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Mathematics · 2023 · 3 marks
CBSE 2023 · Region 1 · Set 1 · Q29
(a) Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠ PTQ =2 ∠ OPQ.
Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that $\displaystyle \angle \mathrm{PTQ}=2 \angle \mathrm{OPQ}$.
In the given figure, a circle is inscribed in a quadrilateral ABCD in which $\displaystyle \angle \mathrm{B}=90^{\circ}$. If $\displaystyle \mathrm{AD}=17 \mathrm{~cm}, \mathrm{AB}=20 \mathrm{~cm}$ and $\displaystyle \mathrm{DS}=3 \mathrm{~cm}$, then find the radius of the circle.
Marking-scheme solution
$\displaystyle \mathrm{TP}=\mathrm{TQ}$
$\displaystyle \Rightarrow \angle \mathrm{TPQ}=\angle \mathrm{TQP}$
Let $\displaystyle \angle \mathrm{PTQ}$ be $\displaystyle \theta$
$\displaystyle \Rightarrow \angle \mathrm{TPQ}=\angle \mathrm{TQP}=\frac{180^{\circ}-\theta}{2}=90^{\circ}-\frac{\theta}{2}$
Now $\displaystyle \angle \mathrm{OPT}=90^{\circ}$
$\displaystyle \Rightarrow \angle \mathrm{OPQ}=90^{\circ}-\left(90^{\circ}-\frac{\theta}{2}\right)=\frac{\theta}{2}$
$\displaystyle \angle \mathrm{PTQ}=2 \angle \mathrm{OPQ}$
$\displaystyle \mathrm{DR}=\mathrm{DS}=3 \mathrm{~cm}$
\[\begin{array}{l}
\therefore \quad A R=A D-D R=17-3=14 \mathrm{~cm}
\Rightarrow A Q=A R=14 \mathrm{~cm}
\end{array}
\]
\[\therefore \quad Q B=A B-A Q=20-14=6 \mathrm{~cm}
\]
Since $\displaystyle \mathrm{QB}=\mathrm{OP}=\mathrm{r} \quad \therefore \quad$ radius $\displaystyle =6 \mathrm{~cm}$
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CBSE Class 10 Mathematics past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.