FIND SURFACE AREAS AND VOLUMES 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. 

Volume of similar figures :

If two solids are similar, then the ratio of their volumes is equal to the cube of the ratio of their corresponding linear measures.

The scale factor between two similar figures is given. The surface area and volume of the smaller figure are given. Find the surface area and volume of the larger figure.

Problem 1 :

Scale factor = 1 : 2

SA = 90 yd²

V = 216 yd³

Solution :

Scale factor (k) = 1 : 2 = 1/2

i)  Surface area of smaller figure / surface area of larger

= (1/2)²

Surface area of smaller figure = 90 yd²

90/Surface area of larger figure = (1/4)

Surface area of larger figure = 90(4)

= 360 yd²

So, surface area of larger figure is 360 yd².

ii)  Volume of smaller figure / Volume of larger = (1/2)³

Volume of smaller figure = 216 yd³

216/volume of larger figure = (1/8)

Volume of larger figure = 216(8)

= 1728 yd³

So, volume of larger figure is 1728 yd³.

Problem 2 :

Scale factor = 4 : 9

SA = 256 km²

V = 1536 km³

Solution :

i)  Surface area of smaller figure / surface area of larger

= (4/9)²

Surface area of smaller figure = 256 km²

256/Surface area of larger figure = (4/9)²

Surface area of larger figure = 256(81/16)

= 1296 yd²

So, surface area of larger figure is 1296 yd².

ii)  Volume of smaller figure / Volume of larger = (4/9)³

Volume of smaller figure = 1536 yd³

1536/volume of larger figure = (64/729)

Volume of larger figure = 1536(729/64)

= 17496 km³

So, volume of larger figure is 17496 km³.

Problem 3 :

The dimensions of the touch tank at an aquarium are doubled. What is the volume of the new touch tank?

A) 150 ft³       B) 4000 ft³    C) 8000 ft³    D) 16,000 ft³ 

Solution :

The dimensions are doubled, so the ratio of the dimensions in the original tank to the dimensions in the new tank is 1 : 2.

(Original volume/new volume) = (original dimension/new dimension)³

(2000/new volume)= (1/2)³

(2000/new volume)= (1/8)

New volume = 2000(8)

= 16000 ft³

The solids are similar. Find the volume of the red solid. Round your answer to the nearest tenth.

Problem 4 :

Solution :

Scale factor (k) = 5/12

Volume of red solid / volume of blue solid = k³

Volume of red solid / 288 = (5/12)³

Volume of red solid / 288 = 125/1728

Volume of red solid = (125/1728)⋅288

= 20.8 cm³

Problem 5 :

Solution :

Scale factor (k) = 3/4

Volume of blue solid / volume of red solid = k³

9 / Volume of red solid = (3/4)³

9 / Volume of red solid = (27/64)

Volume of red solid = 9(64/27)

= 21.3 in³

Problem 6 :

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

i) Ratio between surface area = 3² : 5²

= 9 : 25

ii) Ratio between volumes = 3³ : 5³

= 27 : 125

Problem 7 :

Two solid toys, C and D, are similar.

The volume of toy C is 40 cm³

The surface area of C : surface area of D = 2 : 9

Work out the volume of toy D.

Solution :

Volume of toy C = 40 cm³

Ratio between the surface area of C : surface area of D = 2 : 9

a² : b² = 2 : 9

a : b = √(2 : 9)

a : b = 1.414 : 3

a³ : b³ = 1.414³ : 3³

1.414³ : 3³ = Volume of C : volume of D

1.414³ : 3³ = 40 : volume of D

2.827 : 27 = 40 : volume of D

2.827 / 27 = 40 / volume of D

volume of D = 40 (27)/2.827

= 382.03

= 382 cm³

Problem 8 :

Washing powder is sold in two different sizes, a large box A and a smaller box B. Cuboid boxes A and B are similar.

Surface area of A : Surface area of B = 81 : 4

How many smaller boxes, B, can be completely filled using the contents of a full box A?

Ratio between sides  :

Let a and b be the corresponding sides.

a² : b² =  81 : 4

a : b =  √(81 : 4)

a : b =  9 : 2

Ratio between volume = 9³ : 2³

=729 / 8

= 91.125

Approximately 91 boxes

Problem 9 :

The volumes of two similar shapes are in the ratio 1000 : 27

The surface area of the larger shape is 250 cm² Work out the surface area of the smaller shape

Solution :

Let a and b be the sides of similar solids.

a³ : b³ = 1000 : 27

a³ : b³ = 10³ : 3³

a : b = 10 : 3

Ratio of squares of side lengths = Ratio of surface areas 

a² : b² = Surface area of shape A : Surface area of shape B

10² : 3² = 250 : Surface area of shape B

100/9 = 250/surface area of shape B

Surface area of shape B = 250(9)/100

= 22.5 cm²

Problem 10 :

Shown are similar solids G and H.

The volume of G : the volume of H = 27 : 1000

(a) Find the height of G : the height of H

(b) Find the surface area of G : the surface area

Solution :

Let a and b be the sides lengths of the prism G and H.

The volume of G : the volume of H = 27 : 1000

a³ : b³ = 27 : 1000

a³ : b³ = 3³ : 10³

a : b = 3 : 10

(a) Find the height of G : the height of H = 3 : 10

(b) Find the surface area of G : the surface area = 3² : 10²

= 9 : 100