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Area and Volume of Similar Figures and Solids

Secondary 3 practice

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Area and Volume of Similar Figures and Solids practice questions

Secondary 3 topics from the O-Level Mathematics 4052 syllabus. Every question below has a full worked solution.

When a student gets a question wrong in Math Amigo, it does not show the answer straight away. It reads the working, identifies the step that went wrong, and asks a question that points at the correction. The worked solution is there once a genuine attempt has been made.

Practise Area and Volume of Similar Figures and Solids free

What this topic covers

Worked examples

Example 1

Two similar triangles have areas 1818 cm² and 3232 cm². The corresponding height of the smaller figure is 33 cm. Find the corresponding height of the larger figure.
Show the worked solution
  • A2A1=(l2l1)2area ratio = square of length ratio\frac{A_2}{A_1} = \left(\frac{l_2}{l_1}\right)^2 \quad \scriptsize\textit{area ratio = square of length ratio}
  • 3218=(l23)2\frac{32}{18} = \left(\frac{l_2}{3}\right)^2
  • l23=3218take positive square root\frac{l_2}{3} = \sqrt{\frac{32}{18}} \quad \scriptsize\textit{take positive square root}
  • l23=43\frac{l_2}{3} = \frac{4}{3}
  • l2=3×43l_2 = 3 \times \frac{4}{3}
  • =4= 4 cm

Answer: 44

Example 2

Two similar pyramids have volumes 88 cm³ and 216216 cm³ respectively. A length on the smaller pyramid is 66 cm. Find the corresponding length on the larger pyramid.
Show the worked solution
  • For similar solids, the ratio of volumes equals the cube of the ratio of corresponding lengths.\text{For similar solids, the ratio of volumes equals the cube of the ratio of corresponding lengths.}
  • V2V1=(l2l1)3volume ratio of similar solids\frac{V_2}{V_1} = \left(\frac{l_2}{l_1}\right)^3 \quad \scriptsize\textit{volume ratio of similar solids}
  • l2l1=V2V13take cube root of both sides\frac{l_2}{l_1} = \sqrt[3]{\frac{V_2}{V_1}} \quad \scriptsize\textit{take cube root of both sides}
  • l26=21683\frac{l_2}{6} = \sqrt[3]{\frac{216}{8}}
  • l26=273\frac{l_2}{6} = \sqrt[3]{27}
  • l26=3\frac{l_2}{6} = 3
  • l2=6×3l_2 = 6 \times 3
  • l2=18l_2 = 18 cm

Answer: 1818

Example 3

Two similar prisms have volumes in the ratio 1:641:64. Find the ratio of their total surface areas.
Show the worked solution
  • Volume ratio=1:64=164given\text{Volume ratio} = 1:64 = \frac{1}{64} \quad \scriptsize\textit{given}
  • Length ratio=1643length ratio is cube root of volume ratio\text{Length ratio} = \sqrt[3]{\frac{1}{64}} \quad \scriptsize\textit{length ratio is cube root of volume ratio}
  • =14= \frac{1}{4}
  • Surface area ratio=(14)2area ratio is square of length ratio\text{Surface area ratio} = \left(\frac{1}{4}\right)^2 \quad \scriptsize\textit{area ratio is square of length ratio}
  • =116= \frac{1}{16}
  • =1:16= 1:16

Answer: 1:161:16

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Which syllabus does this follow?

O-Level Mathematics (4052), for Singapore secondary students. Math Amigo covers Secondary 1 to 4, both O-Level Mathematics and Additional Mathematics.

Will it just show my child the answer?

No. On a wrong answer it looks at the working, points to the step that went wrong, and asks a question that helps the student find the correction. The full worked solution is available after a genuine attempt.

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