✓ Board-verified
Mathematics · 2025 · 5 marks
CBSE 2025 · Region 4 · Set 1 · Q33
(A) The corresponding sides of ABC and PQR are in the ratio 3: 5. AD ⊥ BC and PS ⊥ QR as shown in the following figures: (i) Prove that ADC ∼ PSR…
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 \(\displaystyle \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{DE}}{\mathrm{EF}}\)
More from Triangles
- Sides AB and AC and median AM of a ABC are proportional to sides DE and DF and Median DN of another DEF. Show that ABC ∼ DEF. ABCD is a…2023
- In the given figure, DE ∥ BC. If AD=2 units, DB=AE=3 units and EC=x units, then the value of x is:2023
- Which of the following statements is incorrect?2025
- E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that Δ ABE ∼ Δ CFB. OR Sides AB, BC and the median AD of…2024
- If PQR ∼ ABC; PQ=6 cm, AB=8 cm and the perimeter of ABC is 36 cm, then the perimeter of PQR is2023
- In the given figure, ABCD is a quadrilateral. Diagonal BD bisects ∠ B and ∠ D both. Prove that:2024
- E and F are points on the sides AB and AC respectively of a ABC such that AE/EB=AF/FC=1/2. Which of the following relation is true?2025
- If Δ ABC ∼ Δ PQR with ∠ A=32^° and ∠ R=65^°, then the measure of ∠ B is:2023
CBSE Class 10 Mathematics past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.