CBSE Class 10 Mathematics Basic 2024: question 27(a)

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Step-by-step solution

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1. Figure: quadrilateral ABCD circumscribes a circle with centre O. The sides AB, BC, CD, DA touch the circle at P, Q, R, S. Join OA, OB, OC, OD, OP, OQ, OR, OS. Mark $\angle AOP=\angle 1$, $\angle AOS=\angle 2$, $\angle DOS=\angle 3$, $\angle DOR=\angle 4$, $\angle COR=\angle 5$, $\angle COQ=\angle 6$, $\angle BOQ=\angle 7$, $\angle BOP=\angle 8$. To prove: $\angle AOB+\angle COD=180^{\circ }$ and $\angle BOC+\angle AOD=180^{\circ }$.

   ½ mark

2. In $△AOP$ and $△AOS$: $OA=OA$ [common], $OP=OS$ [radii], $AP=AS$ [tangents from an external point are equal]. So $△AOP\cong △AOS$ [SSS], giving $\angle 1=\angle 2$ [CPCT].

   1 mark

3. In the same way, $\angle 3=\angle 4$, $\angle 5=\angle 6$ and $\angle 7=\angle 8$.

   ½ mark

4. The angles round O add up to $360^{\circ }$: $\angle 1+\angle 2+\cdots +\angle 8=360^{\circ }$. So $2(\angle 1+\angle 8+\angle 4+\angle 5)=360^{\circ }$, i.e. $(\angle 1+\angle 8)+(\angle 4+\angle 5)=180^{\circ }$.

   ½ mark

5. Since $\angle 1+\angle 8=\angle AOB$ and $\angle 4+\angle 5=\angle COD$, we get $\angle AOB+\angle COD=180^{\circ }$. Similarly, $(\angle 6+\angle 7)+(\angle 2+\angle 3)=180^{\circ }$, i.e. $\angle BOC+\angle AOD=180^{\circ }$. Hence opposite sides subtend supplementary angles at the centre.

   ½ mark

Answer: ∠AOB + ∠COD = 180° and ∠BOC + ∠AOD = 180°.

Revise the chapter

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