CBSE Class 10 Mathematics Standard 2025: question 26(b)

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: Draw quadrilateral ABCD circumscribing a circle with centre O. The circle touches AB at P, BC at Q, CD at R and DA at S. Join OA, OB, OC, OD and OP, OQ, OR, OS. Name the eight angles at O: $\angle 1=\angle AOP$, $\angle 2=\angle AOS$, $\angle 3=\angle DOS$, $\angle 4=\angle DOR$, $\angle 5=\angle COR$, $\angle 6=\angle COQ$, $\angle 7=\angle BOQ$, $\angle 8=\angle BOP$.

   ½ mark

2. In $△OAP$ and $△OAS$:

   • $OP=OS$ [radii of the same circle]
   • $AP=AS$ [tangents from an external point are equal]
   • $OA=OA$ [common side]

   So $△OAP\cong △OAS$ [SSS], and hence $\angle 1=\angle 2$ [CPCT].

   1 mark

3. In the same way, using the vertices D, C and B, we get $\angle 3=\angle 4$, $\angle 5=\angle 6$ and $\angle 7=\angle 8$.

   ½ mark

4. The angles round the point O add up to $360^{\circ }$:
   $\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8=360^{\circ }$

   Using the equal pairs, this becomes $2(\angle 1+\angle 8+\angle 4+\angle 5)=360^{\circ }$, so $\angle 1+\angle 8+\angle 4+\angle 5=180^{\circ }$.

   ½ mark

5. Now $\angle 1+\angle 8=\angle AOB$ and $\angle 4+\angle 5=\angle COD$.

   So $\angle AOB+\angle COD=180^{\circ }$. Similarly, $\angle BOC+\angle AOD=180^{\circ }$.

   Hence opposite sides subtend supplementary angles at the centre. Proved.

   ½ mark

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

Revise the chapter

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