1 Marks Question

Question 11 Mark

If ΔABC ~ ΔPQR and AB: PQ = 2:3, then fill in the blanks.
$\frac{ A (\triangle ABC )}{ A (\Delta PQR )}=\frac{ AB ^2}{\square}=\frac{2^2}{3^2}=\frac{\square}{\square}$

Answer

∵ Δ ABC~Δ PQR and AB:PQ = 2:3
$\Rightarrow \frac{A(\triangle ABC )}{ A (\triangle PQR )}=\frac{ AB ^2}{ PQ ^2}=\frac{2^2}{3^2}=\frac{4}{9}$

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Question 21 Mark

The ratio of corresponding sides of similar triangles is $3 : 5$; then find the ratio of their areas.

Answer

Theorem: When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides.
$\Rightarrow $ Ratio of areas $= 3^2:5^2$
$\Rightarrow$ Ratio of areas $= 9 : 25$

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Question 31 Mark

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Question 41 Mark

Answer

$\text { : In } \Delta \text { PMN and } \Delta \text { UVW }$
$\frac{\mathrm{PM}}{\mathrm{UV}}=\frac{6}{3}=\frac{2}{1}, \frac{\mathrm{MN}}{\mathrm{VW}}=\frac{10}{5}=\frac{2}{1}$
$\therefore \frac{\mathrm{PM}}{\mathrm{UV}}=\frac{\mathrm{MN}}{\mathrm{VW}}$
$\text { and } \angle \mathrm{M} \cong \angle \mathrm{V}$
$\text { Given }$
$\Delta \mathrm{PMN} \sim \Delta \mathrm{UVW}$
$\text { SAS test of similarity }$

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Question 51 Mark

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Question 61 Mark

In △ PQR, seg RS bisects ∠ R. If PR = 15, RQ = 20 PS = 12 then find SQ.

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Question 71 Mark

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Question 81 Mark

The ratio of the areas of two triangles with the common base is $6: 5$. Height of the larger triangles is 9 cm . Then find the corresponding height of the smaller triangle.

Answer

7.5 cm

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Question 91 Mark

In the adjoining figure, seg $B E \perp \operatorname{seg} A B$ and seg $B A \perp$ seg $A D$. If $B E=6$, $A D=9$, then find $A(\triangle A B E): A (\triangle B A D)$

Answer

$2: 3$

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Question 101 Mark

The ratio of corresponding sides of similar triangles is 3 : 5; then find the ratio of their areas.

Answer

Theorem: When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides.

⇒ Ratio of areas = 3²:5²

⇒ Ratio of areas = 9 : 25

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Question 111 Mark

Answer

$\frac{A(\triangle ABC )}{ A (\Delta DBC )}=\frac{ AE }{ DF } \ldots \ldots \ldots$. bases are equal, hence areas proportional to heights.
$=\frac{4}{6}=\frac{2}{3}$

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Question 121 Mark

Answer

$\begin{array}{l}\text { : In } \Delta \text { PMN and } \Delta \text { UVW } \\ \frac{\mathrm{PM}}{\mathrm{UV}}=\frac{6}{3}=\frac{2}{1}, \frac{\mathrm{MN}}{\mathrm{VW}}=\frac{10}{5}=\frac{2}{1} \\ \therefore \frac{\mathrm{PM}}{\mathrm{UV}}=\frac{\mathrm{MN}}{\mathrm{VW}} \\ \text { and } \angle \mathrm{M} \cong \angle \mathrm{V} \\ \text { Given } \\ \Delta \mathrm{PMN} \sim \Delta \mathrm{UVW} \\ \text { SAS test of similarity } \\\end{array}$

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Question 131 Mark

Answer

: In $\triangle X Y Z$ and $\triangle L M N$,
$\begin{array}{l}
\angle Y =100^{\circ}, \angle M =100^{\circ}, \therefore \angle Y \cong \angle M \\
\angle Z =30^{\circ}, \angle N =30^{\circ}, \therefore \angle Z \cong \angle N \\
\therefore \triangle XYZ \sim \triangle LMN \quad \ldots \ldots \text { by AA test. }
\end{array}$

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Question 141 Mark

In △ PQR, seg RS bisects ∠ R. If PR = 15, RQ = 20 PS = 12 then find SQ.

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Question 151 Mark

Answer

$\begin{array}{l}: \ln \triangle ABC , DE \| BC \\ \therefore \frac{ AD }{ DB }=\frac{ AE }{ EC } \ldots . . \text { Basic proportionality theorem } \\ \therefore \frac{1.8}{5.4}=\frac{ AE }{7.2} \\ \therefore AE \times 5.4=1.8 \times 7.2 \\ \therefore AE =\frac{1.8 \times 7.2}{5.4}=2.4 \\ AE =2.4 cm \end{array}$

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Question 161 Mark

The ratio of the areas of two triangles with the common base is $6: 5$. Height of the larger triangles is 9 cm . Then find the corresponding height of the smaller triangle.

Answer

7.5 cm

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Question 171 Mark

In the adjoining figure, seg $B E \perp \operatorname{seg} A B$ and seg $B A \perp$ seg $A D$. If $B E=6$, $A D=9$, then find $A(\triangle A B E): A (\triangle B A D)$

Answer

$2: 3$

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