CBSE 2025 · Region 4 · Set 1 · Q33 · 5 marks
The corresponding sides of $\displaystyle \triangle \mathrm{ABC}$ and $\displaystyle \triangle \mathrm{PQR}$ are in the ratio $\displaystyle 3$ : 5. $\displaystyle \mathrm{AD} \perp \mathrm{BC}$ and $\displaystyle \mathrm{PS} \perp \mathrm{QR}$ as shown in the following figures :
(i)Prove that $\displaystyle \triangle \mathrm{ADC} \sim \triangle \mathrm{PSR}$(ii)If $\displaystyle \mathrm{AD}=4 \mathrm{~cm}$, find the length of PS.(iii)Using (ii) find ar $\displaystyle (\triangle \mathrm{ABC})$ : ar ( $\displaystyle \triangle \mathrm{PQR}$ )State basic proportionality theorem. Use it to prove the following : If three parallel lines $\displaystyle l, m, n$ are intersected by transversals $\displaystyle q$ and $\displaystyle s$ as shown in the adjoining figure, then $\displaystyle \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{DE}}{\mathrm{EF}}$.
The corresponding sides of $\displaystyle \triangle \mathrm{ABC}$ and $\displaystyle \triangle \mathrm{PQR}$ are in the ratio $\displaystyle 3$ : 5. $\displaystyle \mathrm{AD} \perp \mathrm{BC}$ and $\displaystyle \mathrm{PS} \perp \mathrm{QR}$ as shown in the following figures :
(i)
Prove that $\displaystyle \triangle \mathrm{ADC} \sim \triangle \mathrm{PSR}$
(ii)
If $\displaystyle \mathrm{AD}=4 \mathrm{~cm}$, find the length of PS.
(iii)
Using (ii) find ar $\displaystyle (\triangle \mathrm{ABC})$ : ar ( $\displaystyle \triangle \mathrm{PQR}$ )
State basic proportionality theorem. Use it to prove the following : If three parallel lines $\displaystyle l, m, n$ are intersected by transversals $\displaystyle q$ and $\displaystyle s$ as shown in the adjoining figure, then $\displaystyle \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{DE}}{\mathrm{EF}}$.
Marking-scheme solution
As, \(\displaystyle \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{3}{5}\)
\[\begin{aligned}
& \Rightarrow \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR} \\
& \Rightarrow \angle \mathrm{C}=\angle \mathrm{R}
\end{aligned}
\]
(i)
In \(\displaystyle \triangle \mathrm{ADC}\) and \(\displaystyle \triangle \mathrm{PSR}\),
\[\begin{aligned}
& \angle \mathrm{ADC}=\angle \mathrm{PSR} \\
& \text { and } \angle \mathrm{C}=\angle \mathrm{R} \\
& \therefore \triangle \mathrm{ADC} \sim \triangle \mathrm{PSR}
\end{aligned}
\]
\[\begin{aligned}
& \text { (ii) } \frac{\mathrm{AD}}{\mathrm{PS}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{3}{5} \\
& \Rightarrow \frac{4}{\mathrm{PS}}=\frac{3}{5} \\
& \Rightarrow \mathrm{PS}=\frac{20}{3} \mathrm{~cm}
\end{aligned}
\]
\[\text { (iii) } \begin{aligned}
\frac{\operatorname{ar}(\triangle \mathrm{ABC})}{\operatorname{ar}(\triangle \mathrm{PQR})} & =\frac{\dfrac{1}{2} \times \mathrm{BC} \times \mathrm{AD}}{\dfrac{1}{2} \times \mathrm{QR} \times \mathrm{PS}} \\
& =\frac{3}{5} \times \frac{3}{5}=\frac{9}{25}
\end{aligned}
\]
\[\therefore \operatorname{ar}(\triangle \mathrm{ABC}): \operatorname{ar}(\triangle \mathrm{PQR})=9: 25
\]
Join AF intersecting line \(\displaystyle m\) at G
In \(\displaystyle \triangle \mathrm{ACF}, \mathrm{BG} \| \mathrm{CF}\)
\[\Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{AG}}{\mathrm{GF}} \quad \ldots \text { (i) }
\]
In \(\displaystyle \Delta \mathrm{FDA}, \mathrm{GE} \| \mathrm{AD}\)
\[\Rightarrow \frac{\mathrm{EF}}{\mathrm{DE}}=\frac{\mathrm{GF}}{\mathrm{AG}} \text { or } \frac{\mathrm{DE}}{\mathrm{EF}}=\frac{\mathrm{AG}}{\mathrm{GF}} \quad \ldots \text { (ii) }
\]
From, (i) and (ii), we get \(\displaystyle \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{DE}}{\mathrm{EF}}\)
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