mileage
160 =  4×mileage
160÷4 = (4×mileage)÷4
40 = mileage
The car can travel another 40 miles.

f) If the car travels more than 40 miles, how much gas is left in the tank?
Express your answer in the form of an inequality in terms of x, where x stands for the amount of gas left in the gas tank.
If the distance traveled is more than 40 miles, then .

Answer:
If the car travels more than 40 miles, then x ≥10, where x stands for the amount of gas left in the gas tank.

g) Name the dependent and independent variables.
Answer:
From the above analysis, we can observe that,
x is independent and y is dependent

Question 2.
Sarah plants a seed. After t weeks, the height of the plant, h centimeters, is given by h = 2t. Copy and complete the table. Graph the relationship between t and h. Use 1 unit on the horizontal axis to represent 1 week and 1 unit on the vertical axis to represent 2 centimeters.
a)

Answer:
Given that Sarah plants a seed. After t weeks, the height of the plant, h centimeters, is given by h = 2t
To calculate the height, we can use this relation,  h = 2t
If t=0, h = 2t = 2×0 = 0 cm
If t=4, h = 2t = 2×4 = 8 cm
If t=5, h = 2t = 2×5 = 10 cm

b) What type of graph is it?
Answer:
By joining all the points it forms a straight line graph.
This is also called a linear graph.

c) What is the height of the plant after 3 weeks?
Answer:
From the graph, we can observe that the height of the plant will be 6 cm after 3 weeks.

d) What is the height of the plant after 5 weeks?
Answer:
From the graph, we can observe that the height of the plant will be 10 cm after 5 weeks.

e) What is the height of the plant if less than 4 weeks have passed? Express your answer in the form of an inequality in terms of h, where h stands for the height of the plant in centimeters.
Answer:
The height of the plant if less than 4 weeks have passed will be 8cm.
Therefore, the height in the form of an inequality in terms of h will be h ≥ 8, where h stands for the height of the plant in cm.

f) Name the dependent and independent variables.
Answer:
From the above observation, we can say the dependent variable is h
and the independent variable is t.

Math in Focus Course 1B Practice 9.3 Answer Key

Use graph paper. Solve.

Question 1.
A cyclist took part in a competition. The distance traveled, d meters, after t minutes, is given by d = 700t. Graph the relationship between t and d. Use 2 units on the horizontal axis to represent 1 minute and 1 unit on the vertical axis to represent 350 meters.

a) What type of graph is it?
Answer:
By joining all the points it forms a straight line graph.
This is also called a linear graph.

b) What is the distance traveled in 2.5 minutes?
Answer:
Given that the distance traveled, d meters, after t minutes, is given by d = 700t.
d = 700×2.5
= 1750 m
Thus, 1750 m is the distance traveled in 2.5 min.

c) What is the distance traveled in 3.5 minutes?
Answer:
Given that the distance traveled, d meters, after t minutes, is given by d = 700t.
d = 700×3.5
= 2450 m
Thus, 2450 m is the distance traveled in 3.5 min.

d) What is the average speed of the cyclist?
Answer:
The average speed of the cyclist can be calculated by using the below formula,
Average speed = \(\frac{total distance}{total elapsed time}\)
Total distance = 0+700+1400+2100+2800 = 7000 m
Total time = 0+1+2+3+4 = 10 min
= \(\frac{7000}{10}\)
= 700
The average speed of the cyclist is 700 metre per min.

e) Assuming that the cyclist travels at a constant speed throughout the competition, what distance will he travel in 7 minutes?
Answer:
Let us assume that the cyclist travels at a constant speed of 700m in a min.
So, the distance he travel in 7 mins will be 7×700=4900 m.

f) If the cyclist needs to cycle for at least 2.1 kilometers, how many minutes will he need to cycle? Express your answer in the form of an inequality in terms of t, where t stands for the number of minutes.
Answer:
The cyclist needs to cycle for at least 2.1 kilometers.
1 kilometer equals to 1000 m.
2.1 km will be equal to 2.1×1000 = 2100 m
Given that the distance traveled, d meters, after t minutes, is given by d = 700t.
2100 = 700t
2100÷700 =(700×t)÷700
300 = t
The number of minutes the cyclist will need to cycle 2.1 km will be 300 min.

g) Name the dependent and independent variables.
Answer:
From the above observation, we can say the dependent variable is d
and the independent variable is t.

Question 2.
A bus uses 1 gallon of diesel for every 7 miles traveled. The amount of diesel left in the gas tank, p gallons, after traveling q miles, is given by q = 112 – 7p. Copy and complete the table. Graph the relationship between p and q. Use 1 unit on the horizontal axis to represent 1 gallon and 1 unit on the vertical axis to represent 7 miles.

a)

Answer:
Given that a bus uses 1 gallon of diesel for every 7 miles traveled. The amount of diesel left in the gas tank, p gallons, after traveling q miles, is given by q = 112 – 7p.
If p is 14, then q will be 112 – 7×14 = 112-98 = 14
If p is 8, then q will be 112 – 7×8 = 112-56 = 56

b) How many gallons of diesel were left after the bus has traveled 49 miles?
Answer:
The amount of diesel left in the gas tank, p gallons, after traveling q miles, is given by q = 112 – 7p.
Here q is 49 miles.
49 = 112 – 7p
49 +7p = 112 – 7p +7p
49 +7p = 112
49 +7p – 49 = 112 – 49
7p = 63
7p÷7 = 63÷7
p=9
9 gallons of diesel were left after the bus has traveled 49 miles

c) After the bus has traveled for 56 miles, how much farther can the bus travel before it runs out of diesel?
Answer:
For 56 miles, 8 gallons of diesel will be used
Here p is 8
8 = 112 – 7p
8+7p = 112 – 7p + 7p
8+7p = 112
8+7p-8 = 112-8
7p = 104
7p÷7 = 104÷7
p = 14.8
After the bus has traveled for 56 miles, it will travel 0.8 m before it runs out of diesel.

d) If the bus travels more than 28 miles, how much diesel is left? Express your answer in the form of an inequality in terms of p, where p stands for the amount of diesel left.
Answer:
If the bus travels more than 28 miles, 12 gallons will be utilised.
It can be expressed as p ≥ 12, where p stands for the amount of diesel left.

Question 3.
A kettle of water is heated and the temperature of the water, j°C, after k minutes, is given by j = 5k + 30. Copy and complete the table. Graph the relationship between k and j. Use 1 unit on the horizontal axis to represent 1 minute and 1 unit on the vertical axis to represent 5°C.

a)

Answer:
Given that a kettle of water is heated and the temperature of the water, j°C, after k minutes, is given by j = 5k + 30.
If k = 0 then j will be
j = 5k + 30
j = 5×0 + 30
j = 30
If k = 4 then j will be
j = 5k + 30
j = 5×4 + 30
j = 20+30
j = 50
If j=70
70 = 5k + 30
70-30 = 5k+30-30
40 = 5k
40÷5 = 5k÷5
8 = k

b) What is the temperature of the water after 5 minutes?
Answer:
Given that a kettle of water is heated and the temperature of the water, j°C, after k minutes, is given by j = 5k + 30.
If k = 5 then j will be
j = 5k + 30
j = 5×5 + 30
j = 25+30
j = 55°C

c) What is the average rate of the heating?
Answer:
The average rate of the heating can be calculated by using the below formula,
Average rate = \(\frac{total temp}{total elapsed time}\)
Total temp = 30+40+50+60+70 = 250
Total time = 0+2+4+6+8 = 20 min
= \(\frac{250}{20}\)
= 12.5 rate of heat per min

d) Assuming the temperature of the water rises at a constant rate, what is the temperature of the water after 10 minutes?
Answer:
Given that a kettle of water is heated and the temperature of the water, j°C, after k minutes, is given by j = 5k + 30.
Here, k=10 min
j = 5k + 30
= 5×10 + 30
= 50+30
= 80°C
80°C is the temperature of the water after 10 minutes.

e) The kettle of water needs to be heated till the water boils. For how many minutes does the kettle need to be heated? Express your answer in terms of k, where k stands for the number of minutes. (Hint: Water boils at 100°C.)
Answer:
Given that the kettle of water needs to be heated till the water boils and it will boils at 100°C
j = 5k + 30
100 = 5k + 30
100-30 = 5k + 30 – 30
70 = 5k
70÷5 = 5k÷5
k = 14 min
The kettle needs to be heated for 14 mins.

Brain @ Work

Use graph paper. For each exercise, plot the points on a coordinate plane.

Question 1.
A (-5, 1), B (-3, -3), C (3, 1), and D (-1, 5)
Answer:
Point A (-5, 1):
As the x coordinate is negative, start from origin and move 5 points towards negative x axis.
As the y-coordinate is positive, start from origin and move 1 point towards positive y axis.
Thus, the required point A is plotted.
Point B (-3, -3):
As the x coordinate is negative, start from origin and move 3 points towards negative x axis.
As the y-coordinate is negative, start from origin and move 3 point towards negative y axis.
Thus, the required point B is plotted.
Point C (3,1):
As the x coordinate is positive, start from origin and move 3 points towards positive x axis.
As the y-coordinate is positive, start from origin and move 1 point towards positive y axis.
Thus, the required point C is plotted.
Point D(-1, 5):
As the x coordinate is negative, start from origin and move 1 points towards negative x axis.
As the y-coordinate is positive, start from origin and move 5 point towards positive y axis.
Thus, the required point D is plotted.

Question 2.
J (4, 2), K (-2, 4), L (-4, 0), and M (0, -2)
Answer:
Point J (4, 2):
As the x coordinate is positive, start from origin and move 4 points towards negative x axis.
As the y-coordinate is positive, start from origin and move 2 point towards positive y axis.
Thus, the required point J is plotted.
Point K (-2, 4):
As the x coordinate is negative, start from origin and move 2 points towards negative x axis.
As the y-coordinate is positive, start from origin and move 4 point towards positive y axis.
Thus, the required point K is plotted.
Point L (-4, 0):
As the x coordinate is negative, start from origin and move 4 points towards negative x axis.
As the y-coordinate is zero, mark at origin.
Thus, the required point L is plotted.
Point M (0, -2):
As the x coordinate is zero, mark at origin.
As the y-coordinate is negative, start from origin and move 2 point towards negative y axis.
Thus, the required point M is plotted.

Question 3.
S (-1, 3), T (-3, -1), U (1, -1), and V(5, 3)
Answer:
Point S (-1, 3):
As the x coordinate is negative, start from origin and move 1 points towards negative x axis.
As the y-coordinate is positive, start from origin and move 3 point towards positive y axis.
Thus, the required point A is plotted.
Point T (-3, -1):
As the x coordinate is negative, start from origin and move 3 points towards negative x axis.
As the y-coordinate is negative, start from origin and move 1 point towards negative y axis.
Thus, the required point B is plotted.
Point U (1,-1):
As the x coordinate is positive, start from origin and move 1 points towards positive x axis.
As the y-coordinate is negative, start from origin and move 1 point towards negative y axis.
Thus, the required point C is plotted.
Point V(5, 3):
As the x coordinate is positive , start from origin and move 5 points towards positive x axis.
As the y-coordinate is positive, start from origin and move 3 point towards positive y axis.
Thus, the required point D is plotted.

Question 4.
In questions 1 to 3, what is the figure formed?
Answer:
Join the plotted points A,B,C and D.
It will form a four sided figure or a quadrilateral.

Join the plotted points K,L,M and N.
It will form a four sided figure or a quadrilateral.

Join the plotted points S,T,U and V.
It will form a four sided figure or a quadrilateral.

Question 5.
a) For each figure in questions to mark the middle of each side and connect the points in order.
Answer:
Mark the mid-points of AB, BC, CD and DA line segment.

Mark the mid-points of JK, KL, LM and MJ line segment.

Mark the mid-points of SV, VU, UT and TS line segment.

b) What are the figures formed? Explain your answers.
Answer:
By joining the midpoints on ABCD:
Join the points to form a quadrilateral.
A parallelogram is formed as the opposite sides are parallel to each other.
By joining the midpoints of JLKM:
Join the points to form a quadrilateral.
A rectangle is formed as the opposite sides are equal to each other.
By joining the midpoints of STUV:
Join the points to form a quadrilateral.
A rectangle is formed as the opposite sides are equal to each other.

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 9 Review Test to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 9 Review Test Answer Key

Concepts and Skills

Use the coordinate plane below.

Question 1.
Give the coordinates of points A, B, C, D, and E.

Answer:
Point A:
The first coordinate represents distance along negative x axis which is 4 and the second coordinate represents distance along negative y axis which is 3.
So, the point A will be (-4,-3)
Point B:
The first coordinate represents distance along x axis which is 0 and the second coordinate represents distance along negative y axis which is 6.
So, the point B will be (0,-6)
Point C:
The first coordinate represents distance along x axis which is 2 and the second coordinate represents distance along negative y axis which is 4.
So, the point C will be (2,-4)
Point D:
The first coordinate represents distance along x axis which is 6 and the second coordinate represents distance along y axis which is 3.
So, the point D will be (6,3)
Point E:
The first coordinate represents distance along the negative x axis which is 2 and the second coordinate represents distance along y axis which is 3.
So, the point E will be (-2,3)

Use graph paper. Plot the points on a coordinate plane. In which quadrant is each point located?

Question 2.
A (3, 5), B (-2, 0), C (7, -2), D (0, -5), and E (-3, -8)
Answer:
Point A (3, 5): Since both the x and y coordinate are positive, the point A will be located in quadrant I.
Point B (-2, 0): Since the x coordinate is negative and y coordinate is origin, the point B will not lie in any quadrant. It will be located on x axis.
Point C (7, -2): Since the x coordinate is positive and y coordinate is negative, the point C will lie in quadrant IV.
Point D (0, -5): Since the x coordinate is zero and y coordinate is negative, the point D will not lie in any quadrant. It will be located on y axis.
Point E (-3, -8): Since both the x and y coordinate are negative, the point A will be located in quadrant III.

Use graph paper. Points A and B are reflections of each other about the x-axis. Give the coordinates of point B if the coordinates of point A are the following:

Question 3.
(3, 6)
Answer:
Given that the points A and B are reflections of each other about the x-axis.
Given that B is reflection of point A in the x-axis and point A is (3, 6).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x coordinate will be 3 and y coordinate will be -6.
Therefore, point B will be (3,-6).

Question 4.
(-6, 2)
Answer:
Given that B is reflection of point A in the x-axis and point A is (-6, 2).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x coordinate will be 3 and y coordinate will be -2.
Therefore, point B will be (-6,-2).

Question 5.
(5, -4)
Answer:
Given that B is reflection of point A in the x-axis and point A is (5, -4).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x coordinate will be 5 and y coordinate will be 4.
Therefore, point B will be (5,4).

Question 6.
(-3, -5)
Answer:
Given that B is reflection of point A in the x-axis and point A is (-3, -5).
If a point is reflected across the x-axis, The x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse.
Thus, x coordinate will be -3 and y coordinate will be 5.
Therefore, point B will be (-3,5).

Use graph paper. Points C and D are reflections of each other about the y-axis. Give the coordinates of point D if the coordinates of point C are the following:

Question 7.
(3, 6)
Answer:
Given that the points C and D are reflections of each other about the y-axis.
Given that C is reflection of point D in the y-axis and point C is (3, 6).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be -3 and y coordinate will be 6.
Therefore, point D will be (-3,6).

Question 8.
(-6, 2)
Answer:
Given that B is reflection of point A in the y-axis and point C is (-6, 2).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be 6 and y coordinate will be 2.
Therefore, point D will be (6,2).

Question 9.
(5, -4)
Answer:
Given that B is reflection of point A in the y-axis and point C is (5, -4).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be -5 and y coordinate will be -4.
Therefore, point D will be (-5,-4).

Question 10.
(-3, -5)
Answer:
Given that B is reflection of point A in the y-axis and point C is (-3, -5).
If a point is reflected across the y-axis, The x-coordinate is taken to be the additive inverse and the y-coordinate remains the same.
Thus, x cordinate will be 3 and y coordinate will be -5.
Therefore, point D will be (3,-5).

Use graph paper. For each exercise, plot the given points on a coordinate plane. Then connect the points in order with line segments to form a closed figure. Name each figure formed.

Question 11.
A (2, -4), B (2, 4), C (-6, 4), and D (-6, -4)
Answer:
Plot the points A,B,C and D on a coordinate plane. Connect the points in order with line segments to form a closed figure.
The formed figure has all equal sides. Therefore, it can be a square.

Question 12.
E (0, -2), F (-3, 1), G(-5, -1), and H (-2, -4)
Answer:
Plot the points E,F,G and H on a coordinate plane. Connect the points in order with line segments to form a closed figure.
The formed figure has all four sides different. Therefore, it can be a quadrilateral.

Question 13.
J (0, 1), K(1, 4), and L(-4, 3)
Answer:
Plot the points J,K and L on a coordinate plane. Connect the points in order with line segments to form a closed figure.
The formed figure has all three sides different. Therefore, it can be a triangle.

Question 14.
M (6, 5), N (3, 5), P (3, -3), and Q (6, -3)
Answer:
Plot the points M,N,P and Q on a coordinate plane. Connect the points in order with line segments to form a closed figure.
The formed figure has opposite sides equal. Therefore, it can be a rectangle.

Question 15.
A (6, -3), B (4, 2), C (-1, 2), and D (0, -3)
Answer:
Plot the points A,B,C and D on a coordinate plane. Connect the points in order with line segments to form a closed figure.
The formed figure has a pair of opposite sides parallel and equal. Therefore, it can be a parallelogram.

Question 16.
E (-1, 6), F (-3, 3), G (3, 3), and H (5, 6)
Answer:
Plot the points E,F,G and H on a coordinate plane. Connect the points in order with line segments to form a closed figure.
The formed figure has all different sides. Therefore, it can be a quadrilateral.

Question 17.
J(6, 1), K(8, -2), L (2, -2), and M (0, 1)
Answer:
Plot the points J,K,L and M on a coordinate plane. Connect the points in order with line segments to form a closed figure.
The formed figure has a pair of opposite sides parallel and equal. Therefore, it can be a parallelogram.

Question 18.
P (2, 7), Q (-1, 7), R (-5, 4), and S (4, 4)
Answer:
Plot the points P,Q,R and S on a coordinate plane. Connect the points in order with line segments to form a closed figure.
The formed figure has a pair of opposite sides parallel and equal. Therefore, it can be a trapezium.

Question 19.
T (-2, 1), U (-6, 1), V(-6, -3), and W(-2, -3)
Answer:
Plot the points T,U,V and W on a coordinate plane. Connect the points in order with line segments to form a closed figure.
The formed figure has opposite sides equal. Therefore, it can be a rectangle.

Use graph paper. Plot the points on a coordinate plane and answer the question.

Question 20.
a) Plot points A (1, -1) and B (7, -1) on a coordinate plane. Connect the two points to form a line segment.
Answer:
Point A:
The first coordinate represents distance along positive x axis which is 1 and the second coordinate represents distance along negative y axis which is 1.
So, the point A will be (1,-1)
Point B:
The first coordinate represents distance along positive x axis which is 7 and the second coordinate represents distance along negative y axis which is -1.
So, the point A will be (7,-1)

b) Point C lies above \(\overline{\mathrm{AB}}\), and is 2 units away from the x-axis. If triangle ABC is an isosceles triangle with base \(\overline{\mathrm{AB}}\), find the coordinates of point C.
Answer:
Point C lies above \(\overline{\mathrm{AB}}\), and is 2 units away from the x-axis. If triangle ABC is an isosceles triangle with base \(\overline{\mathrm{AB}}\)
Since it is an isosceles triangle, two sides will be equal. When the line segment is joined through midpoint, the formed line segments will be equal in length.
Midpoint of AB line segment is 3 units, so midpoint will be (4,2).
Thus, coordinates of point C will be (4,2)

c) Points D and E lie below \(\overline{\mathrm{AB}}\) such that ABDE is a rectangle. If BD is 5 units, find the coordinates of points D and E.
Answer:
Given that BD is 5 units.
Points D and E lie below \(\overline{\mathrm{AB}}\) such that ABDE is a rectangle.
Start from point B and move 5 units above and mark it. Thus, the required point D will be formed.
Since it is a rectangle, opposite sides will be equal. Therefore, AB will be equal to DE, which is 6 units.
Start from point D and move 6 units towrads left and mark it is as E.
Thus, the formed D point will be (7,4) and E will be (1,4).

Use graph paper. Plot each pair of points on a coordinate plane. Connect the points to form a line segment and find its length.

Question 21.
A (-1,0) and B (8, 0)
Answer:
To plot point A (-1,0):
Here, the x-coordinate of A is -1 and the y-coordinate is 0. Start at the Origin. As the x coordinate is negative, move 1 units on the negative x-axis and point it.
Thus, the required point (-1,0) is marked.
To plot point B (8, 0):
Here, the x-coordinate of A is 8 and the y-coordinate is 0. Start at the Origin. As the x coordinate is positive, move 8 units along the positive x-axis and point it.
Thus, the required point B (8, 0) is marked.

After connecting the points to form a line segment, the length will be 9 units.

Question 22.
C (-2, 4) and D (6, 4)
Answer:
To plot point C (-2, 4):
Here, the x-coordinate is -2 and the y-coordinate is 4. Start at the Origin. As the x coordinate is negative, move 2 units on the negative x-axis and 4 units on the positive y -axis.
Thus, the required point C (-2,4) is marked.
To plot point D (6, 4):
Here, the x-coordinate is 6 and the y-coordinate is 4. Start at the Origin. As the x coordinate is positive, move 4 units along the positive x-axis and 4 units along positive y -axis.
Thus, the required point D (6, 4) is marked.

After connecting the points to form a line segment, the length will be 8 units.

Question 23.
E (-6, -2) and F(-6, -6)
Answer:
To plot point E (-6, -2):
Here, the x-coordinate is 6 and the y-coordinate is-2. Start at the Origin. As the x coordinate is negative, move 6 units on the negative x-axis and 2 units on the negative y -axis.
Thus, the required point C (-2,4) is marked.
To plot point F(-6, -6):
Here, the x-coordinate is -6 and the y-coordinate is -6. Start at the Origin. As the x coordinate is negative, move 6 units along the negative x-axis and 6 units along negative y -axis.
Thus, the required point D (-6, -6) is marked.

After connecting the points to form a line segment, the length will be 4 units.

Question 24.
G (-5, -4) and H (2, -4)
Answer:
To plot point G (-5, -4):
Here, the x-coordinate is -5 and the y-coordinate is-4. Start at the Origin. As the x coordinate is negative, move 5 units on the negative x-axis and 4 units on the negative y -axis.
Thus, the required point G (-5, -4) is marked.
To plot point H (2, -4):
Here, the x-coordinate is 2 and the y-coordinate is -4. Start at the Origin. As the x coordinate is positive, move 4 units along the positive x-axis and 4 units along negative y -axis.
Thus, the required point H (2, -4) is marked.

After connecting the points to form a line segment, the length will be 7 units.

Question 25.
J(0, -3) and K(0, -8)
Answer:
To plot point J(0, -3):
Here, the x-coordinate is 0 and the y-coordinate is-3. Start at the Origin. As the x coordinate is origin, move 3 units on the negative y -axis.
Thus, the required point J(0, -3) is marked.
To plot point K(0, -8):
Here, the x-coordinate is 0 and the y-coordinate is-8. Start at the Origin. As the x coordinate is origin, move 8 units on the negative y -axis.
Thus, the required point K(0, -8) is marked.

After connecting the points to form a line segment, the length will be 5 units.

Question 26.
M (5, 2) and N (5, -5)
Answer:
To plot point M (5, 2):
Here, the x-coordinate is 5 and the y-coordinate is 2. Start at the Origin. As the x and y coordinates are positive , move 5 units on the positive x-axis and 2 units on the positive y -axis.
Thus, the required point M (5, 2) is marked.
To plot point N (5, -5):
Here, the x-coordinate is 5 and the y-coordinate is 5. Start at the Origin. Move 5 units on the positive x-axis and 2 units on the negative y -axis.
Thus, the required point N (5, -5) is marked.

After connecting the points to form a line segment, the length will be 7 units.

Problem Solving

The diagram shows the plan of a room. The side length of each grid square is 10 feet. Use the diagram to answer questions 27 to 28.

Question 27.
The eight corners of the room are labeled points A to H. Give the coordinates of each of these corners.
Answer:
The first coordinate represents distance along negative x axis which is 40 and the second coordinate represents distance along y axis which is 100.
The coordinates of A will be (-40,100)
Similarly,The coordinates of B will be (-40,20)
The coordinates of C will be (-60,20)
The coordinates of D will be (-60,-40)
The coordinates of E will be (60,40)
The coordinates of F will be (60,20)
The coordinates of G will be (40,20)
The coordinates of H will be (40,100)

Question 28.
The entrance of the room is situated along \(\overline{\mathrm{AH}}\). What is the shortest possible distance in feet between the entrance and \(\overline{\mathrm{DE}}\) of the room?
Answer:
The entrance of the room is situated along \(\overline{\mathrm{AH}}\).
The shortest distance between the the entrance and \(\overline{\mathrm{DE}}\) of the room will be the straight route from AH to DE.
The distance between AH to DE is 14 square grids.
Each square grid equals to 10 feet.
Thus, the distance between AH to DE will be 14×10 = 140 ft.

Question 29.
Diana walks across the room from point B to point G, and then walks from point G to point H. Find the total distance, in feet, that Diana walks.
Answer:
Given that Diana walks across the room from point B to point G, and then walks from point G to point H.
Distance between B and G will be 8 square grids, 8×10=80 ft.
Distance between G to H will be 8 square grids, 8×10=80 ft.
The total distance will be BG+GH = 80+80 = 160 ft

Question 30.
Calculate the floor area of the room in square feet.
Answer:
As the complete floor area is in irregular shape, we will find the area in parts.
To calculate the floor area, we will find the area of rectangle AHGB and area of CFED
Area of rectangle AHGB:
Length of AB will be 8 square grids, 8×10=80 ft.
Length of AH will be 8 square grids, 8×10=80 ft.
Area of the rectangle = length×width
= 80×80
= 6400 sq.ft
Area of rectangle CFED:
Length of CD will be 6 square grids, 6×10=60 ft.
Length of DE will be 12 square grids, 12×10=120 ft
Area of the rectangle = length×width
= 60×120
= 7200 sq.ft
Thus, the total area of floor will be 6400+7200 = 13600 sq.ft

Use graph paper. Solve.

Question 31.
An athlete took part in a race. The distance the athlete ran, v meters, after t minutes, is given by v = 300t. Graph the relationship between t and v. Use 2 units on the horizontal axis to represent 1 minute and 1 unit on the vertical axis to represent 150 meters.

a) What type of graph is it?
Answer:
It is a straight line graph.This is also called a linear graph.

b) What is the distance the athlete ran in 3.5 minutes?
Answer:
An athlete took part in a race. The distance the athlete ran, v meters, after t minutes, is given by v = 300t.
Here, t=3.5 min
v = 300t
v = 300×3.5
= 1050 m
Thus the distance the athlete ran in 3.5 min will be 1050 m.

c) What is the average speed of the athlete?
Answer:
The average speed of the athlete will be,
Average speed = \(\frac{total distance}{total time}\)
Total distance will be 0+300+600+900+1200 = 3000 m
Total time will be 0+1+2+3+4 = 10 min
= \(\frac{3000}{10}\)
= 300 meters per min

d) Assuming the athlete runs at a constant speed, what is the distance she will run in 8 minutes?
Answer:
The distance the athlete ran, v meters, after t minutes, is given by v = 300t.
The distance she will run in 8 min will be, 300×8=2400 m

e) Name the dependent and independent variables.
Answer:
From the graph, we can say v is the dependent variable and t is the independent variable.

Question 32.
A truck uses 1 gallon of diesel for every 12 miles traveled, The amount of diesel left in the gas tank, r gallons, after traveling s miles is given by s = 300 – 12r. Copy and complete the table. Graph the relationship between r and s. Use 1 unit on the horizontal axis to represent 1 gallon and 1 unit on the vertical axis to represent 12 miles. Start your horizontal axis at 17 gallons.
a)

Answer:
A truck uses 1 gallon of diesel for every 12 miles traveled, The amount of diesel left in the gas tank, r gallons, after traveling s miles is given by s = 300 – 12r.
Here s=24,
24 = 300 – 12r
24+12r = 300-12r+12r
24+12r = 300
24+12r-24 = 300-24
12r = 276
r = 276÷12 = 23 gallons
Here r=17,
s = 300 – 12×17
= 300-204
= 96 miles

b) How many gallons of diesel are left after the truck has traveled 60 miles?
Answer:
If 60 miles is travelled,
s = 300 – 12r
60 = 300 – 12r
60 + 12r = 300 – 12r + 12r
60 + 12r = 300
60 + 12r – 60 = 300-60
12r = 240
12r÷12 = 240÷12
r = 20
Thus, 20 gallons of diesel are left after the truck has traveled 60 miles

c) After the truck has traveled for 72 miles, how much farther can the truck travel before it runs out of diesel?
Answer:
After 72 miles, 19 gallons of gas were left.
Distance = 19 × mileage
72 =  19×mileage
72÷19 = (19×mileage)÷19
3.7 = mileage
The car can travel another 3.7 miles

d) If the truck travels more than 48 miles, how much diesel is left in the gas tank? Express your answer in the form of an inequality in terms of r, where r stands for the amount of diesel left.
Answer:
If the truck travels more than 48 miles, r ≥ 21, where r stands for the amount of diesel left.

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 10 Area of Polygons to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 10 Answer Key Area of Polygons

Math in Focus Grade 6 Chapter 10 Quick Check Answer Key

Solve.

Question 1.
The length of a rectangle is 15 meters and its width is 8 meters. Find the area of the rectangle.
Answer:
Given that the length of a rectangle is 15 meters and its width is 8 meters.
Formula to find the area of rectangle will be,
= length × width
Here length = 15m and width = 8m
= 15×8
= 120 sq.m

Solve.

Question 2.
A side length of a square is 10 centimeters. Find the area of the square.
Answer:
Given that a side length of a square is 10 cm.
Formula to find the area of a square
= length×length
= 10×10
= 100sq.cm
Thus, the area of square will be 100sq.cm

Name each figure and identify the pairs of parallel lines.

Question 3.

Answer:
Given figure RSTU is similar to rhombus.
In rhombus opposite sides are parallel to each other.
Here, line segment RU is parallel to ST and RS is parallel to UT.
Therefore, RS and UT, RU and ST forms the pairs of parallel lines.

Question 4.

Answer:
Given figure EFGH is similar to parallelogram.
In parallelogram opposite sides are parallel to each other.
Here, line segment EH is parallel to FG and EF is parallel to HG.
Therefore, EH and FG, EF and HG forms the pairs of parallel lines.

Question 5.

Answer:
Given figure has two opposite sides of different lengths which is similar to the properties of trapezium.
Therefore, we can assume the given figure to be  a trapezium.
In trapezium, a pair of sides will be parallel to each other.
Here, in the given figure JM is parallel to KL, thus forming a pair of parallel lines.

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 10 Lesson 10.1 Area of Triangles to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 10 Lesson 10.1 Answer Key Area of Triangles

Math in Focus Grade 6 Chapter 10 Lesson 10.1 Guided Practice Answer Key

Hands-On Activity

Materials:

• scissors
• graph paper

PROVE THE FORMULA FOR FINDING THE AREA OF A TRIANGLE

Work in pairs.

Triangle PQR is an acute triangle. \(\overline{\mathrm{Q} R}\) is the base and PX is the height.

Step 1.
Draw triangle POR on a piece of graph paper as shown. Then draw and label rectangle AQRD,
Example

Step 2.
Cut up triangle PQR into smaller triangles. Rearrange the triangles to form rectangle EQRF, as shown below.
Example

The orange, blue, and yellow figures form a rectangle.

Step 3.
Find the area of rectangle EQRF. How does its area compare to the area of rectangle AQRD?
Answer:
Area of rectangle EQRF:
Figure EQRF is occupying 8 units wide and 2 units high.
Area of rectangle = length × width
= 8×2
= 16 sq.units.
Area of rectangle AQRD:
Figure AQRD is occupying 8 units wide and 4 units high.
Area of rectangle = length × width
= 8×4
= 32 sq.units.
By observing the area of EQRF and AQRD, we can say that the area of ANPD is twice the area of AQRD.

How does the area of triangle PQR compare to the area of rectangle EQRF?
Answer:
Area of triangle MNP:
Let the base of the triangle be PQR , which is occupying 8 horizontal squares. Thus, the base will be 8 units.
The perpendicular segment between the base and the opposite vertex will be the height of the triangle which is 4 units.
Area of triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×8×4
= \(\frac{1}{2}\)×32
= 16 sq.units
Area of rectangle EQRF:
Figure EQRF is occupying 8 units wide and 2 units high.
Area of rectangle = length × width
= 8×2
= 16 sq.units.
By observing the area of triangle MNP and rectangle ENPF, we can say that both the areas are same.

How does the area of triangle PQR compare to the area of rectangle AQRD?
Answer:
Area of triangle PQR is 16 sq.units.
Area of rectangle AQRD is 32 sq.units
By observing the area of triangle PQR and the area of rectangle AQRD, we can say that the area of AQRD is twice the area of PQR.

Triangle MNP is an obtuse triangle. \(\overline{N P}\) is the base and MF is the height.

Step 1.
Draw triangle MNP on a piece of graph paper as shown. Then draw and label rectangle ANPD.
Example

Step 2.
Cut up triangle MNP into smaller triangles. Rearrange the triangles to form rectangle ENPF, as shown below.

Step 3.
Find the area of rectangle ENPF. How does its area compare to the area of rectangle ANPD?
Answer:
Area of rectangle ENPF:
Figure ENPF is occupying 3 units wide and 3 units high. Since all sides are equal, it is a square.
Area of square = length × length
= 3×3
= 9 sq.units.
Area of rectangle ANPD:
Figure ANPD is occupying 3 units wide and 6 units high, forming a rectangle.
Area of rectangle = length×width
= 3×6
= 18 sq.units
By observing the area of ENPF and ANPD, we can say that the area of ANPD is twice the area of ENPF.

How does the area of triangle MNP compare to the area of rectangle ENPF?
Answer:
Area of triangle MNP:
Let the base of the triangle be NP, which is occupying 3 horizontal squares. Thus, the base will be 3 units.
The perpendicular segment between the base and the opposite vertex will be the height of the triangle which is 6 units.
Area of triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×3×6, multiplying 1×3×6 will give 18 and denominator remains same
= \(\frac{18}{2}\)
= 9 sq.units
By observing the area of triangle MNP and rectangle ENPF, we can say that both the areas are same.

How does the area of triangle MNP compare to the area of rectangle ANPD?
Answer:
Area of triangle MNP is 9 sq.units.
Area of rectangle ANPD is 18 sq.units
By observing the area of triangle MNP and the area of rectangle ANPD, we can say that the area of ANPD is twice the area of MNP.

Complete to find the base, height, and area of each triangle. Each square measures 1 unit by 1 unit.

Question 1.

Base = units
Height = units
Area = \(\frac{1}{2}\)bh
= ∙ ∙
= units²
Answer:
Given that:
The above given triangle is drawn on a graph in which each square measures 1 unit.
Let the horizontal side be the base of triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle.
The triangle drawn is occupying 5 horizontal square boxes which is base and 8 vertical square boxes which is height.
Thus the base length of the triangle will be 5 units and height will be 8 units.
Base = 5 units, Height = 8 units
Formula:
Area of triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×5×8, In numerator, multiplying 1×5×8 will give 40 and denominator remains same
= \(\frac{40}{2}\), 40 divided by 2 gives 20
Therefore, area of the given triangle will be 20 sq.units

Question 2.

Base = units
Height = units
Area = \(\frac{1}{2}\)bh
= ∙ ∙
= units²
Answer:
The above given triangle is drawn on a graph in which each square measures 1 unit.
Let the vertical side be the base of triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which lies on the horizontal axis.
The vertical boxes are occupying 3 square boxes and horizontally they are occupying 4 boxes.
Thus the base length of the triangle will be 3 units and height will be 4 units.
Base = 3 units, Height = 4 units
Formula:
Area of triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×3×4, In numerator, multiplying 1×3×4 will give 12 and denominator remains same
= \(\frac{12}{2}\), 12 divided by 2 gives 6
Therefore, area of the given triangle will be 6 sq.units

Question 3.

Base = cm
Height = cm
Area of triangle ABC
= \(\frac{1}{2}\)bh
= ∙ ∙
= cm²
Answer:
From the given figure, the height of the triangle is 1.8 cm and base length is 2.1 cm.
Base = 2.1cm
Height = 1.8cm
Area of triangle ABC = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×2.1×1.8, multiplying 1×2.1×1.8 gives 3.78
= \(\frac{3.78}{2}\)
= 1.89 sq.cm

Question 4.

Base = ft
Height = ft
Area of triangle PQR
= \(\frac{1}{2}\)bh
= ∙ ∙
= ft²
Answer:
From the given figure, the height of the triangle is 2.7 ft and base length is 3.4 ft.
Base = 2.7 ft
Height = 3.4 ft
Area of triangle PQR
= \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×3.4×2.7
= \(\frac{1}{2}\)×9.18
= 4.59 sq.ft

Complete to find the height of triangle JKL.

Question 5.
The area of triangle JKL is 35 square meters. Find the height of triangle JKL.

Answer:
Given that the area of triangle JKL is 35 sq.m. The base of the triangle is 7m.
Area of triangle JKL = \(\frac{1}{2}\)×b×h
35 = \(\frac{1}{2}\)×b×h
35 = 0.5×7×h
35 = 3.5×h
35÷3.5 = (3.5×h)÷3.5
10 = h
The height of triangle JKL is 10 m.

Complete to find the base of each triangle.

Question 6.
The area of triangle LMN is 36 square inches. Find the base of triangle LMN.

Area of triangle LMN = \(\frac{1}{2}\)bh
= ∙ b ∙
= ∙ ∙ b
= ∙ b
÷ = b ÷
= b
The base of triangle LMN is inches.
Answer:
Given that the area of triangle LMN is 36 sq.in and the height of the triangle is 8 in.
36 = \(\frac{1}{2}\)×b×h
36 = \(\frac{1}{2}\)×b×8
36 = 4 × b
36 ÷ 4 = (4 ×b) ÷ 4
9 = b
The base of triangle LMN is 9 inches.

Question 7.
The area of triangle PQR1s 19.2 square centimeters. Find the base of triangle PQR.

Area of triangle PQR = \(\frac{1}{2}\)bh
= ∙ b ∙
= ∙ ∙ b
= ∙ b
÷ = b ÷
= b
The base of triangle PQR is inches.
Answer:
Given that the area of triangle PQR is 19.2 sq.cm and the height of the triangle is 9.6 cm.
Area of triangle PQR = \(\frac{1}{2}\)×b×h
19.2 = \(\frac{1}{2}\)×b×9.6
19.2 = 4.8 × b
19.2÷4.8 = (4.8×b)÷4.8
4 = b
The base of triangle PQR is 4 inches.

Math in Focus Course 1B Practice 10.1 Answer Key

Identify a base and a height of each triangle.

Question 1.

Answer:
Let the side BC be the base of triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle,which is AB.
Therefore, the base of triangle is BC and height is AB.

Question 2.

Answer:
Let the horizontal side QR be the base of triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle,which is PR.
Therefore, the base of triangle is QR and height is PR.

Copy each triangle. Label a base with the letter b and a height with the letter h.

Question 3.

Answer:

Explanation:
Let the side AB be the base of triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle,which is AC.
Therefore, the side AB can be marked as b and the side AC as h.

Question 4.

Answer:

Explanation:
Let the side EF be the base of triangle and a perpendicular segment is drawn between the base and the opposite vertex which will be the height of the triangle,which is DC.
Therefore, the side EF can be marked as b and the side DC as h.

Question 5.

Answer:

Explanation:
Let the side GH be the base of triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle,which is HI.
Therefore, the side GH can be marked as b and the side HI as h.

Question 6.

Answer:

Explanation:
Let the side JK be the base of triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle,which is JL.
Therefore, the side JK can be marked as b and the side JL as h.

Find the area of each triangle.

Question 7.

Answer:
Let us assume the side measuring 25 cm be the base of the triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 4 cm.
Thus, the base of triangle = 25 cm
and the height of the triangle = 4 cm
Area of triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×25×4
= \(\frac{1}{2}\)×100
= 100÷2
= 50
Therefore, the area of the triangle will be 50 sq.cm

Question 8.

Answer:
Let us assume the horizontal side measuring 15.5 cm be the base of the triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 6 cm.
Thus, the base of triangle = 15.5 cm
and the height of the triangle = 6 cm
Area of triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×15.5×6
= \(\frac{1}{2}\)×93
= 93÷2
= 46.5
Therefore, the area of the triangle will be 46.5 sq.cm

The area of each triangle is 76 square inches. Find the height and round your answer to the nearest tenth of an inch.

Question 9.

Answer:
Given that the area of triangle is 76 sq.in and base is 15.7 in
We now that, Area of triangle = \(\frac{1}{2}\)×b×h
76 = \(\frac{1}{2}\)×15.7×h
76 = (15.7÷2)×h
76 = 7.85 × h
76 ÷ 7.85 = (7.85 × h)÷7.85
9.6 = h
The value nearest to tenth will be 10 in.
Therefore, the height of the given triangle will be 10 in.

Question 10.

Answer:
Given that the area of triangle is 76 sq.in and base is 9.3 in
We now that, Area of triangle = \(\frac{1}{2}\)×b×h
76 = \(\frac{1}{2}\)×9.3×h
76 = (9.3÷2)×h
76 = 4.65 × h
76 ÷ 4.65 = (4.65×h)÷4.65
16.3 = h
The value nearest to tenth will be 16 in.
Therefore, the height of the given triangle will be 16 in.

The area of each triangle is 45 square centimeters. Find the base and round your answer to the nearest tenth of a centimeter if necessary.

Question 11.

Answer:
Given that the area of triangle is 45 sq.cm and height is 7.2 cm
We now that, Area of triangle = \(\frac{1}{2}\)×b×h
45 = \(\frac{1}{2}\)×b×7.2
45 = (7.2 ÷ 2)×b
45 = 3.6 × b
45 ÷ 3.6 = (3.6 × b)÷3.6
12.5 = b
The value nearest to tenth will be 12 in.
Therefore, the base of the given triangle will be 12 in.

Question 12.

Answer:
Given that the area of triangle is 45 sq.cm and height is 16.6 cm
We now that, Area of triangle = \(\frac{1}{2}\)×b×h
45 = \(\frac{1}{2}\)×b×16.6
45 = 8.3 × b
45 ÷ 8.3 = (8.3 × b) ÷ 8.3
5.4 = b
The value nearest to base will be 5 cm.
Therefore, the base of the given triangle will be 5 cm.

Question 13.

Answer:
Given that the area of triangle is 45 sq.cm and height is 7.5 cm
We now that, Area of triangle = \(\frac{1}{2}\)×b×h
45 = \(\frac{1}{2}\)×b×7.5
45 = 3.75 × b
45 ÷ 3.75 = (3.75 × b)÷ 3.75
12 = b
Therefore, the base of the given triangle will be 12 cm.

Question 14.

Answer:
Given that the area of triangle is 45 sq.cm and height is 15 cm
We now that, Area of triangle = \(\frac{1}{2}\)×b×h
45 = \(\frac{1}{2}\)×b×15
45 = 7.5 × b
45 ÷ 7.5 = (7.5×b) ÷ 7.5
6 = b
Therefore, the base of the given triangle will be 6 cm.

Find the area of the shaded region.

Question 15.

Answer:
Total Area:
Given rectangle measures 20m wide and 40m long.
Area of rectangle = length×width
= 20×40
= 800 sq.m
Larger triangle:
Let us assume the 40 m line segment as base and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 20 m.
Area of larger triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×40×20
= \(\frac{1}{2}\)×800
= 800÷2
= 400 sq.m
Smaller triangle:
Let us assume the 20 m line segment as base and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 7 m.
Area of larger triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×20×7
= \(\frac{1}{2}\)×140
= 140÷2
= 70 sq.m
Area of shaded region = Total Area – (Area of larger triangle + Area of smaller triangle)
= 800 – (400+70)
= 800 – 470
= 330 sq.m

Question 16.

Answer:
Larger triangle:
Let us assume the 24 in line segment as base and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures (12-5)=7 in.
Area of larger triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×24×7
= \(\frac{1}{2}\)×168
= 168÷2
= 84 sq.in
Smaller triangle:
Let us assume the 12 in line segment as base and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 10 in.
Area of larger triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×12×10
= \(\frac{1}{2}\)×120
= 120÷2
= 60 sq.m
Area of shaded region = Area of larger triangle + Area of smaller triangle
= 84 + 60
= 144 sq.in

Question 17.

Answer:

First triangle:
Let us assume the base of first triangle be 12 cm and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 34 cm.
Area of first triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×12×34
= \(\frac{1}{2}\)×408
= 408÷2
= 204 sq.cm
Second triangle:
Let us assume the base of second triangle be 12 cm and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 10 cm.
Area of second triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×12×10
= \(\frac{1}{2}\)×120
= 120÷2
= 60 sq.cm
Third triangle:
Let us assume the base of first triangle be 24 cm and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 24 cm.
Area of first triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×24×24
= \(\frac{1}{2}\)×576
= 576÷2
= 288 sq.cm
Total Area:
The rectangle measures 24 cm wide and 34 cm high
Area of rectangle = length×width
= 24×34
= 816 sq.cm
Area of shaded region will be =  Total Area – (Area of first triangle + Area of second triangle + Area of third trinagle)
= 816 – (204+60+288)
= 816 – 552
= 264 sq.cm

Question 18.

Answer:

Let us assume the base of first triangle be 6 ft and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 24 ft.
Area of first shaded triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×6×24
= 144÷2
= 72 sq.ft
Area of second shaded triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×24×12
= 288÷2
= 144 sq.ft
Area of shaded region = Area of first triangle + Area of second triangle
= 72+144
= 216 sq.ft

Question 19.

Answer:

Area of larger triangle:
Let us assume the base of first triangle be 6 cm and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 12 cm.
Area of larger triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×6×12
= 72÷2
= 36 sq.cm
Area of smaller triangle:
Let us assume the base of first triangle be 6 cm and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 6 cm.
= \(\frac{1}{2}\)×6×6
= 36÷2
= 18 sq.cm
Area of shaded region = Area of larger triangle + Area of smaller triangle
= 36 + 18
= 54 sq.cm

Use graph paper. Solve.

Question 20.
The coordinates of the vertices of a triangle are A (4, 7), B (4, 1), and C (8, 1). Find the area of triangle ABC.
Answer:

Given that the vertices of a triangle are A (4, 7), B (4, 1), and C (8, 1).
Let BC be the base of the triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which is AB.
BC measures 4 units and AB measures 6 units.
Area of triangle ABC = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×4×6
= 24 ÷ 2
= 12 sq.units

Question 21.
The coordinates of the vertices of a triangle are D (1, 7), E (-3, 2), and F (6, 2). Find the area of triangle DEF.
Answer:

Given that the vertices of a triangle are D (1, 7), E (-3, 2), and F (6, 2).
Let EF measuring 9 units be the base of the triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 5 units.
Area of triangle DEF = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×9×5
= 45 ÷ 2
= 22.5 sq.units

Question 22.
The coordinates of the vertices of a triangle are J (-5, 2), K (1, -2), and L (5, -2). Find the area of triangle JKL.
Answer:

Given that the vertices of a triangle are J (-5, 2), K (1, -2), and L (5, -2).
Let LK measuring 4 units be the base of the triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 4 units.
Area of triangle JKL= \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×4×4
= 16 ÷ 2
= 8 sq.units

Question 23.
The area of triangle MNP is 1 7.5 square units. The coordinates of M are (-9, 5), and the coordinates of N are (-2, 0). The height of triangle MNP is 5 units and is perpendicular to the x-axis. Point P lies to the right of point N. Given that \(\overline{N P}\) is the base of the triangle, find the coordinates of point P.
Answer:
Given that the area of triangle MNP is 17.5 square units and height of the triangle is 5 units.
Area of triangle = \(\frac{1}{2}\)×b×h
17.5 = \(\frac{1}{2}\)×b×5
17.5 = 2.5×b
(17.5÷2.5) = (2.5xb)÷2.5
7 = b
Thus, the base of the triangle will be 7 units.

Explanation:
From the above calculations, the base of the triangle measures 7 units and lies on the right of the point N.
Move 7 units to the right from the point N and mark the point. The marked point is the required P point.
Thus, the coordinates of P will be (5,0)

Question 24.
The coordinates of the vertices of a triangle are X (1, 2), Y (-6, -2), and Z (1, -4). Find the area of triangle XYZ. (Hint: Use the vertical side as the base.)
Answer:

Given that the vertices of a triangle are X (1, 2), Y (-6, -2), and Z (1, -4).
As given the base of the triangle is the vertical side, XZ which measures 6 units.
The perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 5 units.
Area of triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×6×5
= \(\frac{1}{2}\)×30
= 15 sq.units

Question 25.
The coordinates of the vertices of a triangle are P (-2, 6), Q (-4, 2), and R (5, 1). Find the area of triangle PQR. (Hint: Draw a rectangle around triangle PQR.)
Answer:

Part 1:
From the given figure, we can observe that the rectangle around triangle PQR measures 5 units long and 9 units wide.
Area of rectangle = length× width
= 5×9
= 45 sq.units
The total area of the figure is 45 sq.units
Part 2:
Area of first triangle:
Let us assume the base of first triangle be 2 units and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 4 units
Area of first triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×2×4
= \(\frac{1}{2}\)×8
= 8÷2
= 4 sq.units
Part 3:
Area of second triangle:
Let us assume the base of second triangle be 7 units and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 5 units
Area of first triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×7×5
= \(\frac{1}{2}\)×35
= 35÷2
= 17.5 sq.units
Part 4:
Area of third triangle:
Let us assume the base of third triangle be 9 units and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 1 units
Area of first triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×9×1
= 9÷2
= 4.5 sq.units
Part 5:
Area of the triangle = Total Area-(Area of first triangle+Area of second triangle+Area of third triangle)
= 45-(4+17.5+4.5)
= 45-26
= 19 sq.units

Find the area of the shaded region for questions 26 to 29.

Question 26.
Figure DGHJ is a trapezoid.

Answer:
Given that figure DGHJ is a trapezoid measuring 28 ft high and two parallel sides with 24ft and 56 ft long.
Area of trapezoid = \(\frac{1}{2}\)×(a+b)×h
= \(\frac{1}{2}\)×(24+56)×28
= \(\frac{1}{2}\)×80×28
= 2240÷2
= 1120 sq.ft
The unshaded figure seems to be square as each angle is 90 degrees. Thus, all sides of it will be equal.
Area of square = 24×24
= 576 sq.ft
Area of the shaded region = Total Area – Area of the unshaded region
= 1120-576
= 544 sq.ft

Question 27.
Figure ABCD is a rectangle. The length of \(\overline{Z B}\) is \(\frac{3}{7}\) the length of \(\overline{A B}\).

Answer:
Part 1:
From the given figure, we can observe that the area of shaded region will be the sum of areas of square and triangle.
Given that the figure ABCD is a rectangle and the length of \(\overline{Z B}\) is \(\frac{3}{7}\) the length of \(\overline{A B}\).
Length of AB = 35cm
Length of ZB will be \(\frac{3}{7}\) × AB
= \(\frac{3}{7}\) × 35
= 3×5
= 15 cm.
Area of square = 15×15
= 225 sq.cm
Area of triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×15×35
= 525÷2
= 262.5
Part 2:
Total Area of shaded region = Area of sqaure + Area of triangle
= 225 + 262.5
= 487.5

Question 28.
The area of triangle POS is \(\frac{7}{12}\) of the area of trapezoid PRST.

Answer:
Part 1:
Area of first triangle SQR:
Let us assume the base of first triangle be 14 units and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 22 units
Area of first triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×14×22
= \(\frac{1}{2}\)×308
= 154 sq.in
Part 2:
Area of second triangle PTS:
Let us assume the base of first triangle be 10 units and the perpendicular segment between the base and the opposite vertex will be the height of the triangle which measures 22 units
Area of first triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×10×22
= \(\frac{1}{2}\)×220
= 110 sq.in
Part 3:
Given that the area of triangle PQS is \(\frac{7}{12}\) of the area of trapezoid PRST.Let us assue the area of trapezoid as ‘x’ inches.
Then the area of triangle PQS will be \(\frac{7}{12}\) of the area of trapezoid PRST.
Area of triangle PQS = \(\frac{7}{12}\) × x
Total Area will be the area of trapezoid.
Total Area = Area of triangle PQS + Area of triangle SQR + Area of triangle PTS
x = (\(\frac{7}{12}\) × x)+ 154 + 110
x = (\(\frac{7}{12}\) × x)+ 264
x = \(\frac{7x}{12}\)+ 264
x-\(\frac{7x}{12}\) = \(\frac{7x}{12}\)–\(\frac{7x}{12}\)+264
(\(\frac{12}{12}\)×x)-\(\frac{7x}{12}\)  = 264
(\(\frac{12x}{12}\))-\(\frac{7x}{12}\) = 264
\(\frac{5x}{12}\) = 264
\(\frac{5x}{12}\) × 12 = 264 × 12
5x = 3168
x = 633.6
Thus, the area of trapezoid = 633.6 sq.in
Area of triangle = \(\frac{7}{12}\) × 633.6
= 4435.2÷12
= 369.6 sq.in

Question 29.
Figure EFHL is a parallelogram. The length of \(\overline{F G}\) is \(\frac{5}{8}\) the length of \(\overline{F H}\).

Answer:

Given that:
Part 1:
The length of \(\overline{F G}\) is \(\frac{5}{8}\) the length of \(\overline{F H}\).
Length of FH be ‘x’ ft.Length of FG will be \(\frac{5x}{8}\)
FH length = FG length + GH length
x = \(\frac{5x}{8}\) + 3
x – \(\frac{5x}{8}\) = \(\frac{5x}{8}\)–\(\frac{5x}{8}\)+3
(\(\frac{8}{8}\)×x)- \(\frac{5x}{8}\) = \(\frac{5x}{8}\)–\(\frac{5x}{8}\)+3
\(\frac{3x}{8}\) = 3
\(\frac{3x}{8}\)÷\(\frac{3}{8}\) = 3÷\(\frac{3}{8}\)
x = 8 ft
Length of FH be 8ft and length of FG will be \(\frac{5}{8}\) × 8 = 5 units.
In the figure, we can observe two squares, larger square with 8 ft length and smaller square with length of 3 ft.
Area of larger square = 8×8 = 64 sq.ft
Area of smaller square = 3×3 = 9 sq.ft
Figure EFHL is a parallelogram.
Part 2:
Area of first shaded region = Area of larger sqaure – Area of smaller square
= 64-9
= 55 sq.ft
Area of triangle LEH
Let us assume EL as base and EH.
Area of triangle mesauring 3 ft wide and 3 ft high = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×3×3
= 4.5 sq.ft
There are 2 such triangles = 4.5+4.5 = 9 sq.ft
Part 3:
Area of shaded region will be 55+9 = 64 sq.ft