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Mathematics · 2024 · 3 marks
CBSE 2024 · Region 2 · Set 1 · Q28
(a) In the given figure, PQ is tangent to a circle centred at O and ∠ BAQ =30^°; show that BP = BQ.
In the given figure, PQ is tangent to a circle centred at O and $\displaystyle \angle \mathrm{BAQ}=30^{\circ}$; show that $\displaystyle \mathrm{BP}=\mathrm{BQ}$.
In the given figure, AB, BC, CD and DA are tangents to the circle with centre O forming a quadrilateral ABCD. Show that $\displaystyle \angle \mathrm{AOB}+\angle \mathrm{COD}=180^{\circ}$
Marking-scheme solution
Join OQ
\[\mathrm{OQ}=\mathrm{OA}
\]
\[\Rightarrow \angle 2=30^
\]
\]\[\angle 3=90^{\circ}-30^{\circ}=60^{\circ}
\]
\[\angle 4=90^{\circ}-60^{\circ}=30^{\circ}
\]
\[\angle 6=\angle 1+\angle 2=60^
\]
\]Hence \(\displaystyle \angle 5=90^-60^=30^=\angle 4\)
-60^=30^=\angle 4\)\[\therefore \mathrm{BP}=\mathrm{BQ}
\]
Join OP, OQ, OR and OS \(\displaystyle \triangle \mathrm{POB} \cong \triangle \mathrm{QOB}\)
\(\displaystyle \Rightarrow \angle 1=\angle 2\)
Similarly \(\displaystyle \angle 3=\angle 4, \angle 5=\angle 6, \angle 7=\angle 8\)
Now, \(\displaystyle \angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8=360^{\circ}\)
\(\displaystyle \Rightarrow 2(\angle 1+\angle 8+\angle 4+\angle 5)=360^{\circ}\)
\(\displaystyle \therefore \angle \mathrm{AOB}+\angle \mathrm{COD}=180^{\circ}\)
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CBSE Class 10 Mathematics past-paper question from the 2024 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.