SURFACE AREA OF SIMILAR SOLIDS

If two figures are similar then:

• the figures are equiangular, and
• the corresponding sides are in proportion.

Surface area of similar figures :

If two solids are similar, then the ratio of their surface areas is equal to the square of the ratio of their corresponding linear measures. 

Each pair of figures is similar. Use the information given to find the scale factor of the figure on the left to the figure on the right. 

Problem 1 :

Solution :

Surface area of large shape : Surface area of small shape

= 36 : 25

k² = 36/25

k = √36/25

k = 6/5

So, the scale factor is 6 : 5.

Problem 2 :

Solution :

Surface area of small shape : Surface area of large shape

= 7π : 175π

k² = 7π/175π

k² = 1/25

k = √1/25

k = 1/5

So, the scale factor is 1 : 5.

Problem 3 :

The solids are similar. Find the surface area S of the shaded shape.

Solution :

Let l = side length of small cube

L = Side length of large cube

s = surface area of small cube

S = surface area of large cube

s : S = (l : L)²

198/S = (6/8)²

198/S = 36/64

Doing cross multiplication,

S = (198 ⋅ 64)/36

S = 352 m²

Problem 4 :

The pyramids are similar. What is the surface area of Pyramid A?

Solution :

Let h = height of small pyramid

H = height of large pyramid

s = surface area of small pyramid

S = surface area of large pyramid

s : S = (h : H)²

s : 600 = (6/10)²

s : 600 = 36/100

s = (36/100) ⋅ 600

s = 216 ft²

Problem 5 :

Find the surface area of smaller cuboid.

Solution :

Let w = width of cuboid

W = width of cuboid

s = surface area of small cuboid

S = surface area of large cuboid

s : S = (w : W)²

s : 608 = (5/8)²

s : 608 = 25/64

s = (25/64) ⋅ 608

s = 237.5 m²

Problem 6 :

Find the surface area of large cylinder.

Solution :

Let R = radius of large cylinder = 5/2

r = radius of small cylinder = 2

s = surface area of small cylinder = 110 cm²

S = surface area of large cylinder

s : S = (r : R)²

110 : S = (2/(5/2))²

110 : S = (4/5)²

110 : S = (4/5)²

110/S = 16/25

S = 110(25/16)

S = 171.875 cm²

Problem 7 :

The ratio of the corresponding linear measures of two similar cans of fruit is 4 to 7. The smaller can has a surface area of 220 square centimeters. Find the surface area of the larger can.

Solution :

Ratio between similar cans = 4 : 7

Surface area of smaller can = 220 square cm

Let S be the surface area of the larger can.

4 : 7 = 220 : S

4/7 = 220/S

4S = 220(7)

S = 220(7) / 4

= 385

So, the required surface area of large can is 385 square cm.

Problem 8 :

A and B are mathematically similar.

The height of A : the height of B = 3 : 5

(a) Find the surface area of A : the surface area of B

(b) Find the volume of A : the volume of B

Solution :

The height of A : the height of B = 3 : 5

Ratio between surface area of cylinders is = 3² : 5²

= 9 : 25

Ratio between volume of cylinders is = 3³ : 5³

= 27 : 125

Problem 9 :

A, B and C are similar.

• The volume of A is 729 cm³ and the volume of B is 64 cm³.
• The surface area of B is 25 cm² and the surface area of C is 121 cm².
• Find the ratio length of A : length of B : length of C

Solution :

Let a, b and c be the sides of A, B and C.

(a : b)³ = 729 : 64

a³ : b³ = 729 : 64

a³ / b³ = 729 / 64

a / b = ∛(729/64)

a / b = ∛(7 ⋅ 7 ⋅ 7)/(4 ⋅ 4 ⋅ 4)

a / b = 7 / 4

a : b = 7 : 4 ---(1)

Surface area of B : surface area of C = 25 : 121

(b : c)² = 25 : 121

b : c = √(25/121)

b : c = 5 / 11

a : b : c = 7 : 4  ---(2)

(1) x 7 ==> 49 : 28

(2) x 4 ==> 28 : 16

a : b : c = 49 : 28 : 16

Problem 10 :

Shapes A, B and C are similar.

• The height of shape A is 8 cm.
• The height of shape C is 4cm.
• The ratio of the surface area of shape B to the surface area of shape C is 25:9
• Work out the ratio of the volume of shape A to shape B.

Solution :

Let side length of shape A, shape B and shape C be a, b and c.

b² : c² = 25 : 9

b : c = √25 : √9

b : c = 5 : 3

Ratio between sides a and b

a : c = 8 : 4

a : c = 2 : 1--------(1)

c : b = 3 : 5 --------(2)

(1) x 3 ==> 6 : 3

Ratio between volume of shape A : shape B = 6³ : 3³

= 216 : 27