Math in Focus

This handy Math in Focus Grade 7 Workbook Answer Key Chapter 8 Volume and Surface Area of Solids detailed solutions for the textbook questions.

Math in Focus Grade 7 Course 2 B Chapter 8 Answer Key Volume and Surface Area of Solids

Math in Focus Grade 7 Chapter 8 Quick Check Answer Key

Solve.

Question 1.
A cube has edges measuring 6 centimeters each.
a) Find its volume.
Answer: The volume of the cube V  is 216 cm³

Explanation:
Given, the edge of a cube a = 6 cm each
Formula for volume of a cube V  is  a³
Here, a = 6
V = a³ = 6³
V = 6 × 6 × 6 = 216
So, The volume of the cube V  is 216 cm³.

b) Find its surface area.
Answer: The surface area of the cube A is 216 cm²

Explanation:
Given, the edge of a cube a = 6 cm each
Formula for surface area of a cube  A  is  6a²
Here, a = 6
A = 6a² = 6 × 6²
A = 6 × 6 × 6 = 216 cm²
So, The surface area of the cube A  is  216 cm².

Question 2.
The volume of a cube is 512 cubic centimeters. Find the length of each edge of the cube.
Answer: The length of each edge of the cube a is 8 cm

Explanation:
Given, the volume of a cube is 512 cubic centimeters.
Here, V = a³ = 512 cm³
a³ = 8× 8 × 8 = 512 cm³
Then, a = 8.
So, the length of each edge of the cube a is 8 cm

Question 3.
A pyramid has a square base measuring 10 inches on each side. It has four faces that are congruent isosceles triangles. The height of each triangle is 13 inches. Find the surface area of the pyramid.

Answer: The total surface area of the pyramid is 360 in²

Explanation:
Given, The height of each triangle is 13 inches.
The pyramid has a square base measuring 10 inches on each side.
The area of each triangle is \(\frac{1}{2}\) × bh ,
Here, base b = 10 in. and height h = 13 in.
Area of each triangle = \(\frac{1}{2}\) × 10 × 13
Area of each triangle  = 65 in²
There are total 4 triangles in the pyramid , so the area of 4 triangles are 4 × 65 = 260 in².
The area of a square is a² = 10² = 10 × 10 = 100 in²
So, The total surface area of a pyramid is 260 + 100 = 360 in².

Use 3.14 as an approximation for π.

Question 4.
Shawn makes waffles for breakfast. Each waffle is a circle with a diameter of 6 inches.
a) Find the circumference of a waffle.
Answer: The circumference of a waffle is  18.85 in.

Explanation:
Given, Each waffle is a circle with a diameter of 6 inches
We know that radius r = \(\frac{d}{2}\)
Here, d = 6
then, r = \(\frac{6}{2}\) = 3,
The circumference of a circle  is C = 2πr
C = 2 × π  × 3  , let us take π = 3.14,
C = 6.28 × 3  = 18.85
So, The circumference of a waffle is C = 18.85 in.

b) Find the area of the waffle.
Answer: The area of the waffle is 28.27 in²

Explanation:
Here r = 3 in.
The area of a circle is A =  πr²,  let us take π = 3.14,
A = 3.14  × 3 × 3 = 28.27
So, The area of the waffle is 28.27 in²

Question 5.
The circumference of a wheel is 320.28 centimeters.
a) Find the radius of the wheel.
Answer: The radius of the wheel is 51 cm

Explanation:
Given, The circumference of a wheel is 320.28 centimeters.
We know that the circumference of a circle is  π d
Where d is the diameter of the circle
then,  π d = 320.28
d = \(\frac{320.28}{ π}\) , take  π = 3.14
= \(\frac{320.28}{ 3.14}\)
d = 102,
d = 2r or r = \(\frac{d}{2}\)
r = \(\frac{102}{2}\)
r = 51 cm.
So, The radius of the wheel is 51 cm

b) Find the area of the wheel.
Answer: The area of the wheel is 8167.14 cm²

Explanation:
Here r = 51 cm
The area of a circle is A =  πr²,  let us take π = 3.14,
A = 3.14  × 51 × 51
= 3.14 × 2601
A = 8167.14
So, The area of the wheel is 8167.14 cm²

Match each solid to its net.

Question 6.

Answer:

Explanation:
1. The first solid represents a pyramid with a square base and four triangles having a height perpendicular to the base.
2. The second solid represents a triangular prism .
3. The third solid represents a cuboid.

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 7 Review Test to score better marks in the exam.

Math in Focus Grade 7 Course 2 B Chapter 7 Review Test Answer Key

Concepts and Skills

Construct the angle bisector of ∠ABC on a copy of each figure using a compass and straightedge.

Question 1.

Answer:

 

 

 

 

 

 

Question 2.

Answer:

 

 

 

 

 

 

Use a protractor to draw an angle with the given measure. Then use a compass and straightedge to construct its angle bisector.

Question 3.
m∠XYZ = 37°
Answer:

 

 

 

 

 

Question 4.
m∠PQR = 72°
Answer:

 

 

 

 

 

 

Question 5.
m∠KLM = 128°
Answer:

 

 

 

 

 

 

On a copy of each angle, construct the angle with the given measure by constructing the angle bisector. Use only a compass and straightedge.

Question 6.
Construct a 65° angle whose vertex is point X.

Answer:

 

 

 

 

 

Question 7.
Construct a 66° angle whose vertex is point Y.

Answer:

 

 

 

 

 

 

 

Draw a line segment with the given length. Then construct the perpendicular bisector of the segment using a compass and straightedge.

Question 8.
AB = 6.5 cm
Answer:

 

 

 

 

 

 

 

Question 9.
CD = 4.5 cm
Answer:

 

 

 

 

 

 

 

Question 10.
AD = 10.8 cm
Answer:

 

 

 

 

 

 

 

On a copy of the figure shown, using only a compass and straightedge, draw the perpendicular bisectors of \(\overline{\mathbf{P Q}}\) and \(\overline{\mathbf{P R}}\). Label the point where the two perpendicular bisectors intersect as W.

Question 11.

Answer:

 

 

 

 

 

 

Question 12.

Answer:

 

 

 

 

 

Use the given information to find the number of triangles that can be constructed. Try constructing the triangles to make your decision.

Question 13.
Triangle WXY: WX = 4.5 cm, m∠XWY = 60°, and m∠WXY = 40°.
Answer:

 

 

 

 

 

Question 14.
Triangle ABC: AB = 5 cm, AC = 4.5 cm, and m∠CAB = 60°.
Answer:

 

 

 

 

 

Question 15.
Triangle DEF: DE = 4 cm, EF = 3 cm, and DF = 8 cm.
Answer:

 

 

 

 

 

 

Use the given information to construct each quadrilateral.

Question 16.
Rhombus DEFG with diagonal DF = 4.2 cm and DE = 5 cm.
Answer:

 

 

 

 

 

 

Question 17.
Parallelogram ABCD with AB = 7 cm, DA = 4.5 cm, and m∠ABC = 50°.
Answer:

 

 

 

 

 

 

Question 18.
Quadrilateral ABCD such that AB = 5 cm, AD = 3.5 cm, BC = 4 cm, m∠BAD = 60°, and m∠ABC = 90°.
Answer:

 

 

 

 

 

 

 

 

Solve. Show your work.

Question 19.
A rectangular garden is 15 meters long and 9 meters wide. Use a scale of 1 centimeter to 3 meters, make a scale drawing of the garden.
Answer:
length of the garden is = 45 m
width of the garden is = 27 m

Explanation:
Given the length of the rectangle = 15m
Breadth of the rectangle = 9 m
given scale as 1 centimeter = 3 m
The new length of the garden is = 15 × 3 = 45 m
New breadth of the garden is = 9 × 3 = 27 m

Question 20.
The scale of the floor plan of a room is 1 inch: 6.5 feet. On the floor plan, the room is 8 inches long and 6 inches wide. What are the actual dimensions of the room?
Answer:
1 inch : 6.5 feet means 1 inch on the map represents 6.5 feet on the ground.
Let’s note:
x cm = the actuaL Length
y cm = the actual width.
We have:
1 in. : 6.5 ft = 8 in. : x ft = 6 in. : y ft
Write ratios in fraction form:
\(\frac{1 \mathrm{in} .}{6.5 \mathrm{ft}}=\frac{8 \mathrm{in} .}{x \mathrm{ft}}=\frac{6 \mathrm{in} .}{y \mathrm{ft}}\)
Write without units:
\(\frac{1}{6.5}\) = \(\frac{8}{x}\) = \(\frac{6}{y}\)
Write cross products:
x = 8 · 6.5
y = 6 · 6.5
Simplify
x = 52 ft
y = 39 ft
The actual dimensions of the room are 52 ft and 39 ft.

Question 21.
A model of a car is made using the scale 1 : 25. The actual length of the car is 4.8 meters. Calculate the length of the model of the car in centimeters.

Answer:
1 : 25 means 1 cm on the map represents 25 cm on the ground.
Let x inches be the Length of the car on the map.
We have:
4.8 cm = 4.8 · 100 cm = 480 cm
1 cm : 25 cm = x cm : 480 cm
Write ratios in fraction form:
\(\frac{1 \mathrm{~cm}}{25 \mathrm{~cm}}\) = \(\frac{x \mathrm{~cm}}{480 \mathrm{~cm}}\)
Write without units:
\(\frac{1}{25}\) = \(\frac{x}{480}\)
Write cross products:
25x = 480
Divide both sides by 25:
\(\frac{25x}{25}\) = \(\frac{480}{25}\)
Simplify:
x = 19.2
On the map, the car’s length is 19.2 cm.

Question 22.
The scale on a map is 1 inch : 120 miles. On the map, a highway is 5 inches long. Find the actual length of the highway in miles.
Answer:
Length = 24 miles

Explanation:
Scale on a map is 1 inch: 120 miles
The highway is 5 inches.
The actual length of the highway in miles is 1 × \(\frac{120}{5}\)
length of the highway is 24 miles.

Problem Solving

Solve.

Question 23.
Joe constructed an isosceles triangle WXY such that VVX = WY = 5 cm and XY = 4 cm. Construct another isosceles triangle ABC such that AB = AC = 10 cm and BC = 8 cm. Is triangle ABC an enlargement or a reduction of triangle WXY? Explain your answer and give the scale factor. Justify your answer.
Answer:
We are given △WXY and △ABC:
WX = WY = 5
XY = 4
AB = AC = 10
BC = 8
We construct △WXY:

We construct △ABC:

The sides of △ABC have twice the lengths of the sides of △WXY. Therefore △ABC is an enlargement of △WXY. The scale factor is 2.

Question 24.
James was asked to design a square decorative tile with a side length of 90 millimeters. Construct the square on which James will draw his design.
Answer:
We have to construct the square:
AB = 90 mm
Sketch the square:

Use a ruler to draw \(\overline{A B}\) so that is 90 mm long:

Using a protractor, draw ∠A and ∠B with a measure of 90°.

Because AD = BC = 90 mm, set the compass to a radius of 90 mm. Then using A and B as the center, draw two arcs intersecting the rays drawn in the previous step. Label these points of intersection as D and C.

Draw \(\overline{C D}\).

Question 25.
Harry is designing a theater platform in the shape of a rhombus using a blueprint. The lengths of the diagonals on his blueprint are 4 centimeters and 9 centimeters. Construct the rhombus. Then measure a side length. If the scale of the drawing is 1 centimeter: 2 meters, about what length of skirting does Harry need to go around all four edges of the platform?
Answer:
We are given the rhombus:
AC = 9
BD = 4
We draw the segment of length 20 and label it BD:

We bisect the segment \(\overline{A C}\).
Place the compass point at A. Then draw an arc on each side of \(\overline{A C}\) with a radius greater than half of the length of \(\overline{A C}\).
Using the same radius, set the compass point in C. Draw one arc on each side of \(\overline{A C}\).
Label the points where the arcs intersect as E and F.

Use a straightedge to draw a line through E and F. Label the intersection point of \(\overline{E F}\) and \(\overline{A C}\) by O.

As rhombus diagonals bisect each other, point O is the middle point of \(\overline{B D}\). Place the compass point at O.Then draw an arc on each side of \(\overline{A C}\) with a radius \(\frac{4}{2}\) = 2cm. Label the intersections of these arcs with \(\overline{E F}\) by B and D.

Use a ruler to draw AB, BC, CD, DA.

Measure a side of the rhombus:
AB ≈ 4.9 cm
We are given the scale:
1 cm : 2 m
Let’s note by r the actual length of one side of the rhombus.
\(\frac{1}{2}\) = \(\frac{4.9}{x}\)
We determine x:
x = 2 49
x = 9.8 m
Calculate the perimeter of the rhombus:
4 · 9.8 = 39.2 m

Question 26.
Michael wants to make some kites out of a plastic sheet for a family picnic. Before making the kites he wants to make a \(\frac{1}{4}\) scale model to find the lengths and angles needed for each kite. The diagram shows the measurements of the actual kite. He knows that \(\overline{A C}\) is the perpendicular bisector of \(\overline{B D}\), and that \(\overline{A N}\) should be 6 inches long. Construct the model he will use and find the measures of ∠ABC and the lengths AB and BC in the actual kite.

Answer:
We are given the actual kite dimensions:
AC = 24
BD = 20
AN = 6
Determine the dimensions of the model kite:
\(\frac{1}{4}=\frac{A^{\prime} C^{\prime}}{A C}=\frac{B^{\prime} D^{\prime}}{B D}=\frac{A^{\prime} N^{\prime}}{A N}\)
\(\frac{1}{4}=\frac{A^{\prime} C^{\prime}}{24}=\frac{B^{\prime} D^{\prime}}{20}=\frac{A^{\prime} N^{\prime}}{6}\)
4A’C” = 24 ⇒ A’C” = 6
4B’D’ = 20 ⇒ B’D’ = 5
4A’N’ = 6 ⇒ A’N’ = 1.5
We draw the segment of length 5 and label it B’D’:

We bisect the segment \(\overline{B^{\prime} D^{\prime}}\).
Place the compass point at B’. Then draw an arc on each side of \(\overline{B^{\prime} D^{\prime}}\) with a radius greater than half of the length of \(\overline{B^{\prime} D^{\prime}}\).
Using the same radius, set the compass point in D’. Draw one arc on each side of \(\overline{B^{\prime} D^{\prime}}\).
Label the points where the arcs intersect as E and F
Use a straightedge to draw a line through E and F. Label the intersection point of \(\overline{E F}\)
and \(\overline{B^{\prime} D^{\prime}}\) by N’.

Place the compass point at N’. With a radius of 1.5 in. draw an arc above \(\overline{B^{\prime} D^{\prime}}\). Label the intersection of this arc with \(\overline{E F}\) by A’.
Place the compass point at N’. With a radius of 6 – 1.5 = 4.5 in. draw an arc below \(\overline{B^{\prime} D^{\prime}}\). Label the intersection of this arc with \(\overline{E F}\) by C”.

Use a ruler to draw A’B’, B’C”, C”D’, D’A’.
Measure a side of the rhombus:

Measure ∠ABC:
m∠ABC ≈ 93°
Measure the length5 A’B’ and B’C” in the model kite:
A’B’ ≈ 2.9 in.
B’C” ≈ 5.1 in.
Determine the Lengths AB and BC in the actual kite:
\(\frac{1}{4}=\frac{A^{\prime} B^{\prime}}{A B}=\frac{B^{\prime} C^{\prime}}{B C}\)
\(\frac{1}{4}=\frac{2.9}{A B}=\frac{5.1}{B C}\)
AB = 4 · 2.9 = 11.6 in.
BC = 4 · 5.1 = 20.4 in.

Question 27.
The scale of a map is 1 inch to 5 feet. Find the area of a rectangular region on the map if the area of the actual region is 95 square feet.
Answer:
The area of a rectangular region is 19 square inches.

Explanation:
The scale of a map is 1 inch to 5 feet.
The area of the actual region is 95 square feet.
The area of a rectangular region is = 95 × \(\frac{1}{5}\)
Area of rectangle is = 19 sqaure inches.

Question 28.
The floor plan of a building has a scale of \(\frac{1}{4}\) inch to 1 foot. A room has an area of 40 square inches on the floor plan. What is the actual room area in square feet?
Answer:
10 square feet.

Explanation:
The floor plan of a building has a scale of \(\frac{1}{4}\) inch to 1 foot.
A room has an area of 40 square inches.
The actual room area in square feet is 1× \(\frac{40}{4}\)
= 10 sqaure feet.

Question 29.
The scale of a map is 1 : 2,400. If a rectangular piece of property measures 2 inches by 3 inches on the map, what is the actual area of this piece of property to the nearest tenth of an acre? (1 acre = 43,560 ft²)
Answer:
1: 2,400 means 1 inch on the map represents 2,400 inches on the ground.
Map length : Actual length = 1 in. : 2,400 in.
Map area: Actual area = 1 in.² : 2. 400² in.²
Let y represent the actual area of the rectangle in square inches.
Write a proportion:
\(\frac{\text { Area of rectangle on map }}{\text { Actual area of rectangle }}=\frac{1}{5,760,000}\)
Substitute:
\(\frac{2 \cdot 3}{y}\) = \(\frac{1}{5,760,000}\)
Write the cross products:
y = 6 · 5.760,000
Simplify:
y = 34,560,000 in.²
We convert the actual area to acres:
1 acre = 43,560 ft² = 43,460 · 12² in.²
= 6,258,240 in.²
\(\frac{34,560,000}{6,258,240}\) ≈ 5.52 acres

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 7 Lesson 7.5 Understanding Scale Drawings to score better marks in the exam.

Math in Focus Grade 7 Course 2 B Chapter 7 Lesson 7.5 Answer Key Understanding Scale Drawings

Math in Focus Grade 7 Chapter 7 Lesson 7.5 Guided Practice Answer Key

Calculate the scale factor.

Question 1.
In the diagram, trapezoid B has been enlarged to produce trapezoid A. Find the scale factor.

Answer:
3 cm

Explanation:
Length of A = 7.8 cm
Lenght of B = 2.6 cm
Scaled length = 7.8 cm
Original lenght = 2.6 cm
Scale factor = \(\frac{7.8}{2.6}\)
= 3 cm

 

 

 

 

 

 

 

 

Hands-On Activity

Redraw A Given Figure On Grid Paper At A Different Scale

Materials:

• different sizes of grid paper
• ruler

Work in pairs.

The drawing shows a figure formed by nine squares and a triangle enclosed in a polygon. You can use different-sized grids to produce the same drawing at different scales.

Step 1.
Use two different-sized grids. Redraw the figure on one grid. Have your partner redraw the figure on another grid.

Step 2.
Use a ruler to measure a length in the original drawing. Measure the corresponding length in your scale drawings. Then find the scale factor for each drawing. You can use the following to calculate the scale factor:

Complete.

Question 2.
The scale of a map is 1 inch : 15 miles. If the distance on the map between John’s home and his school is 0.6 inch, find the actual distance in miles.

Answer:
The actual distance is 9 miles.

Explanation:

 

 

 

 

 

 

 

 

 

Complete.

Question 3.
The actual distance between Boston and New York is 220 miles. The scale on a particular map is 1 inch : 25 miles. How far apart on the map are the two cities? 1 inch : 25 miles means 1 inch on the map represents 25 miles on the ground.
Let x inches be the length on the map.

Answer:
8.8 inches

Question 4.
A model car is built using a scale of 1: 18. The length of the model car is 12 inches. Find the actual length of the car in feet.
Answer:
Let x be the actual length of the car in inches.
1: 18 = Scaled length of the car: Actual length of the car
We have:
Substitute values:
1 : 18 = 12 : x
Write ratios in fraction form:
\(\frac{1}{18}\) = \(\frac{12}{x}\)
Write cross products:
x = 18 × 12
Simplify the x.
x = 216
Now convert to feet:
12 in = 1 ft
216 in = \(\frac{1}{12}\) × 216
= 18 ft
The actual length of the car is 18 feet.

Hands-On Activity

Investigate The Relationship Between The Scale Factor And Its Corresponding Area

Work individually.

A square that has a side length of 1 centimeter has an area of 1 square centimeter. In this activity, you will explore how enlarging such a square by a scale factor affects its area.

Step 1.
Suppose you enlarge the square by a factor of 2. Find the side length and the area of the resulting square.

Length of square = cm
(twice the original length)
Area of square = cm²
(increased by a factor of 4)
Answer:
Length of square is 4 cm.
Area of square = 16 cm²

Explanation:
The original length is 2 cm.
Now the lenght of square is 2 + 2 = 4 cm.
Therefore the length of square is 4 cm.
And area of the square is = 4 × 4 = 16 cm²

Step 2.
Suppose you enlarge the square by a factor of 3. Find the side length and the area of the resulting square.
Length of square = cm
(three times the original length)

Area of square = cm²
(increased by a factor of 9)
Answer:
Length of square is 9 cm.
Area of square = 9 × 9 = 18 cm²

Explanation:
Original length of the square is 3 cm.
New length of the square is 3 + 3 = 6 cm.
Area of the square = 6 × 6 = 36 cm²

Step 3.
Suppose you enlarge the square by a factor of 4. Find the side length and the area of the resulting square.
Length of square = cm
(four times the original length)

Area of square = cm²
(increased by a factor of )
Answer:
16 cm, 256 cm²

Explanation:
Given, Original length is 4 cm.
Now the new length of the square is 16 cm.
Area of the square is 16 × 16 = 256 cm²

Step 4.
Copy and complete the table.

 

 

 

 

 

 

 

Copy and complete.

a) Increasing the length of a square by a factor of 5 increases the area by a factor of .
Answer:
Area by a factor is 25 cm²

b) Increasing the length of a square by a factor of k increases the area by a factor of .
Answer:
Area of the square is k² cm²  

Math Journal
Compare the side lengths and the areas for the various scale factors. What pattern do you observe?

From the above activity, the area of a square increases as the square of the scale factor. This property applies to other two-dimensional figures. If you enlarge a figure by a scale factor of 3, its area will be enlarged by a scale of 3² = 9

Complete.

Question 5.
Sylvia makes a map of her yard. On a map, 1 inch represents 8 feet. On the map, the area of a patch of grass is 12 square inches. Find the actual area of the patch of grass.

Answer:
1 inch : 8 feet means 1 inch on the map represents 8 feet on the ground.
Map length : Actual length = 1 in. : 8 ft
Map area: Actual area = 1 in.² : 8² ft²
Let y represent the actual area of the patch of grass in square feet.
Write a proportion:
\(\frac{\text { Area of patch of grass on map }}{\text { Actual area of patch of grass }}=\frac{1}{64}\)
Substitute:
\(\frac{12}{y}\) = \(\frac{1}{64}\)
Write the cross products:
y = 12 · 64
Simplify
y = 768
The actual area of the grass patch is 768 square feet.

Copy and complete.

Question 6.
A blueprint is a type of scale drawing used by architects. An architect is making a blueprint for a conference room that will have a floor area of 196 square feet. If the scale on the blueprint is 1 inch : 7 feet, find the area of the conference room floor on the blueprint.

Let y represent the area of the conference room on the blueprint in square inches.

Answer:
1 inch : 7 feet means 1 inch on the bLueprint represents 7 feet on the ground.
Blueprint length : ActuaL Length = 1 in. : 7 ft
Blueprint area : Actual area = 1 in.² : 7² ft²
Let y represent the area of the conference room on the blueprint in square inches.
Write a proportion:
\(\frac{\text { Area of room on blueprint }}{\text { Actual area of room }}=\frac{1}{49}\)
Substitute:
\(\frac{y}{196}\) = \(\frac{1}{49}\)
Write the cross products:
49y = 196
Divide both sides by 49:
\(\frac{49y}{49}\) = \(\frac{196}{49}\)
Simplify:
y = 4
The gloor area of the conference room on the blueprint is 4 square inches.

Math in Focus Course 2B Practice 7.5 Answer Key

Solve. Show your work.

Question 1.
A model of a ship is 6 inches long. The actual ship is 550 feet (6,600 inches). Find the scale factor used for the model.

Answer:
We are given:
Model length: 6 in.
Actual length: 6,600 in.
Determine the scale factor used for the model:
Scale factor: \(\frac{\text { Model length }}{\text { Actual length }}\) = \(\frac{6}{6,600}\)
= \(\frac{1}{1,100}\)

Question 2.
On a blueprint, the length of a wall is 5 inches. The actual length of the wall is 85 feet. What scale is used for the blueprint?
Answer:
We are given:
Blueprint length: 5 in.
Actual length: 85 ft
Convert the actual length from feet to inches:
1 ft = 12 in.
85 ft = (85 · 12) in. = 1020 in.
Determine the scale factor used for the blueprint:
Scale factor: \(\frac{\text { Blueprint length }}{\text { Actual length }}\) = \(\frac{5}{1020}\)
= \(\frac{1}{204}\)

Question 3.
An artist made a painting of a water pitcher. Then the artist reduced the size of the painting. Find the scale factor of the reduction.

Answer:
We are given:
Big size: 12 in.
Reduced size: 8 in.
Determine the scale factor of reduction:
Scale factor: \(\frac{\text { Reduced size }}{\text { Big size }}\) = \(\frac{8}{12}\)
= \(\frac{2}{3}\)

Question 4.
The height of a building in a drawing is 15 inches. If the actual height of the building is 165 feet, find the scale factor of the drawing.
Answer:
We are given:
Drawing height: 15 in.
Actual height: 165 ft
Convert feet to inches:
1 ft = 12 in.
165 ft = (165 · 12) in. = 1980 in.
Determine the scale factor:
Scale factor: \(\frac{\text { Drawing height }}{\text { Actual height }}\) = \(\frac{15}{1980}\)
= \(\frac{1}{132}\)

Question 5.
In a scale drawing, a sofa is 3 inches long. If the actual length of the sofa ¡s 5 feet long, find the scale factor.
Answer:
We are given:
Drawing length: 3 in.
Actual length: 5 ft
Convert feet to inches:
1ft = 12 in.
5 ft = (5 · 12) in. = 60 in.
Determine the scale factor:
Scale factor: \(\frac{\text { Drawing length }}{\text { Actual length }}\) = \(\frac{3}{60}\)
= \(\frac{1}{60}\)

Question 6.
Daniel is making a scale drawing of his classroom for a project. The length of his classroom is 30 feet long. In his drawing, the length of the classroom is 6 inches. Find the scale factor of Daniel’s drawing.
Answer:
We are given:
Drawing length: 6 in.
Actual length: 30 ft
Convert feet to inches:
1 ft = 12 in.
30 ft = (30 · 12) in. = 360 in.
Determine the scale factor:
Scale factor: \(\frac{\text { Drawing height }}{\text { Actual height }}\) = \(\frac{6}{360}\)
= \(\frac{1}{60}\)

Question 7.
Two cities are 7 inches apart on a map. If the scale of the map is 0.5 inch : 3 miles, what is the actual distance between the two cities?
Answer:
0.5 inch : 3 miles means 0.5 inches on the map represents 3 miles on the ground.
Let x miles be the actual distance.
We have:
0.5 inches : 3 miles = 7 inches: x miles
Write ratios in fraction form:
\(\frac{0.5 \mathrm{in}}{3 \mathrm{mi}}\) = \(\frac{7 \mathrm{in}}{x \mathrm{mi}}\)
Write without units:
\(\frac{0.5}{3}\) = \(\frac{7}{x}\)
Write cross products:
0.5x = 3 · 7
Simplify:
0.5x = 21
Multiply both sides by 2:
x = 42
The actual distance is 42 miles.

Question 8.
A road map of New Orleans uses a scale of 1 inch : 3 miles. If Carlton Avenue is 1 .3 inches long on the map, what is the actual length of the street?
Answer:
1 inch : 3 miles means 1 inch on the map represents 3 miles on the ground
Let x miles be the actual distance.
We have:
1 inch : 3 miles = 13 inches : x miles
Write ratios in fraction form:
\(\frac{1 \mathrm{in} .}{3 \mathrm{mi}}\) = \(\frac{1.3 \mathrm{in} .}{x \mathrm{mi}}\)
Write without units:
\(\frac{1}{3}\) = \(\frac{1.3}{x}\)
Write cross products:
x = 3 · 1.3
Simplify:
x = 3.9
The actual distance is 3.9 miles.

Question 9.
The scale of a map is 1 inch : 85 miles.
a) On the map, the river is 14 inches long. Find the actual length of the river in miles.
Answer:
1 inch : 85 miles means 1 inch on the map represents 85 miles on the ground.
Let x miles be the actuaL length.
We have:
1 inch : 85 miles = 14 inches : x miles
Write ratios in fraction form:
\(\frac{1 \mathrm{in}}{85 \mathrm{mi}}\) = \(\frac{14 \mathrm{in} .}{x \mathrm{mi}}\)
Write without units:
\(\frac{1}{85}\) = \(\frac{14}{x}\)
Write cross products:
x = 85 · 14
Simplify:
x = 1,190
The actual length is 1,190 miles.

b) The actual distance between two towns is 765 miles. Find the distance on the map between these towns.
Answer:
Let y inches be the distance on the map.
We have:
1 inch : 85 miles = y inches : 765 miles
Write ratios in fraction form:
\(\frac{1 \mathrm{in} .}{85 \mathrm{mi}}\) = \(\frac{y \mathrm{in}}{765 \mathrm{mi}}\)
Write without units:
\(\frac{1}{85}\) = \(\frac{y}{765}\)
Write cross products:
85y = 765
Divide both sides by 85.
\(\frac{85y}{85}\) = \(\frac{765}{85}\)
Simplify:
y = 9
The distance on the map is 9 inches.

Question 10.
Goodhope River is 48 miles long. What is the length of the river on a map with a scale of 1 inch : 15 miles?
Answer:
1 inch : 15 miles means 1 inch on the map represents 15 miles on the ground.
Let x inches be the length on the map.
We have:
1 inch : 15 miles = x inches : 48 miles
Write ratios in fraction form:
\(\frac{1 \mathrm{in} .}{15 \mathrm{mi}}\) = \(\frac{x \mathrm{in}}{48 \mathrm{mi}}\)
Write without units:
\(\frac{1}{15}\) = \(\frac{x}{48}\)
Write cross products:
15x = 4
Divide both sides by 15:
\(\frac{15x}{15}\) = \(\frac{48}{15}\)
Simplify:
x = 3.2
On the map the length is 3.2 inches.

Question 11.
A map is drawn using a scale of 1 inch : 165 miles. The length of a road on the map is 12 inches. Find the actual length of the road.
Answer:
1 inch : 165 miles means 1 inch on the map represents 165 mites on the ground.
Let x mites be the actual length.
We have:
1 inch : 165 miles = 12 inches : x miles
Write ratios in fraction form:
\(\frac{1 \mathrm{in}}{165 \mathrm{mi}}\) = \(\frac{12 \mathrm{in} .}{x \mathrm{mi}}\)
Write without units:
\(\frac{1}{165}\) = \(\frac{12}{x}\)
Write cross products:
x = 165 · 12
Simplify:
x = 1980
The actual length of the road is 1980 miles.

Question 12.
On a particular map, 2 inches represents an actual distance of 64 miles. Towns A and B are 608 miles apart. Find the distance between the two towns, in inches, on the map.
Answer:
2 inches : 64 miles means 2 inches on the map represents 64 mites on the ground.
Let x inches be the distance on the map.
We have:
2 inches : 64 miles = x inches : 608 miles
Write ratios in fraction form:
\(\frac{2 \mathrm{in}}{64 \mathrm{mi}}\) = \(\frac{x \text { in. }}{608 \mathrm{mi}}\)
Write without units:
\(\frac{2}{64}\) = \(\frac{x}{608}\)
Write cross products:
64x = 2 · 608
64x = 1,216
Divide both sides by 64.
\(\frac{64x}{64}\) = \(\frac{1,216}{64}\)
Simplify:
x = 19
On the map the two cities are 19 inches apart.

Question 13.
On a particular map, 1 inch represents an actual distance of 2.5 miles. The actual area of a lake is 12 square miles. Find the area of the lake on the map.
Answer:
1 inch : 2.5 miles means 1 inch on the map represents 2.5 miles on the ground.
Map length : Actual length = 1 in. : 2.5 mi.
Map area : Actual area = 1 in.² : 2.5² mi.²
Let y represent the area of the lake on map in square inches.
Write a proportion:
\(\frac{\text { Area of lake on map }}{\text { Actual area of lake }}=\frac{1}{6.25}\)
Substitute:
\(\frac{y}{12}\) = \(\frac{1}{6.25}\)
Write the cross products
6.25y = 12
Divide both sides by 6.25 and simplify:
\(\frac{6.25y}{6.25}\) = \(\frac{12}{6.25}\)
y = 1.92
The area of the lake on the map is 1.92 square inches.

Question 14.
On the map, the area of a nature preserve is 54.2 square inches. If the scale of the map is 1 inch : 8 miles, find the actual area of the nature preserve.
Answer:
1 inch : 8 miles means 1 inch on the map represents 8 miles on the ground.
Map Length : Actual length = 1 in. : 8 mi.
Map area: Actual area = 1 in.² : 8² mi.²
Let y represent the actual area of the nature preserve in square miles.
Write a proportion:
\(\frac{\text { Area of nature preserve on map }}{\text { Actual area of nature preserve }}=\frac{1}{64}\)
Substitute:
\(\frac{54.2}{y}\) = \(\frac{1}{64}\)
Write the cross products:
y = 54.2 × 64
Simplify:
y = 3,468.8
The actual area of the nature preserve is 3,468.8 square miles.

Question 15.
The map shows two roads labeled A and B.

a) Using a ruler, measure, in centimeters, the lengths of roads A and B.

Answer:
We measure the lengths of the roads A and B, in centimeters, using a ruler:
Length A = 24 cm
Length B = 3.3 cm

b) Using the scale given, find, in kilometers, the actual lengths of roads A and B.
Answer:
We are given the scale:
1 : 50,000
We determine the actual length x of Road A in centimeters:
\(\frac{2.4}{x}\) = \(\frac{1}{50,000}\)
x = 2.4 · 50,000
= 120,000 cm
Convert to kilometers:
1 km = 1000 m = 100,000 cm
120,000 cm = \(\frac{120,000}{100,000}\) km = 1.2 km
We determine the actual length y of Road B in centimeters:
\(\frac{3.3}{y}\) = \(\frac{1}{50,000}\)
y = 3.3 · 50,000
= 165,000 cm
Convert to kilometers:
1 km = 1000 m = 100,000 cm
165,000 cm = \(\frac{165,000}{100,000}\) km = 1.65 km

Question 16.
The map shows seven cities in Florida. Using the scale on the map, use a ruler to measure the distance between the following pairs of cities. Then find the actual distance between them in miles.

a) Orlando and West Palm Beach
Answer:
We measure the lengths of the distance between Orlando and West Palm Beach, in centimeters, using a ruler:
Length Orlando-West Pam Beach = 3 cm
Length Fort Myers-Miami Beach = 24 cm
We are given the scale:
1 cm : 50 miles
We determine the actual distance x between Orlando and West Palm Beach in centimeters:
\(\frac{3}{x}\) = \(\frac{1}{50}\)
x = 3 · 50
= 150 miles

b) Fort Myers and Miami Beach
Answer:
We determine the actual distance y between Fort Myers and Miami Beach in centimeters:
\(\frac{2.4}{y}\) = \(\frac{1}{50}\)
x = 2.4 · 50
= 120 miles

Question 17.
Use the scale on the floor plan of a house to find each of the following.
a) The actual length and width of room 1.
Answer:
2 cm : 5 m means 2 cm on the plan represent 5 m on the ground.
Let x, y meters be the actual length and width of room 1.
We have:
2 cm : 5 m = 1.8 cm : x m
Write ratios in fraction form:
\(\frac{2 \mathrm{~cm}}{5 \mathrm{~m}}\) = \(\frac{1.8 \mathrm{~cm}}{x \mathrm{~m}}\)
Write without units:
\(\frac{2}{5}\) = \(\frac{1}{2}\)
Write cross products:
2x = 5 · 1.8
Divide by 2 and simplify:
\(\frac{2x}{x}\) = \(\frac{9}{2}\)
x = 4.5 meters
We do the same to find the actual width:
2 cm : 5 m = 1.4 cm : y m
\(\frac{2 \mathrm{~cm}}{5 \mathrm{~m}}\) = \(\frac{1.4 \mathrm{~cm}}{y \mathrm{~m}}\)
\(\frac{2}{5}\) = \(\frac{1.4}{y}\)
2y = 5 · 1.4
\(\frac{2y}{2}\) = \(\frac{7}{2}\)
y = 3.5 meters

b) The width of the door on the floor plan if its actual width is 0.8 meter.
Answer:
We determine the width z of the door on the floor plan:
\(\frac{2 \mathrm{~cm}}{5 \mathrm{~m}}\) = \(\frac{z \mathrm{~cm}}{0.8 \mathrm{~m}}\)
\(\frac{2}{5}\) = \(\frac{z}{0.8}\)
5z = 2 · 0.8
\(\frac{5z}{5}\) = \(\frac{1.6}{5}\)
z = 0.32 cm

c) The actual area of the floor of the house to the nearest square meter

Answer:
We have:
Length scale: 2 cm : 5 m
Area scale: 2² cm² : 5² m²
We determine the actual area s of the floor:
\(\frac{4}{25}\) = \(\frac{7.6 \cdot 3}{s}\)
4s = 25 · 22.8
\(\frac{4s}{4}\) = \(\frac{570}{4}\)
s ≈ 143 m²

Question 18.
A tower is drawn using a scale of 1 inch : 3 feet. The height of the tower in the drawing is 1 foot 5 inches. Then, an architect decides to make a new scale drawing of the tower. In the new scale, the scale is 1 inch : 5 feet. Find the height of the tower in the new drawing.
Answer:
1 inch : 3 feet means 1 inch on the drawing represents 3 feet on the ground.
Let x miles be the actual height
We have:
1 inch : 3 feet 1 ft 5 in. : x feet
Write ratios in fraction form:
\(\frac{1 \mathrm{in} .}{3 \mathrm{ft}}\) = \(\frac{(12+5) \mathrm{in} .}{x \mathrm{ft}}\)
Write without units:
\(\frac{1}{3}\) = \(\frac{17}{x}\)
Write cross products:
x = 3 · 17
Simplify
x = 51 ft
The actual height of the tower is 51 ft.
The new scale is 1 inch : 5 feet
We determine the height y of the tower in the new drawing:
\(\frac{1 \mathrm{in} .}{5 \mathrm{ft}}\) = \(\frac{y \mathrm{in} .}{51 \mathrm{ft}}\)
\(\frac{1}{5}\) = \(\frac{y}{51}\)
5y = 51
\(\frac{5y}{5}\) = \(\frac{51}{5}\)
y = 10.2 inches

Question 19.
Each student walked in a straight line from one point to another. Use a centimeter ruler to measure distances on the map shown. Use the scale on the map to find the distance each student walked in meters.

We measure the lengths of the roads, in centimeters, using a ruler:
Length DA = 1.8 cm
Length AB = 3.3 cm
Length BF = 2.5 cm
Length CF = 1.4 cm
Length EF = 2.8 cm
Length CE = 3.5 cm

a) Ethan walked from the library to the school, and then to the gym.
Answer:
We are given the scale:
1 : 12,500
We determine the actual length x Ethan walked:
\(\frac{D A+A B}{x}\) = \(\frac{1}{12,500}\)
\(\frac{1.8+3.3}{x}\) = \(\frac{1}{12,500}\)
x = 5.1 · 12,500
= 63,750 cm
Convert to meters:
1 m = 100 cm
63,750 cm = \(\frac{63,750}{100}\) m = 637.5 m

b) Joshua walked from the motel to the restaurant, and then to the movie theater.
Answer:
We determine the actual length y joshua walked:
\(\frac{F C+C E}{y}\) = \(\frac{1}{12,500}\)
\(\frac{1.4+3.5}{y}\) = \(\frac{1}{12,500}\)
x = 4.9 · 12,500
= 61,250 cm
Convert to meters:
1 m = 100 cm
61,250 cm = \(\frac{61,250}{100}\) m = 612.5 m

c) Chloe walked from the gym to the motel, and then to the movie theater.
Answer:
We determine the actual length z Chloe walked:
\(\frac{B F+F E}{z}\) = \(\frac{1}{12,500}\)
\(\frac{2.5+2.8}{z}\) = \(\frac{1}{12,500}\)
x = 5.3 · 12,500
= 66,250 cm
Convert to meters:
1 m = 100 cm
66,250 cm = \(\frac{66,250}{100}\) m = 662.5 m

Brain @ Work

Question 1.
You know you can bisect any angle using a compass and a straightedge. Geometers have known for thousands of years that it is impossible to trisect any given angle using a compass and straightedge. (The word trisect means to divide into three equal parts.) But it is possible to trisect certain angles, including a right angle. Using only a compass and straightedge, show how you can draw a right angle that is trisected.
Answer:
We are given the right angle:
m∠AOB – 90°

Use the compass: with the center in O, draw an arc. Labet its intersection with \(\overline{O A}\) and \(\overline{O B}\) by C and D

With the same radius place the compass in center C, then D and draw arcs which intersect the arc drawn in the previous step in N and M.

Use a ruler to draw \(\overrightarrow{O M}\) and \(\overrightarrow{O N}\).

m∠AOM = m∠MON = m∠NOB
= \(\frac{m \angle A O B}{3}\) = \(\frac{90^{\circ}}{3}\) = 30°

Question 2.
You accidentally broke your mother’s favorite plate. You want to ask an artist to reproduce it. You ask your math teacher to help you find the original size of the plate. She suggests that you locate three points on the rim of the plate and use the points to draw two segments. The point where the two perpendicular bisectors of these segments intersect will be the center of the plate. From that you can measure the radius. Copy and complete the diagram below. Then measure to find the diameter of the plate.

Answer:
We consider 3 points on the rim of the plate:

The center of the circle/plate is situated at the intersection of the perpendicular bisectors of \(\overline{A B}\) and \(\overline{B C}\).

The radius of the circle/plate is OA.

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 7 Lesson 7.4 Constructing Quadrilaterals to score better marks in the exam.

Math in Focus Grade 7 Course 2 B Chapter 7 Lesson 7.4 Answer Key Constructing Quadrilaterals

Math in Focus Grade 7 Chapter 7 Lesson 7.4 Guided Practice Answer Key

Construct the quadrilateral from the given dimension. Use a ruler and protractor.

Question 1.
Rectangle ABCD measuring 7 centimeters by 5 centimeters.
Explanation:

Answer:
we have drawn the rectangle with the measurements of length=5cm and breadth=7cm.

Construct the quadrilateral from the given information. Use a compass, ruler, and protractor.

Question 2.
Parallelogram KLMN: KL = 6.4 cm, LM = 4.8 cm, and m∠KLM = 60°.
Explanation:

Answer:
The measurements of the parallelogram of length= 4.8cm and breadth=6.4cm. The angle of KLM=60.

Construct the quadrilateral from the given
information. Use a ruler, compass, and protractor.

Question 3.
Rhombus PQRS with diagonal PR = 6.2 cm and PQ = 4.5 cm.
Explanation:

Answer:
we drawn the rhombus with the measurement angels of PR=6.2cm and PQ=4.5cm.

Math in Focus Course 2B Practice 7.4 Answer Key

Construct each quadrilateral from the given information.

Question 1.
Rectangle KLMN measuring 5.3 centimeters by 4.7 centimeters.
Answer:
The measuring angles of rectangle is Length = 5.3 cm and breadth= 4.7 cm

Explanation:

Question 2.
Square with side lengths of 7 centimeters.
Explanation:

Answer:
The measuring angles of the square is 7cm.

Question 3.
Rhombus DEFG with diagonal DF = 6 cm and DE = 4.5 cm.
Explanation:

Answer:

The measuring angles of the rhombus DEFG diagonals DF=6cm and DE=4.5cm.

Question 4.
Parallelogram PQRS with PQ = 3.8 cm, QR = 5 cm, and m∠QPS = 70°.
Answer:
Explanation:

The measuring angles of the parallelogram PQRS
PQ=3.8cm and QR=5cm
we got the angle 70°.

Question 5.
Quadrilateral ABCD with AB = 6.7 cm, BC = 7.2 cm, AD = 4.9 cm, CD = 6.2 cm, and m∠ABC = 55°.
Answer:
Explanation:

The measurements of quadrilateral ABCD of AB = 6.7, BC = 7.2, CD = 6.2 and DA = 4.9. We got the angle ABC = 55°.

Solve. Show your work.

Question 6.
Construct quadrilateral ABCD with diagonal AC = 5 cm, AB = CD = 4 cm, BC = 6 cm, and AD = 6 cm.
Answer:
Explanation:

The measurements of the rectangular quadrilateral is AB = CD = 4cm , BC and AD = 6cm. The diagonal AC = 5cm. So this is a quadrilateral rectangular.
We sketch the quadrilateral:

First we construct △ABC.
Use a ruler to draw the segment 6 cm long. Label its endpoints by B and C.

Because AB = 4 cm, use the ruler to set the compass to a radius of 4 cm. Then using B as center draw an arc of radius 4 cm above \(\overline{B C}\).

Because AC = 5 cm, use the ruler to set the compass to a radius of 5 cm. Then using C as center draw an arc of radius 5 cm that intersects the first arc. Label the intersection as A.

Use the ruler to draw \(\overline{A B}\) and \(\overline{A C}\).

Then we construct △ACD.
Because AD = 6 cm, use the ruler to set the compass to a radius of 6 cm. Then using A as center draw an arc of radius 6cm above \(\overline{A C}\).

Because CD = 4 cm, use the ruler to set the compass to a radius of 4 cm. Then using C as center draw an arc of radius 4 cm that intersects the first arc. Label the intersection as D.

Use the ruler to draw \(\overline{A D}\) and \(\overline{C D}\).

a) What type of quadrilateral is ABCD? Explain your reasoning.
Answer:
The quadrilateral ABCD is rectangle because the angles AB and CD are equal with the  =4cm and the angles BC and AD are equal with the = 6cm. So this is rectangular quadrilateral.

b) Draw \(\overline{B D}\) to intersect \(\overline{A C}\) at point E. Find the lengths of \(\overline{A E}\), \(\overline{C E}\), \(\overline{B E}\), and \(\overline{D E}\).
Answer:
Draw \(\overline{B D}\)

We measure AE, CE, BE, DE:
AE = 2.5
CE = 2.5
BE = 4.4
DE = 4.4

c) Do the diagonals \(\overline{A C}\) and \(\overline{B D}\) bisect each other? Justify your answer.
Answer:
The diagonals \(\overline{A C}\) and \(\overline{B D}\) bisect each other This is explained by the fact that the quadrilateral is a parallelogram.

Question 7.
Construct a quadrilateral STUV by following Steps 1 to 5.
STEP1
Draw \(\overline{S T}\) so that ST = 5 cm.
STEP 2
With T as the center, draw \(\overline{T U}\) perpendicular to \(\overline{S T}\) , with TU = 4 cm.
STEP 3
With U as the center, draw an arc of radius 5 centimeters.
STEP 4
With S as the center, draw an arc of radius 4 centimeters to intersect the arc drawn in Step 3. Label this point of intersection as V.
STEP 5
Complete the construction of quadrilateral STUV.
Answer:
Use a ruler to draw \(\overline{S T}\) so that is 5 cm long.

Using a protractor, draw ∠T with a measure of 90°.

Because TU = 4 cm, set the compass to a radius of 4 cm. Then using T as the center, draw an arc intersecting the ray drawn in previous step. Label this point of intersection as U.

With U as center, draw an arc of radius 5 cm.
With S as center, draw an arc of radius 4 cm to intersect the arc drawn in the previous step. Label this point of intersection as V.

Draw \(\overline{V U}\) and \(\overline{V S}\).

a) Find each of the angles in quadrilateral STUV.
Answer:
We measure the angles of the quadrilateral:
m∠T = 90°
m∠U = 90°
m∠V = 90°
m∠S = 90°

b) Name the quadrilateral.
Answer:
As all angles are right angles, the quadrilateral is a rectangle.

c) Find the lengths of the diagonals. What do you notice?
Answer:
We find the lengths of the diagonals:
SU ≈ 6.4
VT ≈ 6.4

The diagonals are congruent.

Question 8.
Construct a quadrilateral ABCD with all sides of length 3 centimeters and diagonal BD = 5.2 cm.
a) What type of quadrilateral is ABCD? Explain your reasoning.
Answer:
Explanation:

The quadrilateral ABCD is a square because the sides of the square AB, BC, CD  and DA are equal with 3 cm and the diagonal BD=5.2cm.

b) Find the measure of each of the angles formed by the intersection of the diagonals.
Answer:
The measuring of each of the angles formed by the intersection of the diagonals is acute angle 60°.

Question 9.
Construct quadrilateral ABCD with diagonal AC = 6 cm, AB = 3 cm, BC = 4 cm, CD = 4.5 cm, and AD = 9 cm. What type of quadrilateral does ABCD seem to be? Explain your reasoning.
Answer:
Explanation:

The quadrilateral ABCD is trapezoid. AB=3cm ,BC=4cm,CD=4.5cm and DA=9cm.
The diagonal AC = 6cm.So this is trapezoid.

Question 10.
Construct the figure below using the given dimensions.

Answer:
Use a ruler to draw \(\overline{A B}\) so that is 6.6 cm long.

Using a protractor, draw ∠B with a measure of 90° and ∠A with a measure of 90°.

Because AD = BC = 3.4 cm, set the compass to a radius of 3.4 cm. Then using A and B as centers, draw arcs intersecting the rays drawn in previous step. Label this points of intersection as D and C.

Because AE = EC = 4 cm, set the compass to a radius of 4 cm. Then using C and D as centers, draw arcs above \(\overline{C D}\). Label the point of intersection of the two arcs as E.

Draw \(\overline{E D}\) and \(\overline{E C}\).

Question 11.
a) Construct quadrilateral ABCD with AB = 6 cm, BC = 3 cm, AD = 4 cm, m ∠BAD = 120°, and m ∠ABC = 100°.
Explanation:

Answer:
The measurements of the quadrilateral ABCD is AB=6cm , BC=3cm , AD =4cm. The angle BAD= 120° angle ABC =100°.

b) Find the length of \(\overline{C D}\).
Answer:
The length of overline CD=10cm.

c) Label the midpoints of the four sides of this quadrilateral as W, X, Y, and Z. Join them to form quadrilateral WXYZ.
Answer:
Explanation:

d) Compare the lengths of \(\overline{W X}\) and \(\overline{Y Z}\) . Compare the lengths of \(\overline{X Y}\) and \(\overline{W Z}\). What do you notice?
Answer:
Measure WX, YZ, XY, WZ:
WX ≈ 3.6 cm
YZ ≈ 3.6 cm
XY ≈ 4.4 cm
WZ ≈ 4.4 cm
We notice that we have:
WX = YZ
XY = WZ

Question 12.
Construct parallelogram PQRS with PQ = 6 cm, a height of 4.5 centimeters and interior angles 45° and 135°.
Explanation:

Answer:
The measurements of the parallelogram PQRS with PQ =6cm and height= 4.5cm.
The interior angles are  45° and 135°.

Question 13.
Draw rhombus ABCD with AC = 5 cm and AB = 6.5 cm. Also draw diagonal \(\overline{B D}\). How are the two diagonals related to each other? Explain.
Answer:
Explanation:

let ABCD be a rhombus whose diagonal AC and BD are intersecting at point O. We know that the diagram of parallelogram bisect at each other.
OA=OC and OB=OD
angle COB= angle COD
angle COB+ angle COD= 180°
angle COB=angle COD= 90°.

Question 14.
Jenny plans to make a trapezoidal bookmark for each of her teachers. The top will be 5 centimeters long and have right angles at either end. The right side will be 12 centimeters long, and the bottom of the bookmark will make a 50° angle with this side. Construct a template for Jenny’s bookmark. How long is the left side of Jenny’s template to the nearest centimeter?
Answer:
Explanation:

The measurements of the trapezoid length is 5cm and breadth is 12 cm. The bottom of the book mark angle is  50°. The left side of the jenny’s template is 4cm.

Question 15.
Martha plans to cut squares of paper from a roll of wrapping paper. She will package the squares to sell as origami papers to raise funds for a charity. If the area of the square paper is 64 square centimeters, construct the square Martha can use as a template.
Answer:
Determine the length of the square side:
AB² = 64
AB = \(\sqrt{64}\) = 8
Sketch the rectangle:

Use a ruler to draw \(\overline{A B}\) so that is 7 cm long.

Using a protractor, draw ∠B with a measure of 90°.

Because BC = 5 cm, set the compass to a radius of 5 cm. Then using B as the center, draw an arc intersecting the ray drawn in previous step. Label this point of intersection as C. Because AD = 5 cm, use the same compass setting. Using A as the center, draw an arc:

Because CD = 7 cm, set the compass to a radius of 7 cm. Then using C as the center draw an arc intersecting the ray drawn in previous step. Label this point of intersection as D.

Draw \(\overline{C D}\) and \(\overline{A D}\)

Question 16.
Jessie plans to make a patchwork pattern from colored paper by repeating a rhombus whose diagonals measure 4 centimeters and 5 centimeters. Use the given dimensions to construct a template for the rhombus.
Answer:
We have to construct the rhombus ABCD.
AC = 4
BD = 5
Sketch the rhombus:
Use a ruler to draw \(\overline{A C}\) so that is 4 cm long.

Draw the perpendicular bisector of \(\overline{A C}\):

Because OB = OD = \(\frac{5}{2}\) = 2.5 cm, use the compass setting 2.5 cm. Then using O as the center, draw an arc to intersect the perpendicular bisector in the points B and D.

Draw \(\overline{A B}\), \(\overline{B C}\), \(\overline{C D}\) and \(\overline{D A}\).

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 7 Lesson 7.3 Constructing Triangles to score better marks in the exam.

Math in Focus Grade 7 Course 2 B Chapter 7 Lesson 7.3 Answer Key Constructing Triangles

Math in Focus Grade 7 Chapter 7 Lesson 7.3 Guided Practice Answer Key

Construct the triangle from the given information. Use a compass and ruler.

Question 1.
Triangle PQR: PQ = 5.6 cm, QR = 4.5 cm, and PR = 8.2 cm.

Answer:
The measurements of the triangle are PQ=5.6cm ,PR=8.2cm and QR=4.5cm.

Eplanation:

 

Construct the triangle from the given information. Use a ruler and compass.

Question 2.
Triangle ABC: BC = 4 cm, m∠ABC = 25°, and m∠ACB = 120°.
Answer:
The measurements of the triangle are PQ=5.6cm ,PR=8.2cm and QR=4.5cm.

Explanation:

Question 3.
Triangle KLM: KL = 8.2 cm, KM = 6.9 cm, and m∠LKM = 75°.
Answer:
The measurements of the triangle KLM are KL=8.2cm ,KM=6.9cm and the angle LKM=75°.

Explanation:

Construct the triangle from the given information. Use a compass, ruler, and protractor.

Question 4.
Triangle KLM: KL = 7cm, KM = 9 cm, and m∠KLM = 125°.
Answer:
The measurements of the triangle KLM are KL=7cm , KM=9cm and m∠KLM= 125°.

Explanation:

Hands-On Activity

Materials

• protractor
• compass
• ruler

Decide Whether Given Measures Can Be Used To Construct One Triangle, More Than One Triangle. Or No Triangles

Work in pairs.

Step 1.
Try to construct triangle ABC with AB = 7 cm, BC = 8 cm, and DE = 4.6 cm, EF = 6 cm, and AC = 11 cm.
Answer:
The measurements of the triangle ABC are  AB = 7 cm, BC = 8 cm, and DE = 4.6 cm, EF = 6 cm, and AC = 11 cm.

Explanation:

Step 2.
Try to construct triangle DEF with DE = 4.6 cm, EF = 6 cm, and DF = 12 cm.
Answer:
The measurements of the triangle DEF are DE = 4.6 cm, EF = 6 cm, and DF = 12 cm.

Explanation:

Step 3.
Try to construct triangle GHI with GH = 6 cm, HI = 5 cm, and JK = 6 cm, JL = 4.7 cm, and m∠GHI = 50°.

Step 4.
Try to construct triangle JKL with JK = 6 cm, JL = 4.7 cm, and m∠JKL = 50°.
Answer:
The measurements of the triangle JKL are JK = 6 cm, JL = 4.7 cm, and m∠JKL = 50°.

Explanation:

Step 5.
Try to construct triangle MNP with MN = 7 cm, m∠MNP = 60°, and m∠PMN = 40°.
Answer:
The measurements of the triangle MNP are MN = 7 cm, m∠MNP = 60°, and m∠PMN = 40°.

Explanation:

Step 6.
Were there any triangles you could not construct? Were there any triangles that you could construct in more than one way? Explain.
Answer:
Yes, at step 3 we could not construct any triangle. At step 1 we can construct more than one triangle because the measurements of the triangle ABC are  AB = 7 cm, BC = 8 cm, and DE = 4.6 cm, EF = 6 cm, and AC = 11 cm.

Math Journal
Use your results to decide whether you can always construct exactly one triangle from the given information. Justify your answer.

a) Given three side lengths
Answer:

b) Given two side lengths and an angle measure
Answer:

Find the number of triangles that can be constructed. Try constructing the triangles to make your decision.

Question 5.
PQ = 4.8 cm, QR = 5.4 cm, and m∠PQR = 100°.
Answer:
The measurements of the triangle PQR = PQ = 4.8 cm, QR = 5.4 cm, and m∠PQR = 100°.

Explanation:

Question 6.
AB = 6.2 cm, BC = 4.8 cm, and m∠BAC = 75°.
Answer:
The measurements of the triangle ABC are AB = 6.2 cm, BC = 4.8 cm, and m∠BAC = 75°.

Explanation:

Question 7.
ST = 7.7 cm, SU = 5.2 cm, and m∠STU = 40°.
Answer:
The measurements of the triangle STU are ST = 7.7 cm, SU = 5.2 cm, and m∠STU = 40°.

Explanation:

Math in Focus Course 2B Practice 7.3 Answer Key

Use the given information to construct each triangle.

Question 1.
In triangle CDE, CD = 7 cm, DE = 4 cm, and CE = 6.5 cm.
Answer:
The measurements of the triangle CDE , CD = 7 cm, DE = 4 cm, and CE = 6.5 cm.

Explanation:

Question 2.
In triangle ABC, BC = 6 cm, m∠ABC = 30°, and m∠ACB = 60°. Find m∠BAC and AC.
Answer:
The measurements of the triangle ABC , BC = 6 cm, m∠ABC = 30°, and m∠ACB = 60°.
B= 30° and C= 60°
A=180°-B-C
A=180°- 30°- 60°
A=90°.
AC=3cm.

Explanation:

Question 3.
In an equilateral triangle, each side length is 6.5 centimeters long.
Answer:
The measurements of the equilateral triangle , each side length is 6.5 cm.

Explanation:

Question 4.
In triangle ABC, AB = 4 cm, AC = 5 cm, and m∠ABC = 40°.
Answer:
The measurements of the triangle ABC, AB = 4 cm, AC = 5 cm, and m∠ABC = 40°.

Explanation:

Question 5.
In triangle ABC, AB = 6 cm, BC = 8 cm, and AC = 10 cm. What kind of triangle is triangle ABC? Classify it by both sides and angles.
Answer:
The measurements of the triangle ABC, AB = 6 cm, BC = 8 cm, and AC = 10 cm. The triangle ABC is a right angle triangle. Right angle triangle has 3 sides and it have one angle  90°.

Explanation:

Question 6.
In triangle XYZ, XV = XZ = 4 cm and YZ = 5 cm. Find m∠XZY.
Answer:
The measurements of the triangle XYZ, XV = XZ = 4 cm and YZ = 5 cm.
The angle of the XZY is 50°.

Explanation:

Solve. Show your work.

Question 7.
Triangle POR has the dimensions shown in the diagram.

a) Construct triangle POR.
Answer:

b) Using a ruler, from your construction, measure the length of \(\overline{P R}\).
Answer:
Using a ruler we measure the length of \(\overline{P R}\).
PR ≈ 6.1 cm

c) Find the measures of ∠P and ∠R without using a protractor. Justify your answer.
Answer:
△PQR is Isosceles because PQ = QR Therefore the angles P and R we congruent:
m∠P = m∠R
Find m∠P and m∠R:
m∠P + m∠R – m∠Q = 180°
2m∠P + 100° = 180°
2m∠P = 180° – 100°
2m∠P = 80°
m∠P = m∠R = \(\frac{80^{\circ}}{2}\) = 40°

Question 8.
Math Journal
Is it possible to construct a triangle PQR in which PQ = 12 cm, PR = 5 cm, and QR = 4cm? Explain.
Answer:
We cannot construct a triangle PQR in which PQ = 12 cm, PR = 5 cm, and QR = 4cm because the lengths of the PR , QR are very small.

Explanation:

Question 9.
Three triangles have angle measures of 500 and 600. In one triangle, the side included between these angles is 2 centimeters. In the second triangle, the included side length is 3 centimeters, and in the third triangle, the included side length is 4 centimeters.

a) Construct the three triangles.
Answer:

b) In each triangle, what is the measure of the third angle?
Answer:
Sum of angles of triangle is 180 so the given triangles have angle measures of 50 degree and 60 degree another angle is 70 degree.

c) Using the triangles constructed to help you, what can you deduce about the number of triangles that can be constructed if you are given three angle measures of a triangle but not the measure of any side length?
Answer:
Given only three angle measures of a triangle we can construct an infinity of triangles, their sides being proportional.

Question 10.
Math Journal
Suppose you are given three angle measures whose sum is 180°, Can you form a triangle given this information? Are there other different triangles you can form? Explain.
Answer:
The angle sum property of a triangle states that the angles of a triangle always adds upto 180°.

Explanation:

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 7 Lesson 7.2 Constructing Perpendicular Bisectors to score better marks in the exam.

Math in Focus Grade 7 Course 2 B Chapter 7 Lesson 7.2 Answer Key Constructing Perpendicular Bisectors

Math in Focus Grade 7 Chapter 7 Lesson 7.2 Guided Practice Answer Key

Hands-On Activity

Materials:

• ruler

Explore The Distance Between Points On The Perpendicular Bisector Of A Segment And The Endpoints Of The Segment

Work in pairs.

Line XY is the perpendicular bisector of line segment AB. Points W, X, Y, and Z are four points on the perpendicular bisector of \(\overline{\mathrm{AB}}\).

Measure and record each length to the nearest tenth of a centimeter.

Step 2.
Compare the lengths of \(\overline{A W}\) and \(\overline{B W}\). Then compare the lengths of each of the following pairs of segments: \(\overline{A X}\) and \(\overline{B X}\), \(\overline{A Y}\) and \(\overline{B Y}\), and \(\overline{A Z}\) and \(\overline{B Z}\). What do you notice about each pair of line segment lengths?
Answer:

Math Journal
Suppose you choose any point on the perpendicular bisector and measure the distances from that point to points A and B. What do you predict about the distances? What conclusion can you make?

From the activity, any point on the perpendicular bisector of a line segment is equidistant from the two endpoints of the line segment.

Complete.

Question 1.
Copy or trace triangle XYZ. Then draw the perpendicular bisector of line segment XY.

Answer:

 

 

 

 

 

Copy the triangle and complete.

Question 2.
A trucking company has most of its business in Cities P, Q, and R. Where should it locate its new facility so that it is equidistant from all three cities? Mark the location of the facility on a copy of the map.

Answer:

 

 

 

 

 

 

Math in Focus Course 2B Practice 7.2 Answer Key

Draw each line segment and construct its perpendicular bisector.

Question 1.
XY = 5 cm
Answer:

 

 

 

 

 

 

 

Question 2.
PQ = 6.8
Answer:

 

 

 

 

 

 

 

Question 3.
AB = 8.8 cm
Answer:

 

 

 

 

 

 

 

Question 4.
MN = 11 cm
Answer:

 

 

 

 

 

 

 

Question 5.
Draw a line segment between 4 inches and 5 inches and label the endpoints M and N. Construct the perpendicular bisector of \(\overline{\mathrm{MN}}\). Explain briefly if you could construct a different perpendicular bisector of \(\overline{\mathrm{MN}}\).
Answer:

 

 

 

 

 

 

 

Question 6.
Math Journal
In the diagram below, point M is the midpoint of \(\overline{\mathrm{AB}}\). Ben drew line segment XY through point M. He labeled it as the perpendicular bisector of \(\overline{\mathrm{AB}}\). Do you agree with Ben? Give a reason for your answer.

Answer:
Yes, I do agree with Ben.

Explanation:
In the given diagram, point M is the midpoint of \(\overline{\mathrm{AB}}\). Ben also drew line segment XY through point M. So Ben labeled it as the perpendicular bisector of \(\overline{\mathrm{AB}}\). A perpendicular bisector is a line that bisects another line at a right angle, through the intersection point. Hence I agree with Ben.

Question 7.

Answer:

 

 

 

 

 

 

Question 8.
Draw the perpendicular bisectors of \(\overline{\mathrm{PQ}}\) and \(\overline{\mathrm{PR}}\) on a copy of each polygon. Label the point where the two perpendicular bisectors meet as W.

Answer:

 

 

 

 

 

Question 9.
The point that is equidistant from points W, X, and Y. On a copy of WXYZ, mark the points that are described, if possible. Otherwise, explain why you cannot.
Answer:
We are given the polygon:

The point equidistant from points W, X and Y is the intersection of the perpendicular bisectors of the segments \(\overline{W X}\), \(\overline{X Y}\) and \(\overline{Y W}\).

It is enough to draw two of the perpendicular bisectors as the three of them intersect in the same point.
We draw the perpendicuLar bisectors on \(\overline{X Y}\) and \(\overline{W X}\):

They intersect in point A. therefore A is the point that is equidistant from points W, X and Y.

Question 10.
The point that is equidistant from \(\overline{W X}\), \(\overline{W Z}\), and \(\overline{X Y}\).
Answer:

The points equidistant from \(\overline{W X}\) and \(\overline{W Z}\) are on the angle bisector of ∠XWZ.
The points equidistant from \(\overline{W X}\) and \(\overline{X Y}\) are on the angle bisector of ∠WXY.
Therefore the point equidistant from \(\overline{W X}\), \(\overline{W Z}\) and \(\overline{X Y}\) is the intersection of the two angle bisectors.
We draw the angle bisector of ∠XYZ:

We draw the angle bisector of ∠WXY:

They intersect in point B, therefore B is the point that is equidistant from \(\overline{W X}\), \(\overline{W Z}\) and \(\overline{X Y}\):

Question 11.
The point that is equidistant from \(\overline{W X}\) and \(\overline{W Z}\), and also from points X and Y.
Answer:
We are given the polygon:

The points equidistant from \(\overline{W X}\) and \(\overline{W Z}\) are on the angle bisector of ∠XYZ.
The points equidistant from the points X and Y are on the perpendicular bisector of \(\overline{X Y}\).
Therefore the point equidistant from \(\overline{W X}\) and \(\overline{W Z}\) and from points X and Y is the intersection between the angle bisector of ∠XWZ and the perpendicular bisector of \(\overline{XY}\).
We draw the angle bisector of ∠XWZ:

We draw the perpendicular bisector of \(\overline{XY}\):

They intersect in point C, therefore C is the point that ¡s equidistant from \(\overline{W X}\) and \(\overline{W Z}\) and points X, Y.

Solve.

Question 12.
Mark the point on \(\overline{Q R}\) that is equidistant from points Q and R on a copy of triangle POR.

Answer:

 

 

 

 

 

 

Question 13.
Mr. Smith wants to put a circular water sprinkler in his triangular-shaped garden. His garden has a tree at each vertex. The water sprinkler is to be equidistant from the trees. Copy or trace the given triangle. Mark point W to show where Mr. Smith should put the water sprinkler.

Answer:
We are given:

The point equidistant from the 3 trees is the intersection of the perpendicuLar bisectors of the triangles edges

It is enough to draw two of the perpendicular bisectors as the three of them intersect in the same point
We draw the perpendicular bisectors on two edges:

They intersect in point A. therefore A is the point that is equidistant from the 3 trees.

Question 14.
Math Journal
Melissa was asked to bisect a very long line segment using a compass and a straightedge. Melissa opened the compass to the widest possible setting. She found that it was not wide enough to do the standard construction that she had learned in math class. Is it possible for Melissa to bisect the very long line segment with only her compass and a straightedge? Explain and give a suggestion.
Answer:
Let \(\overline{A B}\) be the initial very long segment.
In order to be able to use an opening of the compass greater than half of a segment, she should ‘decrease the length of the given long segment as much as possible. For this, she places the compass, with its greatest opening, in point A and draw an arc which intersects \(\overline{A B}\) in X₁. With the same opening she places the compass in B and draw an arc which intersects \(\overline{A B}\) in Y₁.. She got a smaller segment \(\overline{X_{1} Y_{1}}\).

She continues placing the compass in X₁ and Y₁ and drawing arcs until she gets a segment \(\overline{X_{n} Y_{n}}\),  for which she can construct the perpendicular bisector.

She constructs the perpendicular bisector of \(\overline{X_{n} Y_{n}}\), which is also the perpendicular bisector of \(\overline{A B}\) because AX_n = BY_n.

Decrease the length of the segment with equal values until the procedure of constructing the perpendicular bisector can be applied.

Question 15.
Math Journal
Copy the rectangle shown. Use the least possible number of arcs and lines to construct the perpendicular bisectors of each side of the rectangle. Explain how you know you have used the least possible number.

Answer:

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 4 Lesson 4.1 Understanding Equivalent Equations to score better marks in the exam.

Math in Focus Grade 7 Course 2 A Chapter 4 Lesson 4.1 Answer Key Understanding Equivalent Equations

Math in Focus Grade 7 Chapter 4 Lesson 4.1 Guided Practice Answer Key

Copy and complete to state whether each pair of equations are equivalent equations. Give a reason for your answer.

Question 1.

x – 3 + 4x = 5 and 5x = 2
x – 3 + 4x = 5
5x – 3 = 5 Group like terms.
5x – 3 + = 5 + Add to both sides.
5x = Simplify.
x – 3 + 4x = 5 be rewritten as 5x = 2.
So, the equations have solutions and are
Answer:
No, they are not equivalent equations.

Explanation:
Given 1st equation, x – 3 + 4x = 5

It can also be written as 5x – 3 =5

Add 3 to both sides, 5x – 3 + 3 = 5 + 3

5x = 8

So, equation x – 3 + 4x = 5  cannot be written as 5x = 2

Question 2.
x + 7 = 12 and 2x = 10
First solve x + 7 = 12.
x + 7 = 12
x + 7 – = 12 – Subtract from both sides.
x = Simplify.
Then check to see if is the solution of the equation 2x = 10.
If x = , 2x = 2 • Substitute for x.
= a solution.
Because the equations have the solution, they are equations.
Answer:
Yes, Both are equivalent equations.

Explanation:
First, let’s solve x + 7 = 12

Subtract 7 on both sides.

x + 7 – 7 = 12 -7

x = 5

Lets consider second equation 2x = 10

substitute x= 5 which we got by solving  the 1st equation in the 2nd equation.

2.(5) = 10

Hence proved.

Question 3.
1.2x = 2.4 and x – 6 = 8
First solve x – 6 = 8.
x – 6 = 8
x – 6 + = 8 + Add to both sides,
x = Simplify.
Then check to see if is the solution of the equation 1.2x = 2.4.
If x = , 1.2x = 1.2 • Substitute for x.
= a solution.
Because the equations have solutions, they are equations.
Answer:
No, they are not equivalent equations.

Explanation:
First, lets consider x – 6 = 8

Add 6 on both sides

x – 6 + 6 = 8 + 6

x = 14

Lets substitute x = 14  in the first equation.

1.2 x = 2.4

1.2(14) = 16.8

Both the equations are not equivalent equations.

Question 4.
\(\frac{2}{5}\)x = 4 and x = 10.
If x = 10, \(\frac{2}{5}\)x = \(\frac{2}{5}\) • 10 Substitute 10 for x.
= a solution.
Because the equations have the solution, they are equations.
Answer:
Yes, both the equations have the same solution, they are equivalent equations.

Explanation:
Let’s consider first equation, (2/5)x =4

Let us  substitute the second equation, x =10 in first equation.

(2/5) × 10 = 4

Hence proved.

Both the equations have the same solution, they are equivalent equations.

Math in Focus Course 2A Practice 4.1 Answer Key

Tell whether each pair of equations are equivalent. Give a reason for your answer.

Question 1.
2x = 4 and 4x + 5 = 13
Answer:
Yes, Both the equations are equivalent.

Explanation:
Let us consider second equation, 4x + 5 = 13

Subtract 5 on both sides.

4x + 5 – 5 = 13 – 5

4x = 8

x = 8 ÷ 4

x = 2

Substitute x=2 in the first equation, 2x = 4

2(2) = 4

Hence proved.

Both the equations are equivalent.

Question 2.
-2x + 9 = 7 and —2x = 2
Answer:
No, both the equations are not equivalent.

Explanation:
Lets solve the first equation, -2x + 9 = 7

Subtract 9  on both sides.

-2x + 9 – 9 = 7 – 9

-2x = -2

The second equation is -2x = 2

So, both the equations are not equivalent.

Question 3.
5x – 4 + 3x = 8 and 8x = 12
Answer:
Both are equivalent equations.

Explanation:
Lets us consider first equation, 5x – 4 + 3x = 8

8x – 4 = 8

8x = 8 + 4

8x = 12

As we derived second equation from the first equation, both are equivalent equations.

Question 4.
\(\frac{3}{4}\)x – 7 = 2 and x = 12
Answer:
Yes, both are equivalent equations.

Explanation:
Let us consider first equation, (3/4)x = 2

Substitute, the second equation, x = 12 in the first equation.

(3/4)(12) – 7 = (3).(3) – 7

=9 – 7

=2

Hence proved. Both are equivalent equations.

Match each equation with an equivalent equation.

Answer:
5 – d
6 – c
7 – e
8 – a
9 – g
10 – f
11 – b

Explanation:
5.  0.5x + 1 = 1.5

0.5x = 1.5 – 1 = 0.5

x = 0.5 ÷ 0.5

x = 1

We can match 5th equation with equation (d) on the other side.

6. 9 + 3.5x = 16

Let us solve the equation

3.5x = 16 – 9

3.5x = 7

x = 7 ÷ 3.5

x = 2

We can match the equation (3/2)x = 3

x = 3 × (2/3)

x = 2

Hence both are same.

7. (4/5)x =4

4x = 20

x =5

We can match 7th equation with equation (e) on the other side.

2x =10

x =5

Hence Matched.

8.  2x + (1/2) = (7/2)

2x = (7/2) – (1/2)

2x = (6/2)

2x = 3

x = 3/2

As option (a) is 6x =9

It can be also written as 3/2 as both are dividends of 3.

9. x – 8.3 = 1.3

x = 1.3 + 8.3

x = 9.6

Option (g) is  (1/2)x = 4.8

x =9.6

Hence matched.

10. 13.9 = 2.5x

x = 13.9 ÷ 2.5

x = 5.56

Option (f) is 1.2 + x = 6.76

x = 6.76 – 1.2

x =5.56

Hence proved.

11. 4x = (4/9)

x = 1/9

We can match with option (b).

(3/5)x = (1/15)

x = (1/9)

Solve.

Question 12.
Math Journal Max was asked to write to \(\frac{2}{3}\)x = 3 – x. He wrote the following:
\(\frac{2}{3}\)x = 3 – x
\(\frac{2}{3}\)x ∙ 3 = 3 ∙ 3 – x
2x = 9 – x
He concluded that \(\frac{2}{3}\)x = 3 – x and 2x = 9 \(\frac{2}{3}\) x are equivalent equations. Do you agree with his conclusion? Give a reason for your answer.
Answer:
No, I don’t agree with Max, both are not equivalent equations.

Explanation:
Let us consider first equation, (2/3)x = 3 – x

2x = 9 -3x (Max gave the equation 2x = 9 – x which is wrong, if we multiplied 3 on both sides not only it is multiplied with 3 it is also multiplies with x. so it is 2x = 9 -3x)

5x = 9

x = 9/5

x = 1.8

The second equation, 9(2/3)x =2x

substitute x =1.8 in the second equation

2 (1.8) = 9(2/3)(1.8)

3.6 = 10.8

Hence the both the equations are not equivalent.

 

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 4 Algebraic Equations and Inequalities to score better marks in the exam.

Math in Focus Grade 7 Course 2 A Chapter 4 Answer Key Algebraic Equations and Inequalities

Math in Focus Grade 7 Chapter 4 Quick Check Answer Key

Solve each equation.

Question 1.
x + 4 = 10
Answer:
6

Explanation:
x + 4 = 10
x = 10 -4

x = 6

Question 2.
x – \(\frac{1}{2}\) = 2
Answer:
2(1/2)

Explanation:
x – (1/2) = 2

x = 2 + (1/2)

x = 2 (1/2)

Question 3.
\(\frac{1}{5}\)x = 3
Answer:
15

Explanation:
(1/5)x = 3
x=  3 × 5

x = 15

Question 4.
1.2x = 2.4
Answer:
2

Explanation:
1.2x=2.4

x= 2.4 ÷ 1.2

x= 2

State whether each statement is True or False.

Question 5.
x = 1 gives the solution of the algebraic equation 3x + 5 = 8.
Answer:
True

Explanation:
3x + 5 = 8

Given, x = 1. Substitute x= 1 in the equation.

3(1) + 5  = 8

8 = 8

Hence proved.

Question 6.
y = 2 gives the solution of the algebraic equation 6y – 3 = 8.
Answer:
False

Explanation:
6y – 3 = 8

Given y = 2, substitute in the equation.

6(2) – 3 = 8

12 – 3 = 8

9 ≠ 8

So False.

Question 7.
z = 6 gives the solution of the algebraic equation \(\frac{z}{3}\) = 3.
Answer:
False

Explanation:
\(\frac{z}{3}\) = 3

(z/3) = 3

Given, z = 6, Substitute it in the equation.

(6/3) = 3

2 ≠ 3

So  false.

Question 8.
w = 3 gives the solution of the algebraic equation 2w = 6.
Answer:
True

Explanation:
Given equation, 2w = 6.

Given w=3, substitute it in the equation.

2(3) = 6

6 = 6

So the equation is true.

Draw a number line to represent each inequality.

Question 9.
x ≥ 3.5
Answer:

Explanation:
Given equation, x ≥ 3.5

In the equation , x can be any number which is equal to (or) more than 3.5.

It is represented in the number line as given above.

Question 10.
y < \(\frac{1}{2}\)
Answer:

Explanation:
Given equation, y < (1/2)

In the equation, y is a number which is less than (1/2)

It is represented in the number line as given above.

Complete each with =, >, or <.

Question 11.
11 12
Answer:
11 < 12

Question 12.
-9 -7
Answer:
-9  <  -7

Question 13.
25 ∙ (-1) (-1) ∙ 25
Answer:
25 ∙ (-1)  =  (-1) ∙ 25

Question 14.
3 ÷ (-1) (-1) ÷ 3
Answer:
3 ÷ (-1) = (-1) ÷ 3

Use x to represent the unknown quantity. Write an algebraic inequality for each statement.

Question 15.
The box can hold less than 70 pounds.
Answer:
x < 70

Explanation:
Let us consider the box be “x”.

As the box can hold less than 70 pounds.

Hence the equation x < 70 .

Question 16.
You have to be at least 17 years old to qualify for the contest.
Answer:
x ≥ 17

Explanation:
“At least” means it should be minimum of 17 or more than that.

Given, you have to be at least 17 years old to qualify for the contest.

Hence the equation, x ≥ 17.

Question 17.
The width of luggage that you can carry onto the plane is at most 17 inches.
Answer:
y ≤ 17

Explanation:
“At most” means it should be maximum of 17 nor less than that, but not more than 17.

As, The width of luggage that you can carry onto the plane is at most 17 inches.

Hence the equation, y ≤ 17

Question 18.
There are more than 120 people standing in line for the roller coaster.
Answer:
y > 120

Explanation:
The statement states “There are more than 120 people standing in line for the roller coaster” i.e, higher than 120.

Hence the equation, y > 120

Go through the Math in Focus Grade 7 Workbook Answer Key Chapter 3 Review Test to finish your assignments.

Math in Focus Grade 7 Course 2 A Chapter 3 Review Test Answer Key

Concepts and Skills
Simplify each expression.
Question 1.
1.4w – 0.6w
Answer:
1.4w – 0.6w = 0.8w.

Explanation:
1.4w – 0.6w
= 0.8w.

 

Question 2.
\(\frac{3}{4}\)m + \(\frac{4}{5}\)m
Answer:
\(\frac{3}{4}\)m + \(\frac{4}{5}\)m = \(\frac{31}{20}\)m.

Explanation:
\(\frac{3}{4}\)m + \(\frac{4}{5}\)m
LCD of 4 n 5 = 20.
= [(3 × 5) + (4 × 4)]m ÷ 20
= (15 + 16)m ÷ 20
= 31m ÷ 20 or \(\frac{31}{20}\)m.

 

Question 3.
\(\frac{1}{6}\)y + \(\frac{1}{2}\)y + \(\frac{1}{3}\)y
Answer:
\(\frac{1}{6}\)y + \(\frac{1}{2}\)y + \(\frac{1}{3}\)y = y.

Explanation:
\(\frac{1}{6}\)y + \(\frac{1}{2}\)y + \(\frac{1}{3}\)y = y.
LCD of 6 n 2 = 6.
= [(1 × 1) + (1 × 3)]y ÷ 6] +  \(\frac{1}{3}\)y
= [(1 + 3)y ÷ 6] + \(\frac{1}{3}\)y
= (4y ÷ 6) + \(\frac{1}{3}\)y
= \(\frac{2}{3}\)y+ \(\frac{1}{3}\)y
= (2 + 1)y ÷ 3
= 3y ÷ 3
= y.

 

Question 4.
1.8m – 0.2m – 7m
Answer:
1.8m – 0.2m – 7m = -5.4m.

Explanation:
1.8m – 0.2m – 7m
= 1.6m – 7m
= -5.4m.

 

Question 5.
1.3a – 0.8b + 2.2b – a
Answer:
1.3a – 0.8b + 2.2b – a = 0.3a + 1.4b.

Explanation:
1.3a – 0.8b + 2.2b – a
= 1.3a – a – 0.8b + 2.2b
= 0.3a – 0.8b + 2.2b
= 0.3a + 1.4b.

 

Question 6.
1 + \(\frac{1}{5}\)a + \(\frac{3}{5}\)b + \(\frac{4}{5}\)a
Answer:
1 + \(\frac{1}{5}\)a + \(\frac{3}{5}\)b + \(\frac{4}{5}\)a = 1 + a + \(\frac{3}{5}\)b.

Explanation:
1 + \(\frac{1}{5}\)a + \(\frac{3}{5}\)b + \(\frac{4}{5}\)a
= 1 + \(\frac{1}{5}\)a + \(\frac{4}{5}\)a + \(\frac{3}{5}\)b
= 1 + [(1 + 4)a ÷ 5] + \(\frac{3}{5}\)b
= 1 + (5a ÷ 5) + \(\frac{3}{5}\)b
= 1 + a + \(\frac{3}{5}\)b.

 

Expand each expression. Then simplify when you can.
Question 7.
1.2(2p – 3)
Answer:
1.2(2p – 3) = 2.4p – 3.6.

Explanation:
1.2(2p – 3)
= (1.2 × 2p) – (3 × 1.2)
= 2.4p – 3.6.

 

Question 8.
\(\frac{1}{3}\)(12p + 9q)
Answer:
\(\frac{1}{3}\)(12p + 9q) = 4p + 3q.

Explanation:
\(\frac{1}{3}\)(12p + 9q)
= [12p × \(\frac{1}{3}\)] + [ 9q × \(\frac{1}{3}\)]
= 4p + 3q.

 

Question 9.
\(\frac{1}{5}\)\(\left(\frac{t}{3}+\frac{1}{2}\right)\)
Answer:
\(\frac{1}{5}\)\(\left(\frac{t}{3}+\frac{1}{2}\right)\) = \(\frac{t}{15}\) + \(\frac{1}{10}\).

Explanation:
\(\frac{1}{5}\)\(\left(\frac{t}{3}+\frac{1}{2}\right)\)
= [\(\frac{t}{3}\) × \(\frac{1}{5}\)] + [\(\frac{1}{2}\) × \(\frac{1}{5}\)]
= \(\frac{t}{15}\) + \(\frac{1}{10}\).

 

Question 10.
-4(-2q + 2.5)
Answer:
-4(-2q + 2.5) = 8q – 10.

Explanation:
-4(-2q + 2.5)
= (-4 × -2q) + (2.5 × -4)
= 8q – 10.

 

Question 11.
–\(\frac{2}{3}\)(6x + 3)
Answer:
–\(\frac{2}{3}\)(6x + 3) = 4x – 2.

Explanation:
–\(\frac{2}{3}\)(6x + 3)
= (6x × –\(\frac{2}{3}\)) + (3 × –\(\frac{2}{3}\))
= (2x × 2) + (1 × -2)
= 4x – 2.

 

Question 12.
-0.5(2m – 4n)
Answer:
-0.5(2m – 4n) = -m + 0.5n.

Explanation:
-0.5(2m – 4n)
= (2m × -0.5) – (1n × -0.5)
= -1m – (-0.5n)
= -m + 0.5n.

 

Question 13.
3(a + 3) + 2a
Answer:
3(a + 3) + 2a = 5a + 9.

Explanation:
3(a + 3) + 2a
= [(3 × a) + (3 × 3)] + 2a
= 3a + 9 + 2a
= 3a + 2a + 9
= 5a + 9.

 

Question 14.
4(2p – 3) – 3(p + 2)
Answer:
4(2p – 3) – 3(p + 2) = 5p – 6.

Explanation:
4(2p – 3) – 3(p + 2)
= [(2p × 4) – (3 × 4)] – [(3 × p) + (2 × 3)]
= (8p – 12) – (3p + 6)
= 8p – 12 – 3p – 6
= 8p – 3p – 12 – 6
= 5p – 6.

 

Question 15.
2.5(m – 2) + 5.6m
Answer:
2.5(m – 2) + 5.6m = 8.1m – 5.

Explanation:
2.5(m – 2) + 5.6m
= [(m × 2.5) – (2 × 2.5)] + 5.6m
= (2.5m – 5) + 5.6m
= 2.5m – 5 + 5.6m
= 2.5m + 5.6m – 5
= 8.1m – 5.

 

Question 16.
4(0.6n – 3) – 0.2(2n – 3)
Answer:
4(0.6n – 3) – 0.2(2n – 3) = 2n – 12.6.

Explanation:
4(0.6n – 3) – 0.2(2n – 3)
= 2.4n – 12 – 0.4n + 0.6
= 2.4n – 0.4n – 12 – 0.6
= 2n – 12.6.

Factor each expression.
Question 17.
4t – 20s
Answer:
4t – 20s = 4(t – 5s).

Explanation:
4t – 20s
= 4(t – 5s).

Question 18.
-6p – 21q
Answer:
-6p – 21q = -3(2p + 7q).

Explanation:
-6p – 21q
= -3(2p + 7q).

 

Question 19.
8i + 12 + 4j
Answer:
8i + 12 + 4j = 4(2i + 3 + j).

Explanation:
8i + 12 + 4j
= 4(2i + 3 + j)

 

Question 20.
6a + 10b – 20
Answer:
6a + 10b – 20 = 2(3a + 5b – 10).

Explanation:
6a + 10b – 20
= 2(3a + 5b – 10).

 

Question 21.
-9m – 3n – 6
Answer:
-9m – 3n – 6 = -3(3m + n + 2).

Explanation:
-9m – 3n – 6
= -3(3m + n + 2).

 

Question 22.
-15x – 6 – 12y
Answer:
-15x – 6 – 12y = -3(5x + 2 + 4y).

Explanation:
-15x – 6 – 12y
= -3(5x + 2 + 4y).

Translate each verbal description into an algebraic expression. Then simplify when you can.
Question 23.
One-fourth x less than the sum of 7 and 2x.
Answer:
\(\frac{x}{4}\) – (7 + 2x) = – \(\frac{7}{4}\)x – 7.

Explanation:
One-fourth x less than the sum of 7 and 2x.
=> \(\frac{x}{4}\) – (7 + 2x)
=> \(\frac{x}{4}\) – 7 – 2x
=> \(\frac{x}{4}\) – 2x – 7.
=> [(x – 8x) ÷ 4 ] – 7.
=> (-7x ÷ 4) – 7
=> – \(\frac{7}{4}\)x – 7.

 

Question 24.
4 times 5y divided by 18.
Answer:
(4 × 5y ) ÷ 18 = \(\frac{10}{9}\)y.

Explanation:
4 times 5y divided by 18
=> (4 × 5y ) ÷ 18
=> 20y ÷ 18
=> 10y ÷ 9 or \(\frac{10}{9}\)y.

 

Question 25.
Five-ninths of (3p + 1) subtracted from one-third of (q + p).
Answer:
[\(\frac{5}{9}\) × (3p + 1) ] – [\(\frac{1}{3}\) × (q + p)] = 2p – \(\frac{1}{3}\)q + \(\frac{5}{9}\)

Explanation:
Five-ninths of (3p + 1) subtracted from one-third of (q + p)
=> [\(\frac{5}{9}\) × (3p + 1) ] – [\(\frac{1}{3}\) × (q + p)]
=> \(\frac{15}{9}\)p + \(\frac{5}{9}\) – \(\frac{1}{3}\)q + \(\frac{1}{3}\)p.
=> \(\frac{15}{9}\)p + \(\frac{1}{3}\)p – \(\frac{1}{3}\)q + \(\frac{5}{9}\)
= LCD of 9 n 3 = 9.
= [(15 + 3)p ÷ 9 ] – \(\frac{1}{3}\)q + \(\frac{5}{9}\)
= (18p ÷ 9) – \(\frac{1}{3}\)q + \(\frac{5}{9}\)
= (2p ÷ 1) – \(\frac{1}{3}\)q + \(\frac{5}{9}\)
= 2p – \(\frac{1}{3}\)q + \(\frac{5}{9}\)

 

Problem Solving
Solve. Show your work.
Question 26.
After 14 boys leave a concert, the ratio of boys to girls is 3 : 10. If there are p girls at the concert, write an algebraic expression for the number of boys at the beginning of the concert in terms of p.
Answer:
An algebraic expression for the number of boys at the beginning of the concert in terms of p =
0.3 p + 14.

Explanation:
Number of girls in the concert = p.
After 14 boys leave a concert, the ratio of boys to girls is 3 : 10.
Let total number of boys be b.
Ratio of boys after 14 boys leave:
=> \(\frac{b – 14}{p}\) = \(\frac{3}{10}\)
=> 3 × p = 10( b – 14)
=> 3 p = 10 b – 140
=>  10 b = 3 p + 140
= >  b = 0.3 p + 14

Practice the problems of Math in Focus Grade 7 Workbook Answer Key Chapter 6 Lesson 6.4 Interior and Exterior Angles to score better marks in the exam.

Math in Focus Grade 7 Course 2 B Chapter 6 Lesson 6.4 Answer Key Interior and Exterior Angles

Math in Focus Grade 7 Chapter 6 Lesson 6.4 Guided Practice Answer Key

Hands-On Activity

Materials:

• scissors
• ruler

Explore The Sum Of The Angles In A Triangle

Work in pairs.

Step 1.
Draw and cut out a triangle. Label the three angles of the triangle as 1, 2, and 3.

Step 2.
Cut the triangle into three pieces, so that each piece contains a vertex.

Step 3.
Rearrange the cut pieces on a straight line. What do you notice about the sum of the measures of the three interior angles?

Complete.

Question 1.
Find the value of p.

Answer:
p° = 88°

Explanation:

∠RPQ + ∠QRP + ∠PQR = 180°
35° + p° + 57° = 180°
p° + 92° = 180°
p° = 180° – 92°
p° = 88°

Complete.

Question 2.
Triangle XYZ is an isosceles triangle and m∠XYZ = 55°. Find the value of x.

Answer:
∠XYZ + ∠XZY + ∠YXZ = 180°
55° + 55° + x° = 180°
110° +x° = 180°
x° = 180° – 110°
x° = 70°

Find the value of each variable.

Question 3.

Answer:
x° = 70°

Explanation:

x° + 62° = 132°
x° = 132° – 62°
x° = 70°

Question 4.

Answer:
x° = 76°

Explanation:

x° + 35°= 70° + 41°
x° + 35° = 111°
x° = 111° – 35°
x° = 76°

Complete.

Question 5.
Triangle ABC is an isosceles triangle. Find the measures of ∠1 and ∠2.
AB = AC
m∠ACB = m∠ABC

Answer:
∠1 = 255°, ∠2 = 95°

Explanation:
m∠ACB = m∠ABC
∠1 = 75° + 180°
∠1 = 255°
∠2 = 180° – 75°
∠2 = 95°

Question 6.
Triangle EFG is an isosceles triangle and \(\overleftrightarrow{E F}\) is parallel to \(\overleftrightarrow{H I}\). Find the value of x.

Answer:
x° = 120°

Explanation:

∠FEG = ∠IGJ
= 30°
∠FGE = ∠FEG
= 30°
∠FEG + ∠FGE + ∠EFG = 180°
30° + 30° + x° = 180°
x° + 60° = 180°
x° = 180° -60°
x° = 120°

Math in Focus Course 2B Practice 6.4 Answer Key

Find the value of y.

Question 1.

Answer:
y° = 61°

Explanation:
y° + 52° + 67° = 180°
y° + 119° = 180°
y° = 180° -119°
y° = 61°

Question 2.

Answer:
y° + 26° + 18° = 180°
y° + 44° = 180°
y° = 180° – 44°
y° = 136°

Question 3.

Answer:
20° + y° = 90° + 42°
20° + y° = 132°
y° = 132° -20°
y° = 112°

Question 4.

Answer:
90° + 27° + x° = 180°
117° + x° = 180°
x° = 180° -117°
x° = 63°
63° + 51° + y° = 180°
114° + y° = 180°
y° = 180° – 114°
y° = 66°

Question 5.

Answer:
y° = 122°

Explanation:
y° + 28° + 30° = 180°
y° + 58° = 180°
y° = 180° – 58°
y° = 122°

Question 6.

Answer:
x°  = 57° , y° = 67°

Explanation:
x°  + 123°  = 180°
x°  = 180° -123°
x°  = 57°
56°  + 57° + y° = 180°
y°  + 113° = 180°
y° = 180° – 113°
y° = 67°

Find m∠1 and m∠2 in each diagram.

Question 7.

Answer:
62° + 90° + m∠1 = 18o° ∠s sum in triangle:
152° + m∠1 = 180° Simplify:
152° + m∠1 — 152° = 180° — 152° Subtract 152° from both sides:
m∠1 = 28° Simplify:
m∠2 = 90° + m∠1 Exterior ∠ of triangle:
m∠2 = 90° + 28° substitute
m∠2 = 118° simplify
m∠1 = 28°
m∠2 = 118°

Question 8.

Answer:
∠1 = 76°, ∠2 = 114°

Explanation:
∠1 + 28° + 76° = 180°
∠1 + 104° = 180°
∠1 = 180° – 104°
∠1 = 76°
76° + 38° + x° = 180°
114° + x° = 180°
x° = 180° – 114°
x° = 66°
x° + ∠2 = 180°
66° + ∠2 = 180°
∠2 = 180° – 66°
∠2 = 114°

Question 9.

Answer:
50° + 70° + m∠1 = 180° Corresponding ∠s and ∠s sum in triangle:
120° + m∠1 = 180° Simplify:
120° + m∠1 – 120° = 180° – 120° Subtract 120° from both sides:
m∠1 = 60° Simplify:
m∠2 = 50° + 70° Exterior ∠ of triangle:
m∠2 = 120° simplify
m∠1 = 60°
m∠2 = 120°

Question 10.

Answer:
70° = 43° + m∠1 Alternate interior ∠s
70° – 43° = 43° + m∠1 – 43° Subtract 43° from both sides:
m∠1 = 27° Simplify:
m∠2 + m∠1 + 43° = 180° ∠s sum of triangle:
m∠2 + 27° + 43° = 180°
m∠2 + 70° = 180° Simplify:
m∠2 + 70° — 70° = 180° — 70° Subtract 70° from both sides:
m∠2 = 110° Simplify:
m∠1 = 27
m∠2 = 110°

Use an equation to find the value of y.

Question 11.

Answer::
3y° = y° + 80° Exterior ∠ of triangle:
3y = y + 80 The equation for y is:
3y – y = y + 80 – y Subtract y from both sides:
2y = 80 Simplify:
\(\frac{2 y}{2}\) = \(\frac{80}{2}\) Divide by 2
y = 40 simplify
3y = y + 80
y = 40

Question 12.

Answer:
Let’s note by x° the measure of the congruent angles of the isosceles triangle
x° = 2y° Alternate interior ∠s
x° + 3y° = 180° Supplementary ∠s:
2y° + 3y° = 180° Substitute:
5y° = 180° Simplify:
\(\frac{5 y^{\circ}}{5}\) = \(\frac{180^{\circ}}{5}\) Divide by 5
y = 36 simplify
2y + 3y = 180
y = 36

Find the value of x and name the type of triangle that is shown.

Question 13.

Answer:
x° = 60°, x° = 60°

Explanation:
x° + x° + 60° = 180°
2x° + 60° = 180°
2x° = 180° – 60°
2x° = 120°
x° = 120 ÷ 2
x = 60°

Question 14.

Answer:
x° = 36°, x° = 36°

Explanation:
x°+ x°+ 108° = 180°
2x° + 108° = 180°
2x° = 180° – 108°
2x° = 72°
x° = 72÷2
x° = 36°

Question 15.

Answer:
x° = 8°

Explanation:
We all know that all the angles in a triangle add up to 180 degrees.
In the given figure, each triangle has 3 angles.  Thus, we have the sum of three angles as shown:
A° + B° +C° = 180°
62°+ 90°+ x° = 180°
x°+ 172° = 180°
x° = 180° – 172°
x° = 8°

Find the measure of each numbered angle.

Question 16.

Answer:
m∠2 = 148° Corresponding ∠s:
m∠2 = m∠3 + 63° Vertical ∠s and exterior ∠ of triangle:
148° = m∠3 + 63° Substitute:
148° – 63° = m∠3 + 63° – 63° Subtract 63° from both sides:
m∠3 = 85° Simplify:
m∠1 + 63° = 180° supplementary ∠s
m∠1 + 63° – 63° = 180° – 63° subtract 63° from both sides
m∠1 = 117° simplify
m∠1 = 117°
m∠2 = 148°
m∠3 = 85°

Question 17.

Answer:
m∠1 + 140° = 180° Supplementary ∠s
m∠1 + 140° — 140° = 180° – 140° Subtract 140° from both sides:
m∠1 = 40° Simplify:
m∠2 + 15° + 40° = 180° Alternate interior ∠s and ∠s sum of triangle:
m∠2 + 55° = 180° Simplify:
m∠2 + 55° – 55° = 180° – 55° subtract 55° from both sides
m∠2 = 125° simplify
m∠1 = 40°
m∠2 = 125°

Find the measure of each numbered angle.

Question 18.

Answer:

We are given
m∠5 = 50° Corresponding ∠s
m∠3 = m∠5 = 50° Corresponding ∠s
m∠4 = 50° Alternate Interior ∠s
m∠4 + m∠2 + 77° = 180° substitute
m∠2 + 127° = 180° simplify
m∠2 + 127° – 127° = 180° — 127° Subtract 127° from both sides:
m∠2 = 53° Simplify:
114° = m∠1 + m∠4 Exterior ∠ of triangle:
114° = m∠1 + 50° Substitute:
114° – 50° = m∠1 + 50° + 50° Substitute 50° from both sides
m∠1 = 64° simplify
m∠1 = 64°
m∠2 = 53°
m∠3 = 50°

Question 19.

Answer:

we are given
m∠1 + 67° + 53° = 180° Sum of ∠s in triangle:
m∠1 + 120° = 180° Simplify:
m∠1 + 120° – 120° = 180° — 120° Subtract 120° from both sides:
m∠1 = 60° Simplify:
m∠3 = 67° Alternate interior ∠s
m∠2 = m∠1 + m∠3 vertical ∠s
m∠2 = 60° + 67° substitute
m∠2 = 127° simplify
m∠1 = 60°
m∠2 = 127°

Solve.

Question 20.
\(\overleftrightarrow{\mathrm{BE}}\) is parallel to \(\overleftrightarrow{\mathrm{FH}}\). Find the measure of A
∠CAD in terms of ∠1 and ∠2.

Answer:
m∠ADE = m∠2 Corresponding ∠s:
m∠ADE = m∠CAD + m∠1 Exterior ∠ of triangle:
m∠2 = m∠CAD + m∠1 Substitute:
m∠2 — m∠1 = m∠CAD + m∠1 — m∠1 subtract m∠1 from both sides:
m∠CAD = m∠2 – m∠1 Simplify:
m∠CAD = m∠2 – m∠1

Question 21.
Math Journal
Explain why each of the following statement is true.
a) A triangle cannot have two right angles.
Answer:
m∠A = m∠B = 90° a) Let’s assume that a triangle ABC has two right angles:
m∠A + m∠B + m∠C = 180° Sum of ∠s in triangle:
90° + 90° + m∠C = 180° Substitute:
180° + m∠C = 180° Simplify:
180° + m∠C – 180° = 180° – 180° Subtract 180° from both sides:
m∠C = 0° Simplify:
We got a false statement as the triangle does not exist in this case. Therefore we assumed a wrong statement So a triangle cannot have two right angles.

b) The interior angle measures of an isosceles triangle cannot be 96°, 43°, and 43°.
Answer:
Lets assume that a triangle has the measures of its angles 96°, 43°, 43°
96° + 43° + 43° = 182° We compute the sum of the angles’ measures:
But the sum of a triangle’s angles measures is 180°. We got 182° ≠ 180°, therefore the statement that we assumed is false. So the interior measures of a triangle cannot be 96°, 43°, 43°.
See proofs (sum of the measures of the triangle’s angles)

Question 22.
m∠1 = 2x°, m∠2 = (x – 5)°, and m∠3 = 100°. Use an equation to find the value of x and then find the measures of ∠1 and ∠2.

Answer:
∠1 = 80°, ∠2 = 35°

Explanation:
∠1 + ∠3 = 180°
2x° + 100° = 180°
2x° = 180° – 100°
2x° = 80°
x° = 80°÷2
x° = 40°
∠1 = 2×40 = 80°
∠2 = 40 – 5 =35°

Use an equation to find the value of x.

Question 23.

Answer:
x° = 56°

Explanation:
x° + 94°+ 30° = 180°
x° + 124° = 180°
x° = 180° – 124°
x° = 56°

Question 24.

Answer:
x° = 59°, 2x° = 118°

Explanation:
2x° + 62° = 180°
2x° = 180° – 62°
2x° = 118°
x° = 118°÷2
x° = 59°
x + 13 = 59° + 13° = 72°

Use an equation to find the value of x.

Question 25.

Answer:
x° = 50°, y° = 65°

Explanation:
y° + 115° = 180°
y° = 180° -115°
y° = 65°
65° + 65° + x° = 180°
x° + 130° = 180°
x° = 180° – 130°
x° = 50°

Question 26.

Answer:
x° + x° + 12° = 180°
2x° + 12° = 180°
2x° = 180° – 12°
2x° = 68°
x° = 68 ÷ 2
x° = 34°

Brain@Work

ABCD is a rhombus, and the measure of ∠BCD is 68°. BDE is a triangle where the measure of ∠BED is 36° and the measure of ∠BDE is 73°. Find m∠EBC. Show how you obtain your answer.

Answer:
m∠EBD + 36° + 73° = 180° Sum of ∠s in triangle:
m∠EBD + 109° = 180° Simplify:
m∠EBD + 109° — 109° = 180° — 109° Subtract 109° from both sides:
m∠EBD = 71° Simplify:
\(\overline{B C}\) = \(\overline{D C}\) ABCD rhombus
m∠CDB = m∠CBD ∆BCD isosceles:
68° + m∠CDB + m∠CBD = 180° Sum of ∠s in triangle:
68° + 2m∠CBD = 180°
68° + 2m∠CBD — 68° = 180° — 68° Subtract 68° from both sides:
2m∠CBD = 112° Simplify:

Divide by 2
m∠CBD = 56° Simplify:
m∠EBC = m∠EBD — m∠CBD Substitute:
m∠EBC = 71° — 56°
= 15°