Square or cube a length scale only for the matching dimension.
Extended · Syllabus E4.4
Notes and key methods
For geometrically similar figures with positive length scale k, corresponding areas scale by \(\displaystyle k^{2}\). Similar solids have volumes in ratio \(\displaystyle k^{3}. A\) square root recovers a length ratio from an area ratio; a cube root recovers it from a volume ratio.
Surface area uses the area scale, while capacity uses the volume scale. Keep the order of the figures consistent in every ratio. These formulas require geometric similarity; a single pair of proportional lengths alone is insufficient.
Check the conditions
Use the stated units and precision.
Show the method as well as the final answer.
Three worked examples
Example 1 · Similarity with areas and volumes
Two geometrically similar solids S and T have corresponding lengths 3 cm and 11 cm. S has surface area \(\displaystyle 10 \mathrm{cm}^{2}\) and volume \(\displaystyle 13 \mathrm{cm}^{3}\).
Find the surface-area multiplier from S to T.
Find the volume multiplier from S to T.
Find the surface area of T in \(\displaystyle \mathrm{cm}^{2}\).
Find the volume of T in \(\displaystyle \mathrm{cm}^{3}\).
Show the reasoning
(a)
Use dimension two for an area ratio.
Square the length scale \(\displaystyle \frac{11}{3}\) to obtain \(\displaystyle \frac{121}{9}\).
Answer: \(\displaystyle \frac{121}{9}\).
(b)
Use dimension three for a volume ratio.
Cube the length scale \(\displaystyle \frac{11}{3}\) to obtain \(\displaystyle \frac{1331}{27}\).
Answer: \(\displaystyle \frac{1331}{27}\).
(c)
Apply the area scale in the stated direction.
Multiply the original area by \(\displaystyle \frac{121}{9}\) to obtain \(\displaystyle \frac{1210}{9}\) \(\displaystyle \mathrm{cm}^{2}\).
Answer: \(\displaystyle \frac{1210}{9}\).
(d)
Apply the volume scale in the stated direction.
Multiply the original volume by \(\displaystyle \frac{1331}{27}\) to obtain \(\displaystyle \frac{17303}{27}\) \(\displaystyle \mathrm{cm}^{3}\).
Answer: \(\displaystyle \frac{17303}{27}\).
Example 2 · Similarity with areas and volumes
Two geometrically similar solids S and T have corresponding lengths 4 cm and 7 cm. S has surface area \(\displaystyle 11 \mathrm{cm}^{2}\) and volume \(\displaystyle 14 \mathrm{cm}^{3}\).
Find the surface-area multiplier from S to T.
Find the volume multiplier from S to T.
Find the surface area of T in \(\displaystyle \mathrm{cm}^{2}\).
Find the volume of T in \(\displaystyle \mathrm{cm}^{3}\).
Show the reasoning
(a)
Use dimension two for an area ratio.
Square the length scale \(\displaystyle \frac{7}{4}\) to obtain \(\displaystyle \frac{49}{16}\).
Answer: \(\displaystyle \frac{49}{16}\).
(b)
Use dimension three for a volume ratio.
Cube the length scale \(\displaystyle \frac{7}{4}\) to obtain \(\displaystyle \frac{343}{64}\).
Answer: \(\displaystyle \frac{343}{64}\).
(c)
Apply the area scale in the stated direction.
Multiply the original area by \(\displaystyle \frac{49}{16}\) to obtain \(\displaystyle \frac{539}{16}\) \(\displaystyle \mathrm{cm}^{2}\).
Answer: \(\displaystyle \frac{539}{16}\).
(d)
Apply the volume scale in the stated direction.
Multiply the original volume by \(\displaystyle \frac{343}{64}\) to obtain \(\displaystyle \frac{2401}{32}\) \(\displaystyle \mathrm{cm}^{3}\).
Answer: \(\displaystyle \frac{2401}{32}\).
Example 3 · Similarity with areas and volumes
Two geometrically similar solids S and T have corresponding lengths 5 cm and 8 cm. S has surface area \(\displaystyle 12 \mathrm{cm}^{2}\) and volume \(\displaystyle 15 \mathrm{cm}^{3}\).
Find the surface-area multiplier from S to T.
Find the volume multiplier from S to T.
Find the surface area of T in \(\displaystyle \mathrm{cm}^{2}\).
Find the volume of T in \(\displaystyle \mathrm{cm}^{3}\).
Show the reasoning
(a)
Use dimension two for an area ratio.
Square the length scale \(\displaystyle \frac{8}{5}\) to obtain \(\displaystyle \frac{64}{25}\).
Answer: \(\displaystyle \frac{64}{25}\).
(b)
Use dimension three for a volume ratio.
Cube the length scale \(\displaystyle \frac{8}{5}\) to obtain \(\displaystyle \frac{512}{125}\).
Answer: \(\displaystyle \frac{512}{125}\).
(c)
Apply the area scale in the stated direction.
Multiply the original area by \(\displaystyle \frac{64}{25}\) to obtain \(\displaystyle \frac{768}{25}\) \(\displaystyle \mathrm{cm}^{2}\).
Answer: \(\displaystyle \frac{768}{25}\).
(d)
Apply the volume scale in the stated direction.
Multiply the original volume by \(\displaystyle \frac{512}{125}\) to obtain \(\displaystyle \frac{1536}{25}\) \(\displaystyle \mathrm{cm}^{3}\).
Answer: \(\displaystyle \frac{1536}{25}\).
Common mistakes
Keep intermediate values exact until the final step.