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

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

**Concepts and Skills**

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

Question 1.

Answer:

Question 2.

Answer:

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

Question 3.

m∠XYZ = 37°

Answer:

Question 4.

m∠PQR = 72°

Answer:

Question 5.

m∠KLM = 128°

Answer:

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

Question 6.

Construct a 65° angle whose vertex is point X.

Answer:

Question 7.

Construct a 66° angle whose vertex is point Y.

Answer:

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

Question 8.

AB = 6.5 cm

Answer:

Question 9.

CD = 4.5 cm

Answer:

Question 10.

AD = 10.8 cm

Answer:

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

Question 11.

Answer:

Question 12.

Answer:

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

Question 13.

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

Answer:

Question 14.

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

Answer:

Question 15.

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

Answer:

**Use the given information to construct each quadrilateral.**

Question 16.

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

Answer:

Question 17.

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

Answer:

Question 18.

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

Answer:

**Solve. Show your work.**

Question 19.

A rectangular garden is 15 meters long and 9 meters wide. Use a scale of 1 centimeter to 3 meters, make a scale drawing of the garden.

Answer:

length of the garden is = 45 m

width of the garden is = 27 m

Explanation:

Given the length of the rectangle = 15m

Breadth of the rectangle = 9 m

given scale as 1 centimeter = 3 m

The new length of the garden is = 15 × 3 = 45 m

New breadth of the garden is = 9 × 3 = 27 m

Question 20.

The scale of the floor plan of a room is 1 inch: 6.5 feet. On the floor plan, the room is 8 inches long and 6 inches wide. What are the actual dimensions of the room?

Answer:

1 inch : 6.5 feet means 1 inch on the map represents 6.5 feet on the ground.

Let’s note:

x cm = the actuaL Length

y cm = the actual width.

We have:

1 in. : 6.5 ft = 8 in. : x ft = 6 in. : y ft

Write ratios in fraction form:

\(\frac{1 \mathrm{in} .}{6.5 \mathrm{ft}}=\frac{8 \mathrm{in} .}{x \mathrm{ft}}=\frac{6 \mathrm{in} .}{y \mathrm{ft}}\)

Write without units:

\(\frac{1}{6.5}\) = \(\frac{8}{x}\) = \(\frac{6}{y}\)

Write cross products:

x = 8 · 6.5

y = 6 · 6.5

Simplify

x = 52 ft

y = 39 ft

The actual dimensions of the room are 52 ft and 39 ft.

Question 21.

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

Answer:

1 : 25 means 1 cm on the map represents 25 cm on the ground.

Let x inches be the Length of the car on the map.

We have:

4.8 cm = 4.8 · 100 cm = 480 cm

1 cm : 25 cm = x cm : 480 cm

Write ratios in fraction form:

\(\frac{1 \mathrm{~cm}}{25 \mathrm{~cm}}\) = \(\frac{x \mathrm{~cm}}{480 \mathrm{~cm}}\)

Write without units:

\(\frac{1}{25}\) = \(\frac{x}{480}\)

Write cross products:

25x = 480

Divide both sides by 25:

\(\frac{25x}{25}\) = \(\frac{480}{25}\)

Simplify:

x = 19.2

On the map, the car’s length is 19.2 cm.

Question 22.

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

Answer:

Length = 24 miles

Explanation:

Scale on a map is 1 inch: 120 miles

The highway is 5 inches.

The actual length of the highway in miles is 1 × \(\frac{120}{5}\)

length of the highway is 24 miles.

**Problem Solving**

**Solve.**

Question 23.

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

Answer:

We are given △WXY and △ABC:

WX = WY = 5

XY = 4

AB = AC = 10

BC = 8

We construct △WXY:

We construct △ABC:

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

Question 24.

James was asked to design a square decorative tile with a side length of 90 millimeters. Construct the square on which James will draw his design.

Answer:

We have to construct the square:

AB = 90 mm

Sketch the square:

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

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

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

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

Question 25.

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

Answer:

We are given the rhombus:

AC = 9

BD = 4

We draw the segment of length 20 and label it BD:

We bisect the segment \(\overline{A C}\).

Place the compass point at A. Then draw an arc on each side of \(\overline{A C}\) with a radius greater than half of the length of \(\overline{A C}\).

Using the same radius, set the compass point in C. Draw one arc on each side of \(\overline{A C}\).

Label the points where the arcs intersect as E and F.

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

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

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

Measure a side of the rhombus:

AB ≈ 4.9 cm

We are given the scale:

1 cm : 2 m

Let’s note by r the actual length of one side of the rhombus.

\(\frac{1}{2}\) = \(\frac{4.9}{x}\)

We determine x:

x = 2 49

x = 9.8 m

Calculate the perimeter of the rhombus:

4 · 9.8 = 39.2 m

Question 26.

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

Answer:

We are given the actual kite dimensions:

AC = 24

BD = 20

AN = 6

Determine the dimensions of the model kite:

\(\frac{1}{4}=\frac{A^{\prime} C^{\prime}}{A C}=\frac{B^{\prime} D^{\prime}}{B D}=\frac{A^{\prime} N^{\prime}}{A N}\)

\(\frac{1}{4}=\frac{A^{\prime} C^{\prime}}{24}=\frac{B^{\prime} D^{\prime}}{20}=\frac{A^{\prime} N^{\prime}}{6}\)

4A’C” = 24 ⇒ A’C” = 6

4B’D’ = 20 ⇒ B’D’ = 5

4A’N’ = 6 ⇒ A’N’ = 1.5

We draw the segment of length 5 and label it B’D’:

We bisect the segment \(\overline{B^{\prime} D^{\prime}}\).

Place the compass point at B’. Then draw an arc on each side of \(\overline{B^{\prime} D^{\prime}}\) with a radius greater than half of the length of \(\overline{B^{\prime} D^{\prime}}\).

Using the same radius, set the compass point in D’. Draw one arc on each side of \(\overline{B^{\prime} D^{\prime}}\).

Label the points where the arcs intersect as E and F

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

and \(\overline{B^{\prime} D^{\prime}}\) by N’.

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

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

Use a ruler to draw A’B’, B’C”, C”D’, D’A’.

Measure a side of the rhombus:

Measure ∠ABC:

m∠ABC ≈ 93°

Measure the length5 A’B’ and B’C” in the model kite:

A’B’ ≈ 2.9 in.

B’C” ≈ 5.1 in.

Determine the Lengths AB and BC in the actual kite:

\(\frac{1}{4}=\frac{A^{\prime} B^{\prime}}{A B}=\frac{B^{\prime} C^{\prime}}{B C}\)

\(\frac{1}{4}=\frac{2.9}{A B}=\frac{5.1}{B C}\)

AB = 4 · 2.9 = 11.6 in.

BC = 4 · 5.1 = 20.4 in.

Question 27.

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

Answer:

The area of a rectangular region is 19 square inches.

Explanation:

The scale of a map is 1 inch to 5 feet.

The area of the actual region is 95 square feet.

The area of a rectangular region is = 95 × \(\frac{1}{5}\)

Area of rectangle is = 19 sqaure inches.

Question 28.

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

Answer:

10 square feet.

Explanation:

The floor plan of a building has a scale of \(\frac{1}{4}\) inch to 1 foot.

A room has an area of 40 square inches.

The actual room area in square feet is 1× \(\frac{40}{4}\)

= 10 sqaure feet.

Question 29.

The scale of a map is 1 : 2,400. If a rectangular piece of property measures 2 inches by 3 inches on the map, what is the actual area of this piece of property to the nearest tenth of an acre? (1 acre = 43,560 ft^{2})

Answer:

1: 2,400 means 1 inch on the map represents 2,400 inches on the ground.

Map length : Actual length = 1 in. : 2,400 in.

Map area: Actual area = 1 in.^{2} : 2. 400^{2} 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 ft^{2} = 43,460 · 12^{2} in.^{2}

= 6,258,240 in.^{2}

\(\frac{34,560,000}{6,258,240}\) ≈ 5.52 acres