CBSE Class 10 Mathematics Basic 2026: question 31

Hints

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Step-by-step solution

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1. Let O be the centre of the inscribed circle. Join OA, OB, OC, OD and OP, OQ, OR, OS.

2. In $△OAP$ and $△OAS$: $AP=AS$ [tangents from an external point are equal], $OP=OS$ [radii] and $OA=OA$ [common]. So $△OAP\cong △OAS$ [SSS], which gives $\angle AOP=\angle AOS$.

   1 mark

3. Similarly, $\angle BOP=\angle BOQ$, $\angle COQ=\angle COR$ and $\angle DOR=\angle DOS$.

   ½ mark

4. The angles round the point O add up to $360^{\circ }$: $\angle AOP+\angle BOP+\angle BOQ+\angle COQ+\angle COR+\angle DOR+\angle DOS+\angle AOS=360^{\circ }$, so $2(\angle AOP+\angle BOP+\angle COR+\angle DOR)=360^{\circ }$.

   1 mark

5. Hence $(\angle AOP+\angle BOP)+(\angle COR+\angle DOR)=180^{\circ }$, i.e. $\angle AOB+\angle COD=180^{\circ }$.

   ½ mark

Answer: $\angle AOB+\angle COD=180^{\circ }$ (proved)

Revise the chapter

• Chapter: Circles (notes, formulas, where marks are lost and every past question)
• Topic: Solving Problems Using Tangent Properties