Chapter 9: Similar Figures – Exercise 9.1

Ratio of corresponding sides are given as follows;

\[\frac{4}{6} = \frac{3}{4.5} = \frac{5}{7.5} = \frac{2}{3}\]

Ratio of corresponding sides are given as follows;

\[\frac{6}{10.5} = \frac{9}{15.75} = \frac{12}{21} = \frac{4}{7}\]

Since the ratios of the corresponding sides are equal

\[\frac{m\overline{AB}}{m\overline{DE}} = \frac{m\overline{BC}}{m\overline{EF}} = \frac{m\overline{AC}}{m\overline{DF}}\]

As \(\Delta ABC \sim \Delta DEF\), so corresponding sides have same ratio.

By using

\[\begin{aligned} \frac{m\overline{AB}}{m\overline{DE}} &= \frac{m\overline{BC}}{m\overline{EF}} \\[6pt] \frac{12}{6} &= \frac{16}{m\overline{EF}} \\[6pt] (12)(m\overline{EF}) &= (16)(6) \\[4pt] m\overline{EF} &= \frac{(16)(6)}{12} = \frac{96}{12} \\[4pt] \boldsymbol{m\overline{EF}} &= \boldsymbol{8\text{ cm}} \end{aligned}\]

Similarly,

\[\begin{aligned} \frac{m\overline{AB}}{m\overline{DE}} &= \frac{m\overline{AC}}{m\overline{DF}} \\[6pt] \frac{12}{6} &= \frac{20}{m\overline{DF}} \\[6pt] (12)(m\overline{DF}) &= (20)(6) \\[4pt] m\overline{DF} &= \frac{(20)(6)}{12} = \frac{120}{12} \\[4pt] \boldsymbol{m\overline{DF}} &= \boldsymbol{10\text{ cm}} \end{aligned}\]

Since the triangles are similar, their corresponding sides are proportional. By using

\[\begin{aligned} \frac{m\overline{AB}}{m\overline{CD}} &= \frac{m\overline{AE}}{m\overline{CE}} \\[6pt] \frac{x}{1.2} &= \frac{7.5}{3} \\[6pt] (x)(3) &= (7.5)(1.2) \\[4pt] x &= \frac{(7.5)(1.2)}{3} = \frac{9}{3} \\[4pt] \boldsymbol{x} &= \boldsymbol{3\text{ cm}} \end{aligned}\]

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Since \(\overline{EF} \parallel \overline{CD}\), the segments of the sides \(\overline{BC}\) and \(\overline{BD}\) are in the same ratio:

\[\begin{aligned} \frac{m\overline{BE}}{m\overline{EC}} &= \frac{m\overline{BF}}{m\overline{FD}} \\[6pt] \frac{6}{x} &= \frac{8}{3} \\[6pt] (6)(3) &= (8)(x) \\ 18 &= 8x \\ \frac{18}{8} &= x \\ 2.25 &= x \\ \boldsymbol{x} &= \boldsymbol{2.25\text{ cm}} \end{aligned}\]

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As \(m\angle C = m\angle F\) and \(m\angle CED = m\angle FEG\), so \(\Delta CED \sim \Delta FEG\)

By using

\[\begin{aligned} \frac{m\overline{CD}}{m\overline{FG}} &= \frac{m\overline{ED}}{m\overline{EG}} \\[6pt] \frac{x}{2.1} &= \frac{2.5}{2.4} \\[6pt] (x)(2.4) &= (2.5)(2.1) \\[4pt] x &= \frac{(2.5)(2.1)}{2.4} = \frac{5.25}{2.4} \\[4pt] \boldsymbol{x} &= \boldsymbol{2.19\text{ cm}} \end{aligned}\]

From given figure,

Large Triangle: Height = 16 cm and Base = 20 cm

Small Triangle: Height = 8 cm and Base = \((20-x)\) cm

Since the triangles are similar because they share the same angles, so

\[\begin{aligned} \frac{Base\ of\ Small\ Triangle}{Base\ of\ Large\ Triangle} &= \frac{Height\ of\ Small\ Triangle}{Height\ of\ Large\ Triangle} \\[6pt] \frac{20-x}{20} &= \frac{8}{16} \\[6pt] (16)(20-x) &= (8)(20) \\ 320 - 16x &= 160 \\ -16x &= 160 - 320 \\ -16x &= -160 \\ x &= \frac{-160}{-16} \\[4pt] \boldsymbol{x} &= \boldsymbol{10\text{ cm}} \end{aligned}\]

From given data,

Man: Height = 1.8 m and Shadow = 0.76 m

Pole: Height = \(h\) (Unknown) and Shadow = 3 m

Since the triangles are similar because the sun's angle is the same, so

\[\begin{aligned} \frac{Height\ of\ Pole}{Height\ of\ Man} &= \frac{Shadow\ of\ Pole}{Shadow\ of\ Man} \\[6pt] \frac{h}{1.8} &= \frac{3}{0.76} \\[6pt] (h)(0.76) &= (3)(1.8) \\[4pt] h &= \frac{(3)(1.8)}{0.76} = \frac{5.4}{0.76} \\[4pt] \boldsymbol{h} &= \boldsymbol{7.1\text{ m}} \end{aligned}\]

\[\begin{aligned} (H)^2 &= (B)^2 + (P)^2 \\ (10)^2 &= (6)^2 + (y)^2 \\ 100 &= 36 + y^2 \\ 100 - 36 &= y^2 \\ 64 &= y^2 \\ y^2 &= 64 \\ \sqrt{y^2} &= \sqrt{64} \\ \boldsymbol{y} &= \boldsymbol{8\text{ cm}} \end{aligned}\]

We compare the small top triangle to the large overall triangle. These two triangles are similar, so

\[\begin{aligned} \frac{Large\ hypotenuse}{Small\ hypotenuse} &= \frac{Large\ base}{Small\ base} \\[6pt] \frac{6+x}{10} &= \frac{10}{6} \\[6pt] (6)(6+x) &= (10)(10) \\ 36 + 6x &= 100 \\ 6x &= 100 - 36 = 64 \\ x &= \frac{64}{6} \\[4pt] \boldsymbol{x} &= \boldsymbol{10.67\text{ cm}} \end{aligned}\] \[\begin{aligned} (H)^2 &= (B)^2 + (P)^2 \\ (z)^2 &= (8)^2 + (10.67)^2 \\ z^2 &= 64 + 113.85 \\ z^2 &= 177.85 \\ \sqrt{z^2} &= \sqrt{177.85} \\ \boldsymbol{z} &= \boldsymbol{13.33\text{ cm}} \end{aligned}\]

In \(\Delta ABE\) and \(\Delta CDE\), \(\overline{AB} \parallel \overline{CD}\), so

• \(\angle CDE = \angle ABE\)  (Alternate interior angles)
• \(\angle DCE = \angle BAE\)  (Alternate interior angles)
• \(\angle CED = \angle AEB\)  (Vertical angles)

Since three corresponding angles of the triangles are equal, so \(\Delta ABE \sim \Delta CDE\).  (Proved)

As \(\Delta ABE \sim \Delta CDE\)

\[\begin{aligned} \frac{m\overline{AB}}{m\overline{CD}} &= \frac{m\overline{AE}}{m\overline{CE}} \\[6pt] \frac{8}{4} &= \frac{3}{m\overline{CE}} \\[6pt] (8)(m\overline{CE}) &= (3)(4) \\[4pt] m\overline{CE} &= \frac{12}{8} \\[4pt] \boldsymbol{m\overline{CE}} &= \boldsymbol{1.5\text{ cm}} \end{aligned}\]

We know that a regular dodecagon has 12 equal sides, so

\[\begin{aligned} P &= 12 \times L \\ 72 &= 12 \times L \\ \frac{72}{12} &= L \\ L &= 6\text{ cm} \end{aligned}\]

Since length of original dodecagon is reduced by a factor \(\dfrac{1}{\sqrt{2}}\), so

\[\begin{aligned} Length\ of\ scaled\ dodecagon &= \frac{1}{\sqrt{2}} \times 6 \\[6pt] &= \frac{6}{\sqrt{2}} \\[6pt] &= \frac{6}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \\[6pt] &= \frac{6\sqrt{2}}{(\sqrt{2})^2} \\[6pt] &= \frac{6\sqrt{2}}{2} \\[4pt] &= \boldsymbol{3\sqrt{2}\text{ cm}} \end{aligned}\]