﻿ Area and Perimeter Word Problems - Page 4 | Problems & Solutions

# Area and Perimeter Word Problems - Page 4

Area and Perimeter Word Problems
• Page 4
31.
Find the perimeter of the shaded region in the parallelogram ABCD.

 a. 12 cm b. 18 cm c. 11 cm d. 15 cm

#### Solution:

From the figure, ΔBEC is the shaded area in the parallelogram ABCD.

From the figure, BC = 5 cm, BE = 3 cm, AB = 7 cm, and DE = 3 cm.

EC = DC - DE = AB - DE = 7 - 3 = 4 cm.
[Lengths of the parallel sides in a parallelogram are equal, AB = DC.]

Perimeter of the ΔBEC = BC + CE + EB
[Formula.]

= 5 + 4 + 3
(Substitute the values)

= 12

So, the perimeter of the shaded region of the figure is 12 cm.

32.
Find the perimeter of the figure.

 a. 12.5 in. b. 14.7 in. c. 17.7 in. d. 15.7 in.

#### Solution:

From the figure, BA = 5 in., AC = 5.2 in., CD = 2.5 in., BC = 3 in., and DB = 2 in.

Perimeter of a figure = sum of all the sides of the figure
[Formula.]

Perimeter of the figure = AC + CD + DB + BA

= 5.2 + 2.5 + 2 + 5
[Substitute the values.]

= 14 .7

So, the perimeter of the figure is 14.7 in.

33.
What is the area of the triangle?

 a. 11.5 square units b. 1.5 square units c. 4.5 square units d. 5 square units

#### Solution:

From the figure, the base length of the triangle = 3 units.

Height of the triangle = 3 units.

Area of a triangle with base b and height h = 1 / 2 × b × h

= 12 × 3 × 3
[Substitute the values.]

= 4.5
[Simplify.]

So, the area of the triangle is 4.5 square units.

34.
What is area of the shaded region?

 a. 105 in.2 b. 120 in.2 c. 95 in.2 d. None of the above

#### Solution:

Area of the shaded region is equal to the area of the triangle with measures of 15 in. height and 14 in. as base.

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

= 12 x 14 x 15
[Substitute the values.]

= 105 in.2
[Simplify.]

Area of the shaded region in the figure = 105 in.2