﻿ Area and Perimeter Word Problems - Page 3 | Problems & Solutions

# Area and Perimeter Word Problems - Page 3

Area and Perimeter Word Problems
• Page 3
21.
What is the height of the triangle, if the base of the triangle is twice the height, and its area is 225 ft2?
 a. 16 ft b. 19 ft c. 15 ft d. None of the above

#### Solution:

Let h is the height of the triangle.

Given, area of the triangle = 225 ft2
The base = twice the height = 2 × h.

Area of a triangle = 1 / 2 × base × height
[Formula.]

225 = 1 / 2 × 2 × h × h
[Substitute the values.]

225 = h2
[Simplify.]

225 = h2
[Taking square root on both sides.]

15 = h

The height of the triangle = 15 ft.

22.
What is the height of the triangle, if the triangle and the parallelogram have the same base and the same area?

 a. $h$ = 2H b. H = $h$ c. 2$h$ = H d. None of the above

#### Solution:

From the figure, the triangle and the parallelogram have same base.
Height of the parallelogram = H
Height of the triangle = h.

Area of the triangle = 1 / 2× base × height

Area of the parallelogram = base × height

Given, area of the triangle = area of the parallelogram

1 / 2 × base × height = base × height
[Substitute the area formulae.]

1 / 2 × height = height
[Bases are equal.]

1 / 2 × h = H
[Substitute the height values.]

h = 2H

23.
The area of a parallelogram is half of the area of the triangle. What is the height of the triangle, if they have the same base length of 6 cm and the area of the parallelogram is 39 cm2?
 a. 26 cm b. 21 cm c. 31 cm d. None of the above

#### Solution:

The area of the parallelogram = 39 cm2

The area of the parallelogram = 1 / 2 x area of the triangle
[Given.]

39 = 1 / 2 x area of the triangle
[Substitute the area of parallelogram.]

78 = area of the triangle
[Multiply each side by 2.]

Area of the triangle = 78 cm2

Base length of the parallelogram = base length of the triangle = 6cm
[Given.]

Area of the triangle = 1 / 2 x base x height
[Formula.]

78 = 1 / 2 x 6 x height
[Substitute the values.]

78 = 3 x height
[Multiply 3 with 1 / 2.]

78 / 3 = 3xheight / 3
[Divide each side by 3.]

Height of the triangle = 26 cm
[Simplify.]

24.
What is the base length of the parallelogram, if the height is half of the base and its area is 128 in.2?
 a. 21 in b. 18 in c. 16 in d. None of the above

#### Solution:

Let b be the base length of the parallelogram.

Height of the parallelogram = half of the base = b2

Area of the parallelogram = base × height
[Formula.]

128 = b × b2
[Substitute the values.]

128 × 2 = b22 × 2
[Multiply each side by 2.]

256 = b2
[Simplify.]

256 = b2
[Taking square root on both sides.]

16 = b

The base length of the parallelogram = 16 in.

25.
What is the area of ADB in the figure?

 a. 48 in.2 b. 12 in.2 c. 24 in.2 d. None of the above

#### Solution:

Base of the triangle ADB = 6 in. and the height of the triangle ADB = 4 in.

Area of the triangle ADB = 1 / 2 × base × height
[Formula.]

1 / 2 × 6 × 4
[Substitute the values.]

= 12
[Simplify.]

The area of ADB = 12 in.2

26.
Split the trapezoid into a triangle and a parallelogram and find the area of the trapezoid.

 a. 20 m.2 b. 30 m.2 c. 25 m.2 d. 15 m.2

#### Solution:

Draw a line parallel to CB from the point D.

Let the line touch AB on the point E.

Now the trapezoid is split into a parallelogram DEBC and a Δ ADE.

Base length of the parallelogram = BE = DC = 6 m., height = 4 m.

Area of a parallelogram DEBC = base × height = 6 × 4 = 24 m.2

Base length of the Δ ADE = AE = AB - DC = 9 - 6 = 3 m.

Height of the Δ ADE = 4 m.

Area of the Δ ADE = 1 / 2 × base × length = 1 / 2 × 3 × 4 = 6 m.2

Area of the trapezoid ABCD = Area of a parallelogram DEBC + Area of the Δ ADE.

Area of the trapezoid ABCD = 24 + 6 = 30 m.2

27.
Find the perimeter of the figure.

 a. 24 in. b. 48 in. c. 12 in. d. 10 in.

#### Solution:

Perimeter of a figure = sum of all the sides of the figure.

From the figure AB = 8 in., BC = 10 in., AC = 6 in.

Perimeter of ΔABC = AB + BC + CA

= 8 + 10 + 6
[Substitute the values.]

= 24

So, the perimeter of the figure is 24 in.

28.
Find the perimeter of ΔABC in the figure.

 a. 12 in. b. 18 in. c. 21 in. d. 3 in.

#### Solution:

Perimeter of a triangle is the sum of all the sides of the triangle.

From the figure, AB = 3 in., BC = 10 in. and AC = 8 in.

Perimeter of ΔABC = AB + BC + CA

= 3 + 10 + 8
[Substitute the values.]

= 21

The perimeter of the ΔABC is 21 in.

29.
Find the perimeter of the parallelogram shown.

 a. 10 in. b. 20 in. c. 5 in. d. 16 in.

#### Solution:

Perimeter of a parallelogram = sum of all the sides of the parallelogram

From the figure, AB = 5 in. and AD = 3 in.

As the lengths of the parallel sides are equal, AB = CD = 5 in., AD = BC = 3 in.

Perimeter of the parallelogram = AB + BC + CD + DA

= 5 + 3 + 5 + 3
[Substitute the lengths.]

= 16

So, the perimeter of the parallelogram given in the figure is 16 in.

30.
Find the length of NO in the given triangle, if the perimeter is 35 ft.

 a. 7 ft b. 11 ft c. 6 ft d. 8 ft

#### Solution:

From the figure, MN = 15 ft, OM = 13 ft

Perimeter of a triangle = sum of all sides
[Formula.]

Perimeter of the triangle MNO = MN + NO + OM
[Formula.]

35 = 15 + NO + 13
[Substitute the values.]

35 = 28 + NO