CBSE Class 10 Mathematics Standard sample paper 2026-27: question 37

Hints

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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. (i) $\angle OAP=90^{\circ }$ [tangent ⟂ radius at the point of contact]

   1 mark

2. (ii) In right $△OAP$: $O{P}^2=O{A}^2+A{P}^2\Rightarrow 18^2=9^2+A{P}^2$

   ½ mark

3. $A{P}^2=324-81=243\Rightarrow AP=\sqrt{243}=9\sqrt{3}$ m $\approx 15.6$ m

   ½ mark

4. (iii)(a) Let $\angle AOP=\theta$. In right $△OAP$:

   $\cos \theta =\frac{OA}{OP}=\frac{9}{18}=\frac{1}{2}=\cos 60^{\circ }$

   1 mark

5. $\Rightarrow \theta =60^{\circ }$

   ½ mark

6. OP bisects $\angle AOB$ [$△OAP\cong △OBP$ by RHS].

   So $\angle AOB=2\theta =120^{\circ }$.

   ½ mark

7. (iii)(b) If $\angle APB=60^{\circ }$, then OP bisects it [tangents from an external point are equally inclined to OP].

   So $\angle OPA=\frac{1}{2}\times 60^{\circ }=30^{\circ }$.

   ½ mark

8. In right $△OAP$: $\frac{OA}{AP}=\tan 30^{\circ }\Rightarrow \frac{9}{AP}=\frac{1}{\sqrt{3}}$

   1 mark

9. $\Rightarrow AP=9\sqrt{3}$ m, and $PB=PA=9\sqrt{3}$ m [tangents from an external point are equal]

   ½ mark

Answer: (i) 90°; (ii) 9√3 m; (iii)(a) 120°; (iii)(b) 9√3 m each

Revise the chapter

• Chapter: Circles (notes, formulas, where marks are lost and every past question)
• Topic: Lengths of Tangents from an External Point