CBSE Class 10 Mathematics Standard 2026: question 31(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. In right $△OBA$ (right angle at B):
   $OB=\sqrt{O{A}^2-A{B}^2}=\sqrt{10^2-8^2}=\sqrt{36}=6$ cm

   ½ mark

2. Let $\angle AOB=\theta$. In $△OBA$: $\tan \theta =\frac{AB}{OB}=\frac{8}{6}\text{...(i)}$

   ½ mark

3. $\angle OAP=90^{\circ }$ [tangent ⟂ radius]. So in right $△OAP$:
   $\tan \theta =\frac{AP}{OA}=\frac{AP}{10}$
   Using (i): $\frac{AP}{10}=\frac{8}{6}$

   ½ mark

4. $\Rightarrow AP=\frac{80}{6}=\frac{40}{3}$ cm

   ½ mark

5. By Pythagoras in $△OAP$:
   $OP=\sqrt{A{P}^2+O{A}^2}=\sqrt{\frac{1600}{9}+100}=\sqrt{\frac{2500}{9}}=\frac{50}{3}$ cm

   ½ mark

6. $PB=OP-OB=\frac{50}{3}-6=\frac{32}{3}$ cm $\approx 10.67$ cm.

   ½ mark

Answer: $PB=\frac{32}{3}$ cm ≈ 10.67 cm

Revise the chapter

• Chapter: Circles (notes, formulas, where marks are lost and every past question)