Prepared by Muhammad Tayyab, Subject Specialist Mathematics, Govt Christian High School Daska
📌 Based on National Curriculum / PECTA 2025 Syllabus
📖 What's Inside: This exercise covers properties of similar triangles, scale factors, and proportionality. Perfect for Punjab Boards exam preparation.
📚 Related Resources – Chapter 9: Similar Figures
1
Find the ratio of the areas of similar figures if the ratio of their corresponding lengths are:
(i) \(l_1 : l_2 = 1 : 3\)
\[\begin{aligned}
\frac{A_1}{A_2} &= \left(\frac{l_1}{l_2}\right)^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \\[4pt]
\boldsymbol{A_1 : A_2} &= \boldsymbol{1 : 9}
\end{aligned}\]
(ii) \(l_1 : l_2 = 3 : 4\)
\[\begin{aligned}
\frac{A_1}{A_2} &= \left(\frac{3}{4}\right)^2 = \frac{9}{16} \\[4pt]
\boldsymbol{A_1 : A_2} &= \boldsymbol{9 : 16}
\end{aligned}\]
(iii) \(l_1 : l_2 = 2 : 7\)
\[\begin{aligned}
\frac{A_1}{A_2} &= \left(\frac{2}{7}\right)^2 = \frac{4}{49} \\[4pt]
\boldsymbol{A_1 : A_2} &= \boldsymbol{4 : 49}
\end{aligned}\]
(iv) \(l_1 : l_2 = 8 : 9\)
\[\begin{aligned}
\frac{A_1}{A_2} &= \left(\frac{8}{9}\right)^2 = \frac{64}{81} \\[4pt]
\boldsymbol{A_1 : A_2} &= \boldsymbol{64 : 81}
\end{aligned}\]
(v) \(l_1 : l_2 = 6 : 5\)
\[\begin{aligned}
\frac{A_1}{A_2} &= \left(\frac{6}{5}\right)^2 = \frac{36}{25} \\[4pt]
\boldsymbol{A_1 : A_2} &= \boldsymbol{36 : 25}
\end{aligned}\]
2
Find the unknowns in the following figures (similar figures).
(a) \(A_1 = 240\ cm^2,\; l_1 = 10\ cm,\; l_2 = 6\ cm,\; A_2 = ?\)
\[\begin{aligned}
\frac{A_1}{A_2} &= \left(\frac{l_1}{l_2}\right)^2 \\[6pt]
\frac{240}{A_2} &= \left(\frac{10}{6}\right)^2 = \frac{100}{36} \\[6pt]
240 \times 36 &= 100 \times A_2 \\
8640 &= 100\,A_2 \\
A_2 &= \frac{8640}{100} \\[4pt]
\boldsymbol{A_2} &= \boldsymbol{86.4\ cm^2}
\end{aligned}\]
(b) \(A_1 = 60\ cm^2,\; l_1 = 15\ cm,\; l_2 = 20\ cm,\; A_2 = ?\)
\[\begin{aligned}
\frac{A_1}{A_2} &= \left(\frac{l_1}{l_2}\right)^2 \\[6pt]
\frac{60}{A_2} &= \left(\frac{15}{20}\right)^2 = \frac{225}{400} \\[6pt]
60 \times 400 &= 225 \times A_2 \\
24000 &= 225\,A_2 \\
A_2 &= \frac{24000}{225} \\[4pt]
\boldsymbol{A_2} &= \boldsymbol{106.67\ cm^2}
\end{aligned}\]
(c) \(A_2 = 18\ cm^2,\; l_1 = 3.6\ cm,\; l_2 = 5.76\ cm,\; A_1 = ?\)
\[\begin{aligned}
\frac{A_1}{A_2} &= \left(\frac{l_1}{l_2}\right)^2 \\[6pt]
\frac{A_1}{18} &= \left(\frac{3.6}{5.76}\right)^2 = \frac{12.96}{33.1776} \\[6pt]
A_1 &= \frac{12.96}{33.1776} \times 18 \\[4pt]
\boldsymbol{A_1} &= \boldsymbol{7.03\ cm^2}
\end{aligned}\]
(d) \(A_2 = 96\ cm^2,\; l_1 = 15\ cm,\; l_2 = 12\ cm,\; A_1 = ?\)
\[\begin{aligned}
\frac{A_1}{A_2} &= \left(\frac{l_1}{l_2}\right)^2 \\[6pt]
\frac{A_1}{96} &= \left(\frac{15}{12}\right)^2 = \frac{225}{144} \\[6pt]
A_1 &= \frac{225}{144} \times 96 \\[4pt]
\boldsymbol{A_1} &= \boldsymbol{150\ cm^2}
\end{aligned}\]
(e) \(A_1 = 3.57\ cm^2,\; A_2 = 63\ cm^2,\; l_1 = 3\ cm,\; l = ?\)
\[\begin{aligned}
\frac{A_1}{A_2} &= \left(\frac{l_1}{l_2}\right)^2 \\[6pt]
\frac{3.57}{63} &= \left(\frac{3}{l}\right)^2 = \frac{9}{l^2} \\[6pt]
3.57\,l^2 &= 9 \times 63 = 567 \\[4pt]
l^2 &= \frac{567}{3.57} = 158.82 \\[4pt]
l &= \sqrt{158.82} \\[4pt]
\boldsymbol{l} &= \boldsymbol{12.60\ cm}
\end{aligned}\]
3
Given that area of \(\Delta ABC = 36\ cm^2\), \(m\overline{AB} = 6\ cm\), \(m\overline{BD} = 4\ cm\). Find:
\(Area\ of\ \Delta ABC = A_1 = 36\ cm^2\)
\(m\overline{AB} = l_1 = 6\ cm\)
\(m\overline{AD} = m\overline{AB} + m\overline{BD} = l_2 = 6 + 4 = 10\ cm\)
(a) the area of \(\Delta ADE\)
Since \(\Delta ABC \sim \Delta ADE\). So
\[\begin{aligned} \frac{A_1}{A_2} &= \left(\frac{l_1}{l_2}\right)^2 \\[6pt] \frac{36}{A_2} &= \left(\frac{6}{10}\right)^2 = \frac{36}{100} \\[6pt] 36 \times 100 &= 36 \times A_2 \\ A_2 &= \frac{3600}{36} \\[4pt] \boldsymbol{A_2} &= \boldsymbol{100\ cm^2} \end{aligned}\](b) the area of trapezium \(BCED\)
\[\begin{aligned}
Area\ of\ trapezium &= Area\ of\ \Delta ADE - Area\ of\ \Delta ABC \\
&= 100\ cm^2 - 36\ cm^2 \\
&= \boldsymbol{64\ cm^2}
\end{aligned}\]
(a) Area of \(\Delta ADE\) = 100 cm² | (b) Area of trapezium = 64 cm²
4
Given that \(\Delta ABC \sim \Delta DEF\) with scale factor \(k = 3\). If the area of \(\Delta ABC\) is \(50\ cm^2\), find the area of \(\Delta DEF\).
\(Scale\ factor = k = 3\)
\(Area\ of\ \Delta ABC = A_1 = 50\ cm^2\)
\(Area\ of\ \Delta DEF = A_2 = ?\)
Since \(\Delta ABC \sim \Delta DEF\). So
\[\begin{aligned} \frac{A_1}{A_2} &= k^2 \\[6pt] \frac{50}{A_2} &= 3^2 = 9 \\[6pt] A_2 &= \frac{50}{9} \\[4pt] \boldsymbol{A_2} &= \boldsymbol{5\tfrac{5}{9}\ cm^2} \end{aligned}\]Area of \(\Delta DEF\) = \(5\tfrac{5}{9}\ cm^2\)
5
Quadrilaterals \(ABCD\) and \(EFGH\) are similar, with scale factor \(k = \frac{1}{4}\). If the area of \(ABCD\) is \(64\ cm^2\), find the area of \(EFGH\).
\(Scale\ factor = k = \frac{1}{4}\)
\(Area\ of\ ABCD = A_1 = 64\ cm^2\)
\(Area\ of\ EFGH = A_2 = ?\)
Since \(ABCD \sim EFGH\). So
\[\begin{aligned} \frac{A_1}{A_2} &= k^2 \\[6pt] \frac{64}{A_2} &= \left(\frac{1}{4}\right)^2 = \frac{1}{16} \\[6pt] 64 \times 16 &= 1 \times A_2 \\[4pt] \boldsymbol{A_2} &= \boldsymbol{1024\ cm^2} \end{aligned}\]Area of \(EFGH\) = 1024 cm²
6
The areas of two similar triangles are \(16\ cm^2\) and \(25\ cm^2\). What is the ratio of a pair of corresponding sides?
\(Area\ of\ first\ triangle = A_1 = 16\ cm^2\)
\(Area\ of\ second\ triangle = A_2 = 25\ cm^2\)
Let \(l_1\) and \(l_2\) be the corresponding lengths of both triangles respectively. Since triangles are similar:
\[\begin{aligned} \frac{A_1}{A_2} &= \left(\frac{l_1}{l_2}\right)^2 \\[6pt] \left(\frac{l_1}{l_2}\right)^2 &= \frac{16}{25} \\[6pt] \frac{l_1}{l_2} &= \sqrt{\frac{16}{25}} = \frac{4}{5} \\[4pt] \boldsymbol{l_1 : l_2} &= \boldsymbol{4 : 5} \end{aligned}\]Ratio of corresponding sides = 4 : 5
7
The areas of two similar triangles are \(144\ cm^2\) and \(81\ cm^2\). If the base of the larger triangle is 30 cm, find the corresponding base of the smaller triangle.
\(Area\ of\ large\ triangle = A_1 = 144\ cm^2\)
\(Area\ of\ small\ triangle = A_2 = 81\ cm^2\)
\(Base\ of\ large\ triangle = l_1 = 30\ cm\)
\(Base\ of\ small\ triangle = l_2 = ?\)
Since triangles are similar. So,
\[\begin{aligned} \frac{A_1}{A_2} &= \left(\frac{l_1}{l_2}\right)^2 \\[6pt] \frac{144}{81} &= \left(\frac{30}{l_2}\right)^2 \\[6pt] \frac{12}{9} &= \frac{30}{l_2} \\[6pt] 12\,l_2 &= 30 \times 9 = 270 \\[4pt] l_2 &= \frac{270}{12} \\[4pt] \boldsymbol{l_2} &= \boldsymbol{22.5\ cm} \end{aligned}\]Base of smaller triangle = 22.5 cm
8
A regular heptagon is inscribed in a larger regular heptagon and each side of the larger heptagon is 1.7 times the side of the smaller heptagon. If the area of the smaller heptagon is \(100\ cm^2\), find the area of the larger heptagon.
\(Length\ of\ side\ of\ smaller\ heptagon = l_1 = x\)
\(Length\ of\ side\ of\ larger\ heptagon = l_2 = 1.7x\)
\(Area\ of\ smaller\ heptagon = A_1 = 100\ cm^2\)
\(Area\ of\ larger\ heptagon = A_2 = ?\)
Since the given heptagons are similar. So
\[\begin{aligned} \frac{A_1}{A_2} &= \left(\frac{l_1}{l_2}\right)^2 \\[6pt] \frac{100}{A_2} &= \left(\frac{x}{1.7x}\right)^2 = \left(\frac{1}{1.7}\right)^2 = \frac{1}{2.89} \\[6pt] 100 \times 2.89 &= 1 \times A_2 \\[4pt] \boldsymbol{A_2} &= \boldsymbol{289\ cm^2} \end{aligned}\]Area of larger heptagon = 289 cm²
📈 Key Concepts – Similar Figures
- Similar Figures: Same shape, different size. Corresponding angles are equal, corresponding sides are proportional.
- Scale Factor: Ratio of corresponding sides. Used to enlarge or reduce figures.
- Triangle Similarity Conditions: AAA, SSS, SAS.
- Area Ratio: Ratio of areas = (scale factor)².
- Applications: Shadow problems, map scales, model making, indirect measurement.