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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
[Add.]

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


Correct answer : (1)
 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
[Add.]

So, the perimeter of the figure is 14.7 in.


Correct answer : (2)
 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.


Correct answer : (3)
 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


Correct answer : (1)

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