GEOMETRIC SOLIDS
Similar Solids
1
SIMILAR SOLIDS
Definition: Two solids of the same type with equal
ratios of corresponding linear measures (such as
heights or radii) are called similar solids.
2
SIMILAR SOLIDS
Similar solids
NOT similar solids
3
SIMILAR SOLIDS & CORRESPONDING
LINEAR MEASURES
To compare the ratios of corresponding side or other
linear lengths, write the ratios as fractions in simplest terms.
6
12
Length: 12 = 3
8 2
3
width: 3
2
4
8
2
height: 6 = 3
4 2
** Notice that all ratios for corresponding measures are
equal in similar solids. The reduced ratio is called the
“scale factor”.
4
EXAMPLE:
Are these solids similar?
Solution:
16
length :

12
w idth :
8

6
height :
4
12
3
6
4
12
8
3
12
9
9

4
16
3
All of the corresponding lengths have the same scale
factor, therefore the figures are similar.
5
Example:
Are these solids similar?
Solution:
height :

2
4
1
18
3
6

18
6
8
radius :
8
4
1
Corresponding ratios are not equal, so the figures are not
similar.
6
SIMILAR SOLIDS AND RATIOS OF
AREAS
• If two similar solids have a scale factor
of a : b, then corresponding areas have
a ratio of a2: b2.
• This applies to:
• Lateral Area
• Surface Area
• Base Area
SIMILAR SOLIDS AND RATIOS OF
AREAS
Ratio of sides = 2: 1
8
3.5
7
4
4
5
2
10
Surface Area =
base + lateral =
40 + 108 = 148
Surface Area =
base + lateral =
10 + 27 = 37
Ratio of surface areas = 148:37 = 4:1 = 22: 12
8
SIMILAR SOLIDS AND RATIOS OF
VOLUMES
• If two similar solids have a scale
factor of a : b, then their volumes
have a ratio of a3 : b3.
SIMILAR SOLIDS AND RATIOS OF
VOLUMES
Ratio of heights = 3:2
9
6
15
V = p r2 h = p (92) (15) =
1215
10
V= p r2 h =
360
p (62)(10) =
Ratio of volumes= 1215 : 360 = 27 : 8 = 33 : 23
10
SIMILAR SQUARES EXAMPLE
11
CHANGE IN DIMENSION EXAMPLE
• The dimensions of a water touch tank at the local
aquarium are doubled. What is the volume of the
new tank?
original volume æ original dimension ö
=ç
new volume
è new dimension ÷ø
3
3
2000 æ 1 ö
=ç ÷
è 2ø
V
2000 1
=
V
8
V = 16, 000 ft.3
12
FINDING THE SCALE FACTOR OF
SIMILAR SOLIDS
• To find the scale factor
of the two cubes, find
the ratio of the two
volumes.
a 3 512
=
3
b 1728 Write ratio of volumes.
a 8
=
Use a calculator to take the cube root.
b 12
8 2
Simplify.
=
12 3
So, the two cubes have a scale factor of 2:3.
COMPARING SIMILAR SOLIDS
• Swimming pools. Two
swimming pools are similar
with a scale factor of 3:4.
The amount of chlorine
mixture to be added is
proportional to the volume
of water in the pool. If two
cups of chlorine mixture are
needed for the smaller pool,
how much of the chlorine
mixture is needed for the
larger pool?
SOLUTION:
• Using the scale factor, the ratio of the volume of the
smaller pool to the volume of the larger pool is as
follows:
a
3
27 1
= 3=
=
3
b
4
64 2.4
3
3
• The ratio of the volumes of the mixture is 1:2.4. The
amount of the chlorine mixture for the larger pool can
be found by multiplying the amount of the chlorine
mixture for the smaller pool by 2.4 as follows: 2(2.4) =
4.8 c.
• So the larger pool needs 4.8 cups of the chlorine
mixture.