AREA OF SIMILAR SOLIDS

If the sides of a rectangle are multiplied by k, a similar rectangle is obtained.

The new area = ka × kb

= k² ab

= k² × the old area

If an object or figure is enlarged by a scale factor of k, then

 the area of the image = k² × the area of the object

Consider the following similar shapes. Find

i) the scale factor

ii) the length or area marked by the unknown. 

Problem 1 :

Solution :

i) scale factor  :

10 = k² × 40

k² = 10/40

k² = 1/4

k = 1/2

ii) the length :

Length of smaller figure = Scale factor x length of small figure

x = k × 10

x = 1/2 × 10

x = 5 cm

Problem 2 :

Solution :

i) scale factor  :

9 = k × 6

k = 9/6

k = 3/2

Here, we find area of the large figure.

ii) Area :

Area of large shape = k² × Area of smaller shape

x² = k² × 8

By applying the value of k, we get

x² = (3/2)² × 8

= 18 cm²

Area of the large shape is 18 cm².

Problem 3 :

Solution :

i) scale factor :

5 = k × 2

k = 5/2

ii)  Area :

Area of large shape = k² × Area of smaller shape

25 = (5/2)² × x

25 = 25x/4

x = 25(4/25)

x = 4 cm²

Problem 4 :

Solution :

i) scale factor :

6 = k² × 20

k² = 6/20

k² = 3/10

k = √3/10

k = √0.3

k ≈ 0.548

ii) the length :

Length of smaller shape = k (length of larger shape)

= k × 8

= 0.548 × 8

≈ 4.38 cm

Problem 5 :

Solution :

i) scale factor :

31.8 = k² × 55.4

k² = 31.8/55.4

k² = 0.574

k = √0.574

k ≈ 0.758

ii) the length or area :

x = k × 4.2

x = 0.758 × 4.2

x ≈ 3.18 m

Problem 6 :

Solution :

i) scale factor :

7.25 = k × 5

k = 7.25/5

k ≈ 1.45

ii)  Area :

Area of the large shape = k² x area of small shape

70 = k² (x)

Applying the value of k, we get

70 = (1.45)² x

x = 70/(1.45)²

x = 70/2.1025

x = 33.29

Approximately 33.3 cm²

Problem 7 :

Two similar triangles have a pair of corresponding sides of length 12 meters and 8 meters. The larger triangle has a perimeter of 48 meters and an area of 180 square meters. Find the perimeter and area of the smaller triangle.

Solution :

Ratio of similar side of the triangles = 12 : 8

Perimeter of the larger triangle = 48 meters

Let P be the perimeter of the required smaller triangle.

12 : 8 = 48 : P

12/8 = 48/P

12P = 48(8)

P = 48(8) / 12

P = 4(8)

= 32

So, the perimeter of the smaller triangle is 32 meter.

Area of the larger triangle = 180 square meter

Let A be the area of the smaller triangle.

12 : 8 = 180 : A

12/8 = 180/A

12A = 180(8)

A = 180(8)/12

A = 120 square meters

So, the area of the smaller rectangle is 120 square meters.

Problem 8 :

You are making a scale model of a rectangular park for a school project. Your model has a length of 2 feet and a width of 1.4 feet. The actual park is 800 yards long. What are the perimeter and area of the actual park?

Solution :

Let x be the width of the rectangular park.

length : width = new length : new width

2 : 1.4 = 800 : x

2/1.4 = 800/x

2x = 800(1.4)

x = 800(1.4)/2

= 400(1.4)

x = 560

Perimeter of actual park = 2(length + width)

= 2(800 + 560)

= 2(1360)

= 2720 feet

Area of rectangular park = 800(560)

= 448000 square feet

Problem 9 :

Figure A has an area of 48 square feet and one of the side lengths is 6 feet. Figure B has an area of 75 square feet. When the figures are similar, find the missing corresponding side length.

Solution :

Let x be the missing side of the smaller figure.

(Side length of figure A)² : (side length of figure B)² = area of figure A : area of figure B

6² : x² = 48 : 75

36/x² = 48/75

48x² = 75(36)

x² = 75(36)/48

x² = 56.25

x = √56.25

x = 7.5 feet

Problem 10 :

Figure A has an area of 18 square feet. Figure B has an area of 98 square feet and one of the side lengths is 14 feet. When the figures are similar, find the missing corresponding side length.

Solution :

Let x be the missing side of figure A.

x² : 14² = 18 : 98

x² / 196 = 18 / 98

98x² = 18(196)

x² = 196(18)/96

x² = 36

x = 6

So, the missing side is 6 feet.

Problem 11 :

In the following exercises, JKLM ∼ EFGH.

a)  Find the scale factor of JKLM to EFGH.

b) Find the scale factor of EFGH to JKLM.

c) Find the values of x, y, and z.

Solution :

a) Scale factor of JKLM to EFGH

= 8 : 20

= 8/20

= 2/5

= 2 : 5

a) Scale factor of EFGH to JKLM

= 20 : 8

= 20/8

= 5 / 2

= 5 : 2

c) 

z = 65