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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

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 10 Lesson 10.2 Area of Parallelograms and Trapezoids to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 10 Lesson 10.2 Answer Key Area of Parallelograms and Trapezoids

Math in Focus Grade 6 Chapter 10 Lesson 10.2 Guided Practice Answer Key

Complete to find the sum of the bases, height, and area of each trapezoid. Each square measures 1 unit by 1 unit.

Question 1.

Base = units
Height = units
Area = bh
= ∙
= units²
Answer:
The base in the above figure is occupying 4 units, so it will measure 4 units and the height is occupying 7 units, so it will measure 7 units.
Base = 4 units
Height = 7 units
Area of rectangle = base×height
= 4×7
= 28 sq.units

Question 2.

Base = units
Height = units
Area = bh
= ∙
= units²
Answer:
The base in the above figure is occupying 3 units, so it will measure 4 units and the height is occupying 7 units, so it will measure 7 units.
Base = 3 units
Height = 7 units
Area of rectangle = base×height
= 3×7
= 21 sq.units

Question 3.

Base = units
Height = units
Area = bh
= ∙
= units²
Answer:
The base in the above figure is occupying 4 units, so it will measure 4 units and the height is occupying 8 units, so it will measure 7 units.
Base = 4 units
Height = 8 units
Area of rectangle = base×height
= 4×8
= 32 sq.units

Question 4.

Base = units
Height = units
Area = bh
= ∙
= units²
Answer:
The base in the above figure is occupying 2 units, so it will measure 4 units and the height is occupying 8 units, so it will measure 7 units.
Base = 4 units
Height = 8 units
Area of rectangle = base×height
= 4×8
= 32 sq.units

Complete to find the area of each trapezoid.

Question 5.

Base = units
Height = units
Area = bh
= ∙
= cm²
Answer:
In the given figure, base measures 21cm and the height measures 12cm.
Base = 21cm
Height = 12cm
Area = base × height
= 21 × 12
= 252 sq.cm

Question 6.

Base = in.
Height = in.
Area = bh
= ∙
= in²
Answer:
In the given figure, base measures 24 in and the height measures 14.5 in.
Base = 24 in
Height = 14.5 in
Area = base × height
= 24 × 14.5
= 348 sq.in

Complete to find the sum of the bases, height, and area of each trapezoid. Each square measures 1 unit by 1 unit.

Question 7.

Height = units
Sum of bases = +
= units
Area = \(\frac{1}{2}\)h(sum of bases)
= ∙ ∙
= units²
Answer:
The lengths will be equal to the number of squares each occupies because each square has length of 1 unit.
Given figure has two bases measuring 7 units and 3 units each. The height is measuring 3 units.
Height = 3 units
Sum of bases = (7+3)
= 10 units
Area = \(\frac{1}{2}\)h(sum of bases)
= \(\frac{1}{2}\)×3×10
= \(\frac{1}{2}\)×30
= 30÷2
= 15 sq.units

Question 8.

Height = units
Sum of bases = +
= units
Area = \(\frac{1}{2}\)h(sum of bases)
= ∙ ∙
= units²
Answer:
The lengths will be equal to the number of squares each occupies because each square has length of 1 unit.
Given figure has two vertical bases measuring 1 unit and 3 units each. The horizontal height is measuring 4 units.
Height = 4 units
Sum of bases = 1+3
= 4 units
Area = \(\frac{1}{2}\)h(sum of bases)
= \(\frac{1}{2}\)×4×4
= \(\frac{1}{2}\)×16
= 16÷2
= 8 sq.units

Question 9.

Height = units
Sum of bases = +
= units
Area = \(\frac{1}{2}\)h(sum of bases)
= ∙ ∙
= units²
Answer:
The lengths will be equal to the number of squares each occupies because each square has length of 1 unit.
Given figure has two vertical bases measuring 4 units and 7 units each. The horizontal height is measuring 7 units.
Height = 7 units
Sum of bases = 4+7
= 11 units
Area = \(\frac{1}{2}\)h(sum of bases)
= \(\frac{1}{2}\)×7×11
= \(\frac{1}{2}\)×77
= 77÷2
= 38.5 sq.units

Question 10.

Height = units
Sum of bases = +
= units
Area = \(\frac{1}{2}\)h(sum of bases)
= ∙ ∙
= units²
Answer:
The lengths will be equal to the number of squares each occupies because each square has length of 1 unit.
Given figure has two vertical bases measuring 5 units and 3 units each. The horizontal height is measuring 7 units.
Height = 7 units
Sum of bases = 5+3
= 8 units
Area = \(\frac{1}{2}\)h(sum of bases)
= \(\frac{1}{2}\)×7×8
= \(\frac{1}{2}\)×56
= 56÷2
= 28 sq.units

Complete to find the area of each trapezoid.

Question 11.

Height = ft
Sum of bases = +
= ft
Area = \(\frac{1}{2}\)h(sum of bases)
= ∙ ∙
= ft²
Answer:
Given figure has two vertical parallel sides measuring 25ft and 13ft each. The height of the trapezoid is 39ft.
Height = 39ft
Sum of bases = 25+13
= 38 ft
Area = \(\frac{1}{2}\)h(sum of bases)
= \(\frac{1}{2}\)×39×38
= 1482÷2
= 741 sq.ft

Question 12.

Height = cm
Sum of bases = +
= cm
Area = \(\frac{1}{2}\)h(sum of bases)
= ∙ ∙
= cm²
Answer:
Given figure has two horizontal parallel sides measuring 10.6cm and 21cm each. The height of the trapezoid is 13cm.
Height = 13cm
Sum of bases = 10.6+21
= 31.6 cm
Area = \(\frac{1}{2}\)h(sum of bases)
= \(\frac{1}{2}\)×31.6×13
= 410.8÷2
= 205.4 sq.cm

Complete to find the area of triangle ABD.

Question 13.
The area of trapezoid ABCD is 1,248 square centimeters.

Area of trapezoid ABCD = \(\frac{1}{2}\)h(b₁ + b₂)
= \(\frac{1}{2}\) ∙ h ∙ ( + )
= \(\frac{1}{2}\) ∙ h ∙
= \(\frac{1}{2}\) ∙ ∙ h
= ∙ h
Since area of trapezoid = cm²
• h = area of trapezoid
• h =
h =
= cm
Height of trapezoid =
= cm
Area of triangle ABD = \(\frac{1}{2}\)bh
= \(\frac{1}{2}\) • •
= cm²
The area of triangle ABD is square centimeters.

Answer:
Given that the area of trapezoid ABCD is 1,248 sq.cm
From the figure, we can observe two horizontal parallel sides measuring 22 cm and 38 cm.
Sum of the sides = 22+38 = 60cm
Area of trapezoid ABCD = \(\frac{1}{2}\)×h×(b₁ + b₂)
= \(\frac{1}{2}\) × h × 60
= \(\frac{60}{2}\) × h
= 30×h
Since area of trapezoid = 1,248 cm²
30×h = 1248
(30×h)÷30 = 1248÷30
h = 41.6 cm
Height of trapezoid = 41.6 cm
Area of triangle ABD = \(\frac{1}{2}\)×b×h
Let us assume the base as AD measuring 22cm and the height is 41.6cm
= \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×22×41.6
= 915.2÷2
= 457.6 sq.cm
The area of triangle ABD is 457.6 square centimeters.

Math in Focus Course 1B Practice 10.2 Answer Key

Copy each parallelogram. Label a base and a height for each. Use b and h.

Question 1.

Answer:

Explanation:
In parallelogram, any side can be the base of the paralleogram
But here, Let us assume the side BC as base of the paralleogram and label it as ‘b’.
The height of a parallelogram is the perpendicular distance between the base side and the parallel side opposite to it, the height is labelled as ‘h’.

Question 2.

Answer:

Explanation:
In parallelogram, any side can be the base of the paralleogram
But here, Let us assume the side XY as base of the paralleogram and label it as ‘b’.
The height of a parallelogram is the perpendicular distance between the base side and the parallel side opposite to it, the height is labelled as ‘h’.

Question 3.

Answer:

Explanation:
In parallelogram, any side can be the base of the paralleogram
But here, Let us assume the side QR as the base of the paralleogram and label it as ‘b’.
The height of a parallelogram is the perpendicular distance between the base side and the parallel side opposite to it, the height is labelled as ‘h’.

Question 4.

Answer:

Explanation:
In parallelogram, any side can be the base of the paralleogram
But here, Let us assume the side MN as the base of the paralleogram and label it as ‘b’.
The height of a parallelogram is the perpendicular distance between the base side and the parallel side opposite to it, the height is labelled as ‘h’.

Find the area of each parallelogram.

Question 5.

Answer:
From the given figure the base length is 28 in and the height is 15 in.
Base = 28 in
Height = 15 in
Area of the parallelogram = base × height
= 28×15
= 420 sq.in

Question 6.

Answer:
From the given figure the base length is 18 in and the height is 10 in.
Base = 18 in
Height = 10 in
Area of the parallelogram = base × height
= 18×10
= 180 sq.in

Copy each trapezoid. Label the height and bases. Use h₁, b₁ and b₂.

Question 7.

Answer:

Explanation:
The parallel sides are called the bases of the trapezium.
Let b1​ and b2​ be the lengths of these bases.
The distance between the bases is called the height of the trapezium. Let $h$ be this height.

Question 8.

Answer:

Explanation:
The parallel sides are called the bases of the trapezium.
Let b1​ and b2​ be the lengths of these bases.
The distance between the bases is called the height of the trapezium. Let $h$ be this height.

Question 9.

Answer:

Explanation:
The parallel sides are called the bases of the trapezium.
Let b1​ and b2​ be the lengths of these bases.
The distance between the bases is called the height of the trapezium. Let $h$ be this height.

Question 10.

Answer:

Explanation:
The parallel sides are called the bases of the trapezium.
Let b1​ and b2​ be the lengths of these bases.
The distance between the bases is called the height of the trapezium. Let $h$ be this height.

Find the area of each trapezoid.

Question 11.

Answer:
Given figure has two horizontal bases measuring 15 cm and 7 cm each. The height is measuring 10 cm.
Height = 10 cm
Sum of bases = 15+ 7
= 22 cm
Area = \(\frac{1}{2}\)h(sum of bases)
= \(\frac{1}{2}\) × 10 × 22
= \(\frac{1}{2}\) × 220
= 220÷2
= 110 sq.cm

Question 12.

Answer:
Given figure has two horizontal bases measuring 12 ft and 24 ft each. The height is measuring 17 ft.
Height = 17 ft
Sum of bases = 12+ 24
= 36 ft
Area = \(\frac{1}{2}\)h(sum of bases)
= \(\frac{1}{2}\) × 17 × 36
= \(\frac{1}{2}\) × 612
= 612÷2
= 306 sq.ft

The area of each parallelogram is 64 square inches. Find the height. Round your answer to the nearest tenth of an inch.

Question 13.

Answer:
Given that the area of parallelogram is 64 sq.in and the base is of 7 in length.
Area of parallelogram = base × height
64 = 7×height
64÷7 = (7×height)÷7
9.1 = height
Rounding off 9.1 to the nearest tenth of an inch
Height of the given parallelogram is 9 in.

Question 14.

Answer:
Given that the area of parallelogram is 64 sq.in and the base is of 6 in length.
Area of parallelogram = base × height
64 = 6×height
64÷6 = (6×height)÷6
10.6 = height
Rounding off 10.6 to the nearest tenth of an inch
Height of the given parallelogram is 11 in.

The area of each trapezoid is 42 square centimeters. Find the height. Round your answer to the nearest tenth of a centimeter.

Question 15.

Answer:
Given that the area of trapezoid is 42 sq.cm. The two parallel sides are measuring 5 cm and 7.5 cm.
Sum of the sides = 5+7.5 = 12.5 cm
Area of trapezoid = \(\frac{1}{2}\)×h×(b₁ + b₂)
42 = \(\frac{1}{2}\) × h × 12.5
42 = 6.25 × h
42 ÷6.25 = (6.25 × h)÷6.25
6.7 = h
Rounding off 6.7 to the nearest tenth of an cm.
Height of the given trapezoid is 7 cm.

Question 16.

Answer:
Given that the area of trapezoid is 42 sq.cm. The two parallel sides are measuring 5 cm and 7.5 cm.
Sum of the sides = 5.7+10 = 15.7 cm
Area of trapezoid = \(\frac{1}{2}\)×h×(b₁ + b₂)
42 = \(\frac{1}{2}\) × h × 15.7
42 = 7.8 × h
42 ÷7.8 = (7.8 × h)÷7.8
5.3 = h
Rounding off 5.3 to the nearest tenth of an cm.
Height of the given trapezoid is 5 cm.

Solve.

Question 17.
The area of trapezoid ABCO is 503.25 square centimeters. Find the length of \(\overline{B C}\).

Answer:
Given that the area of trapezoid ABCD is 503.25 sq.cm
From the given figure, length of the height measures 18.3 cm and length of a side is 34 cm
Area of trapezoid = \(\frac{1}{2}\)×h×(b₁ + b₂)
503.25 = \(\frac{1}{2}\) × 18.3 × (34 + b₂)
503.25 = 9.1× (34 + b₂)
503.25÷9.1 = 9.1× (34 + b₂)÷9.1
55.3 = 34 + b₂
55.3 – 34 = 34 – 34 + b₂
21.3 = b₂
The length of second side BC will be 21.3 cm

Question 18.
The area of trapezoid EFGH is 273 square centimeters. Find the area of triangle EGH.

Answer:
Given that the area of trapezoid EFGH is 273 sq.cm. The length of two parallel bases will be 14 cm and 25 cm.
Area of trapezoid = \(\frac{1}{2}\)×h×(b₁ + b₂)
Sum of bases = 14+25 = 39cm
273 = \(\frac{1}{2}\)×h×39
273 = 19.5×h
273 ÷ 19.5 = (19.5×h)÷ 19.5
14 = h
The height of the trapezoid is 14 cm.
Area of triangle EGH = \(\frac{1}{2}\)×b×h
The length of base is 14cm
= \(\frac{1}{2}\)×14×14
= 196÷2
= 98 sq.cm

Solve. Use graph paper.

Question 19.
The coordinates of the vertices of a parallelogram are P (0, 5), Q (-3, 0), R (2, 0), and S (5, 5). Find the area of parallelogram PQRS.
Answer:

From the figure, we can get the height of the parallelogram which is 5 units.
Let us consider QR as base which measures 5 units.
Area of parallelogram = base×height
= 5×5
= 25 sq.units

Question 20.
Three out of the four coordinates of the vertices of parallelogram WXYZ are W(0, 1), X(-4, -4), and Y (-1, -4). Find the coordinates of Z. Then find the area of the parallelogram.
Answer:

Given that three vertices of parallelogram WXYZ are W(0, 1), X(-4, -4), and Y (-1, -4).
Let XY be the base of parallelogram,since XY is measuring 3 units. The distance between W and Z will also be 3 units.
Thus, the point Z will be (3,1)
The height of the parallelogram is 5 units.
Area of parallelogram = base × height
= 3×5
= 15 sq.units

Question 21.
The coordinates of the vertices of trapezoid EFGH are E (-3, 3), F (-3, 0), G (1, -4), and H (1, 4). Find the area of the trapezoid.
Answer:

The two parallel sides EF and HG are measuring 3 units and 8 units respectively. The height is of 4 units length.
Sum of the sides = 3+8 = 11 units
Area of trapezoid = \(\frac{1}{2}\)×h×(b₁ + b₂)
= \(\frac{1}{2}\)×4×11
= 44 ÷2
= 22 sq.units

Question 22.
Three out of the four coordinates of the vertices of trapezoid ABCD are A (0, 1), B (-4, -4), and C(-1, -4). \(\overline{\mathrm{AD}}\) is parallel to \(\overline{\mathrm{BC}}\). AD is 6 units. The point D lies to the right of point A. Find the coordinates of point D. Then find the area of the trapezoid.
Answer:

Given that the three vertices of trapezoid ABCD are A (0, 1), B (-4, -4), and C(-1, -4). AD is parallel to BC and length of AD is 6 units.
Therefore, the point D will be 6 units away from A which can be pointed at (-6,1)
The two parallel sides DA and BC are measuring 6 units and 3 units respectively. The height is of 5 units length.
Sum of the sides = 6+3 = 9 units
Area of trapezoid = \(\frac{1}{2}\)×h×(b₁ + b₂)
= \(\frac{1}{2}\)×5×9
= 45÷2
= 22.5 sq.units

Solve.

Question 23.
Parallelogram PQRT is made up of isosceles triangle PST and trapezoid PQRS. Find the area of parallelogram PQRT.

Answer:
Given that, PST is an isosceles triangle. Therefore the length of PT and TS will be 15cm.
Area of triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×15×15
= \(\frac{1}{2}\)×225
= 225÷2
= 112.5 sq.cm
Length of TR = Length of RS + Length of ST
Length of TR = 5+15 = 20 cm
Since PQRT is parallelogram, TR will be parallel to PQ. Thus, PQ = 20cm.
Area of trapezoid = \(\frac{1}{2}\)×h×(b₁ + b₂)
Sum of bases = SR + PQ
= 5 +15
= 20 cm
The height is of length 14.5 cm.
Area of trapezoid = \(\frac{1}{2}\)×14.5×20
= \(\frac{1}{2}\)×290
= 290÷2
= 145 sq.cm
Area of parallelogram PQRT = Area of isosceles triangle + Area of trapezoid
= 112.5 + 145
= 257.5 sq.cm

Go through the Math in Focus Grade K Workbook Answer Key Chapter 20 Money to finish your assignments.

Math in Focus Kindergarten Chapter 20 Answer Key Money

Lesson 1 Coin Values

Match.

Answer:

Explanation:
1 quarter dollar is equal to 25 cents
1 dime is equal to 10 cents
1 nickel is equal to 5 cents
1 penny is equal to 1 cent

Lesson 2 Counting Coins

How many pennies do you need? Color.

Question 1.

Answer:

Explanation:
The cost of the toy airplane is 6 cents
so, colored 6 cents

Question 2.

Answer:

Explanation:
The cost of the toys are 6 + 2 = 8
so, colored the 8 cents

Question 3.

Answer:

Explanation:
The cost of the toys are 3 + 5 = 8 cents
so, colored 8 cents

Question 4.

Answer:

Explanation:
The cost of the toys are 2 + 4 + 3 = 9 cents
so, colored 9 cents

How much is needed? Circle the purse.

Question 1.

Answer:

Explanation:
The cost of the toy house is 9 cents
5 + 4 = 9 cents
so, colored the first purse

Question 2.

Answer:

Explanation:
The cost of the toys are 4 cents and 2 cents
4 + 2 = 6 and also
5 + 1 = 6
so circled 6 cents purse

Go through the Math in Focus Grade 6 Workbook Answer Key Chapter 5 Lesson 5.1 Rates and Unit Rates to finish your assignments.

Math in Focus Grade 6 Course 1 A Chapter 5 Lesson 5.1 Answer Key Rates and Unit Rates

Math in Focus Grade 6 Chapter 5 Lesson 5.1 Guided Practice Answer Key

State whether each statement is expressed as a unit rate.

Question 1.
A monkey plucked 4 coconuts per minute.
Answer:
yes, it can be expressed in unit ratio.
Explanation:
Statement is expressed as a unit rate.
per minute is unit

Question 2.
Mary paid $2 for a bottle of orange juice.
Answer:
Yes, it can be expressed in unit ratio.
Explanation:
$2 for a bottle of orange juice.
$ is unit per bottle juice.

Question 3.
A basketball team scored 294 points in 6 games.
Answer:
yes, it can be expressed in unit ratio.
Explanation:
points are the units for games

Question 4.
Douglas reads 3 books in a week.
Answer:
Yes, it can be expressed in unit ratio.
Explanation:
Week is unit for reading a book.

Solve.

Question 5.
A photocopy machine can print 405 copies in 9 minutes. What is the rate at which the machine prints the copies?

Answer:
45 copies per min
Explanation:
Question 6.
A sports center buys 1.25 acres of land for a new swimming complex. What is the cost per acre if the sports center pays $100,000 for the land?
Cost per acre = Total cost ÷ Total number of acres
= $100,000 ÷
= $\(\frac{100,000}{?}\)
= $\(\frac{?}{?}\)
= $
The unit cost of the piece of land is $ per acre.
Answer:
$ 80,000 per acre.
Explanation:
Cost per acre = Total cost ÷ Total number of acres
= $100,000 ÷ 1.25
= $\(\frac{100,000}{1.25}\)
= $\(\frac{1,000,000}{125}\)
= $ 80,000
The unit cost of the piece of land is $ 80000 per acre.

Question 7.
A few years ago, when the fuel tank in Sally’s car was completely empty, she paid $63 to fill the tank with 22.5 gallons of gasoline. What was the cost per gallon?
Cost per gallon = Cost of gasoline ÷ Volume of gasoline filled
=$ ÷
= $
The unit cost of the gasoline was $ per gallon.
Answer:
$2.8
Explanation:
Cost per gallon = Cost of gasoline ÷ Volume of gasoline filled
=$ 63 ÷ 22.5
= $ 2.8
The unit cost of the gasoline was $ 2.8 per gallon.

Question 8.
The table shows the costs of some food items Billy bought from a supermarket.

Answer:

Which type of food costs the most per pound?
Cost of potatoes per pound = Cost of potatoes ÷ Weight of potatoes
= $ ÷
= $
The unit cost of the potatoes is $ per pound.
Cost of carrots per pound = Cost of carrots ÷ Weight of carrots
= $ ÷
= $
The unit cost of the carrots is $ per pound.
Cost of onions per pound = Cost of onions ÷ Weight of onions
= $ ÷
= $
The unit cost of the onions is $ per pound.
Comparing the unit costs of the food items, the unit cost of the is the greatest.
Unit cost of < Unit cost of < Unit cost of
$ < $ < $
So, the cost the most per pound.
Answer:
$1.25 cost the most per pound.
Explanation:
Cost of potatoes per pound = Cost of potatoes ÷ Weight of potatoes
= $ 4 ÷ 5
= $ 0.8
The unit cost of the potatoes is $ 0.8 per pound.
Cost of carrots per pound = Cost of carrots ÷ Weight of carrots
= $3 ÷ 5
= $0.6
The unit cost of the carrots is $ 0.6 per pound.
Cost of onions per pound = Cost of onions ÷ Weight of onions
= $2.5 ÷ 2
= $ 1.25
The unit cost of the onions is $ 1.25 per pound.
Comparing the unit costs of the food items, the unit cost of the 1.25 is the greatest.
Unit cost of  Carrot < Unit cost of Potatoes < Unit cost of Onions
$0.6 < $0.8 < $1.25
So, the $1.25 cost the most per pound.

Question 9.
A truck traveled a distance of 280 kilometers in 4 hours. Find the speed of the truck.
Method 1

The speed of the truck is 70kilometers per hour.

Method 2
Speed of truck = \(\frac{\text { Distance }}{\text { Time }}\)
= \(\frac{?}{?}\)
= km/h
The speed of the truck is kilometers per hour.
Answer:
70 kilometers per hour.
Explanation:

Speed of truck = \(\frac{\text { Distance }}{\text { Time }}\)
= \(\frac{280}{4}\)
= 70 km/h
The speed of the truck is 70 kilometers per hour.

Question 10.
A car travels \(\frac{1}{4}\) kilometer in \(\frac{1}{2}\) minute. Find the speed of the car in kilometers per minute.
Answer:
\(\frac{1}{2}\) kilometers per minute.
Explanation:
Speed of car = \(\frac{\text { Distance }}{\text { Time }}\)
Speed of car = \(\frac{1}{4}\) ÷ \(\frac{1}{2}\)
= \(\frac{1}{4}\) X \(\frac{2}{1}\)
= \(\frac{1}{2}\) kilometers per minute.

Math in Focus Course 1A Practice 5.1 Answer Key

Solve. Show your work.

Question 1.
A machine can print 300 T-shirts in 10 minutes. How many T-shirts can the machine print in 1 minute?
Answer:
30 T-shirts
Explanation:
A machine can print 300 T-shirts in 10 minutes.
Number of T-shirts can the machine print in 1 minute
= \(\frac{300}{10}\)
= 30 T-shirts can the machine print in 1 minute

Question 2.
Alisa types 900 words in 20 minutes. What is her typing speed in words per minute?
Answer:
45 words per minute
Explanation:
Alisa types 900 words in 20 minutes.
Her typing speed in words per minute is,
= \(\frac{900}{20}\)
= 45 words per minute

Question 3.
A 2-liter bottle is filled completely with water from a faucet in 10 seconds. How much water is filled into the bottle each second?
Answer:
200 ml bottle each second
Explanation:
A 2-liter bottle is filled completely with water from a faucet in 10 seconds.
water is filled into the bottle each second
= \(\frac{2000}{10}\)
= 200 ml bottle each second

Question 4.
Bill is paid $200 for 5 days of work. How much is he paid per day?
Answer:
$40 per day
Explanation;
Bill is paid $200 for 5 days of work.
Total amount he paid per day
= \(\frac{200}{5}\)
= $40 per day

Question 5.
Janice swims 450 meters in 5 minutes. Find her swimming speed in meters per minute.
Answer:
90 meters per minute.
Explanation:
Janice swims 450 meters in 5 minutes.
Her swimming speed in meters per minute.
= \(\frac{450}{5}\)
= 90 meters per minute.

Question 6.
A garden snail moves \(\frac{1}{6}\) foot in \(\frac{1}{3}\) hour. Find the speed of the snail in feet per hour.
Answer:
\(\frac{1}{2}\) feet per hour
Explanation:
A garden snail moves \(\frac{1}{6}\) foot in \(\frac{1}{3}\) hour.
The speed of the snail in feet per hour.
\(\frac{1}{6}\) ÷ \(\frac{1}{3}\)
= \(\frac{1}{6}\) X \(\frac{3}{1}\)
= \(\frac{1}{2}\) feet per hour

Question 7.
Math Journal A plumber pays $3.60 for 60 centimeters of pipe. Explain how the plumber can use the unit cost of the pipe to find the cost of buying 100 meters of the same kind of pipe. Show the calculations the plumber needs to make.
Answer:
The cost of 100 meters pipe = $600
Explanation:
A plumber pays $3.60 for 60 centimeters of pipe.
1 cm pipe = 3.60 ÷ 60
= 0.6 cm
1m = 100 cm
S0, 0.6 cm = 600

Question 8.
Rovan can make 48 tarts per hour. Copy and complete the table.

Answer:

a) At this rate, how many tarts can Rovan make in 6 hours?
Answer:
288 Tarts
Explanation:
From the above given information,
Number of tarts Rovan can make in 6 hours
6 x 48 = 288 tarts

b) At this rate, how long will she take to make 120 tarts?
Answer:
2.5 hours
Explanation:
From the above given information,
Number of hours she take to make tarts = 120 ÷ 48
= 2.5 hours

Question 9.
A sprinkler system is designed to water \(\frac{5}{8}\) acres of land in \(\frac{1}{4}\) hour. How many acres of land can it water in 1 hour?
Answer:
\(\frac{5}{2}\) per hour
Explanation:
A sprinkler system is designed to water \(\frac{5}{8}\) acres of land in \(\frac{1}{4}\) hour. Total acres of land can it water in 1 hour?
= \(\frac{5}{8}\) ÷ \(\frac{1}{4}\)
= \(\frac{5}{8}\) X \(\frac{4}{1}\)
= \(\frac{20}{8}\)
= \(\frac{5}{2}\)

Question 10.
The table shows the costs of three types of meat John bought at a supermarket. Copy and complete the table.

Which type of meat costs the least per pound?
Answer:
Chicken.
Explanation:

From the above table Chicken costs the least amount per pound.

Question 11.
Math Journal The table shows the data about distances and times for three sprinters.

Who is the fastest sprinter? Justify your answer.
Answer:
Smith is the fastest sprinter
Explanation:

Question 12.
A supermarket sells the three brands of rice shown in the table below.

Raimondo wants to buy 30 kilograms of rice.
a) Which brand of rice should he buy to get the best deal, assuming that all three brands are of the same quality?
Answer:
Brand A
Explanation:

b) How much will he save if he buys the cheapest brand of rice as compared to the most expensive one?
Answer:
$306
Explanation:
Cheapest brand price = $72
Most expensive brand price = $378
378 – 72 = 306