# Math in Focus Grade 7 Chapter 7 Review Test Answer Key

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

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

Concepts and Skills

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

Question 1.

Question 2.

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

Question 3.
m∠XYZ = 37°

Question 4.
m∠PQR = 72°

Question 5.
m∠KLM = 128°

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

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

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

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

Question 8.
AB = 6.5 cm

Question 9.
CD = 4.5 cm

Question 10.

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

Question 11.

Question 12.

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

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

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

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

Use the given information to construct each quadrilateral.

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

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

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

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

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

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

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

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

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

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

Problem Solving

Solve.

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

We construct △ABC:

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

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

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

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

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

Draw $$\overline{C D}$$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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