CBSE Class 10 Mathematics Basic 2025: question 31

Hints

Open one hint at a time and try again before the next.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. Figure: quadrilateral ABCD circumscribes a circle with centre O, touching AB, BC, CD, DA at P, Q, R, S respectively. Join OA, OB, OC, OD and OP, OQ, OR, OS. Name the angles at O: $\angle 1=\angle AOP$, $\angle 2=\angle POB$, $\angle 3=\angle BOQ$, $\angle 4=\angle QOC$, $\angle 5=\angle COR$, $\angle 6=\angle ROD$, $\angle 7=\angle DOS$, $\angle 8=\angle SOA$. To prove: $\angle AOB+\angle COD=180^{\circ }$.

   ½ mark

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

   ½ mark

3. Hence $\angle 1=\angle 8$ [CPCT]. Similarly $\angle 2=\angle 3$, $\angle 4=\angle 5$, $\angle 6=\angle 7$.

   1 mark

4. Angles round O: $\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8=360^{\circ }$, so $2(\angle 1+\angle 2+\angle 5+\angle 6)=360^{\circ }$, giving $\angle 1+\angle 2+\angle 5+\angle 6=180^{\circ }$.

   ½ mark

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

   ½ mark

Answer: $\angle AOB+\angle COD=180^{\circ }$ and $\angle BOC+\angle AOD=180^{\circ }$

Revise the chapter

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