﻿ Area and Perimeter Word Problem Worksheets | Problems & Solutions

# Area and Perimeter Word Problem Worksheets

Area and Perimeter Word Problem Worksheets
• Page 1
1.
The figure shows two regular pentagons which are similar. If the area of the larger pentagon is 45 cm2 then find the area of the shaded portion of the smaller pentagon. [Given $a$ = 8 and $b$ = 12.]

 a. 20 cm2 b. 27 cm2 c. 12 cm2 d. 8 cm2

#### Solution:

Ratio of the coresponding sides = a : b

ab = 812 = 2 / 3
[Replace a =8 and b = 12.]

If the ratio of the two similar figures is a : b then the ratio of their areas is a2 : b2

a²b² = 2²3² = 4 / 9
[Sides of two similar figures.]

Area of the larger pentagon = 45 cm2
[Given.]

Let A be the area of smaller pentagon.

49 = A45
[From step 3 and step 4.]

A = 45 × 49 = 20 cm2
[Simplify.]

Area of shaded portion = 25 × 20 = 8 cm2
[Pentagon is divided into 5 congruent triangles.]

2.
The area of the smaller regular hexagon is 120 cm2. Find the area of the shaded portion of larger regular hexagon if the two hexagons are similar. [Given $a$ = 6 and $b$ = 12.]

 a. 120 cm2 b. 240 cm2 c. 250 cm2 d. 480 cm2

#### Solution:

Ratio of the corresponding sides = a : b

ab = 612 = 12

If the similarity ratio of the two similar figures is a : b then the ratio of their areas is a2 : b2.

a²b² = 1²2² = 1 / 4
[Two hexagons are similar.]

Area of the smaller hexagon = 120 cm2
[Given.]

Let A be the area of larger hexagon.

14 = 120A
[Write in proportion.]

A = 120 × 4 = 480
[Simplify.]

Area of the shaded region = 3 / 6× 480 = 240 cm2

3.
Trapezoid ABCD is similar to trapezoid MNOP. Find the ratio of their areas. [Given $a$ = 8 units and $b$ = 14 units.]

 a. 7 : 4 b. 16 : 49 c. 4 : 7 d. 49 : 16

#### Solution:

Trapezoid ABCD ~ Trapezoid MNOP
[Given.]

If the similarity ratio of two similar figures is a : b then the ratio of their areas is a2 : b2.

ab = 814 = 4 / 7
[Substitute.]

Area of trapezoid ABCD / Area of trapezoid MNOP = a²b²
[Trapezoid ABCD ~ Trapezoid MNOP.]

a²b² = 4²7² = 1649

Therefore ratio of the areas = 16 : 49

4.
Two figures are similar if ______
 a. they have the same size but different shape b. they have the different size and different shape c. they have some portion in common d. they have the same shape but not necessarily the same size

#### Solution:

Two figures are similar if they have the same shape. The figures need not be of same size.

5.
The area of trapezoid ABCD is 9 cm2 and trapezoid PQRS is 16 cm2. If the two trapezoids are similar, then find the value of $x$. [Given AB = 4 cm and CD = 8 cm.]

 a. 6 cm b. 8 cm c. 16 cm d. 2.67 cm

#### Solution:

Trapezoid ABCD ~ Trapezoid PQRS.
[Given.]

Area of trapezoid ABCD / Area of trapezoid PQRS = 916
[Given.]

If a and b are the sides of two similar figures then the ratio of their areas is equal to the square of their corresonding sides.

a²b² = 916

ab = 34
[Simplify.]

Consider the sides AB and PQ.

Length of the side AB is 4 cm and let the length PQ be m.
[Given.]

4m = 34
[From step 5.]

m = 4 × 43 = 163
[Simplify.]

Consider the sides CD and RS.

Length of the side CD is 8 cm and let the length RS be n.
[Given.]

8n = 34

n = 8 × 43 = 323
[Simplify.]

x = 1 / 2(PQ + RS)
[Formula.]

= 1 / 2(16 / 3+ 32 / 3)
[From steps 9 and 13.]

= 12 ×483 = 8

Hence, the value of x is 8 cm.

6.
Two regular pentagons ABCDE and PQRST are similar. The length of a side of ABCDE is 5 cm. Find the length of a side of PQRST if its area is 16 times the area of ABCDE.
 a. 20 cm b. 5 cm c. 80 cm d. 400 cm

#### Solution:

Let each side of the pentagon PQRST be x cm.

The two pentagons are similar.
[Given.]

If the ratio of corresponding sides of two similar figures is a : b, then the ratio of their areas is a2 : b2

Area of pentagon ABCDE / Area of pentagon PQRST = 5²x².
[Pentagons are similar.]

116 = 25x²
[Area of pentagon PQRST = 16 × Area of pentagon ABCDE.]

x2 = 25 × 16 = x2 = 400
[Cross - multiply.]

x = 20
[Simplify.]

The length of each side of a pentagon is 20 cm.

7.
The circle R is similar to the circle S. The ratio of their radii is 2 : 3. If the area of the semicircle S is 9$\pi$ sq.units,then find the area of the semicircle with center R.

 a. 3$\pi$ sq.units b. 9$\pi$ sq.units c. 4$\pi$ sq.units d. 2$\pi$ sq.units

#### Solution:

The ratio of the radii = 2 : 3
[Given.]

Area of the semicircle (S) = 9π
[Given.]

Let the area of the semicircle (R) be x.

If the similarity ratio of two similar figures is a : b, then the ratio of their areas is a2 : b2.

area of the semicircle (R)area of semicircle (S) = a²b²
[a and b are the radii of two circles.]

area of the semicircle (R)area of semicircle (S) = 2²3² = 4 / 9

4 / 9= x9π
[From step 2.]

x = 4×9π / 9 = 4π

Area of the semicircle with center R = 4π sq.units

8.
Franklin paid a carpenter $4 to polish his wooden table which is 12 ft long. His neighbour Samson has a similar wooden table of length 18 ft. How much Samson has to pay to have his wooden table polished by the same carpenter?  a.$6 b. $4 c.$14 d. $9 #### Solution: Length of Franklin's wooden table = 12 ft Length of Samson's wooden table = 18 ft ab = 12 / 18= 2 / 3 [Similarity ratio.] If the similarity ratio of the two similar figures is a : b then the ratio of their areas is a2 : b2. a²b² = 2²3² = 4 / 9 Cost of polishing is proportional to the area of the table. 4 / 9 = 4x [Write in proportion.] x = 9×4 / 4 = 9 [Simplify.] Hence, Samson has to pay$9 to have his wooden table polished by the same carpenter.

9.
If the areas of two similar decagons are 81 ft2 and 225 ft2, then find the ratio of their perimeters.
 a. 3 : 5 b. 25 : 9 c. 9 : 25 d. 5 : 3

#### Solution:

Areas of two similar decagons are 81 ft2 and 225 ft2.
[Given.]

If the similarity ratio of two similar figures is a : b then the ratio of their areas is a2 : b2.

b²a² = 81 / 225
[From steps 1 and 2.]

b²a² = 9²15²
[The ratio of the areas b²a², is the square of the similarity ratio.]

ba = 9 / 15 = 3 / 5
[Simplify.]

3 : 5 is the similarity ratio.
[From step 2.]

The ratio of the perimeters is the same as the similarity ratio.
Hence, the ratio of their perimeters = 3 : 5

10.
A light $a$ m above the ground causes a boy $b$ m tall to cast a shadow $s$ meters long measured along the ground. Express $s$ interms of $d$, where $d$ is the distance between the boy and the base of the light in meters. [Given $a$ = 4, $b$ = 0.8.]
 a. $s$ = 4 $d$ b. $s$ = 0.25 $d$ c. $s$ = 4.8 $d$ d. $s$ = 0.8 $d$

#### Solution:

Draw the figure.

s is the shadow of the boy and 'd' is the boy's distance in meters from the base of the light.

We can observe two similar triangles from the figure.

4d+s = 0.8s
[Write the proportion.]

4s = 0.8(d + s)
[Cross multiply.]

4s = 0.8d + 0.8s
[Simplify.]

3.2s = 0.8d

s = 0.25d
[Solve for s.]