CBSE Class 10 Mathematics Standard 2023: question 38

Hints

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Step-by-step solution

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1. (a) $OA\perp AB$ [tangent $\perp$ radius at the point of contact], so $△OAB$ is right-angled at $A$.
   $\tan 30^{\circ }=\frac{OA}{AB}$, which gives $\frac{1}{\sqrt{3}}=\frac{75}{AB}$

   ½ mark

2. (a) $AB=75\sqrt{3}$ cm ($\approx 129.9$ cm)

   ½ mark

3. (b) $\sin 30^{\circ }=\frac{OA}{OB}$, which gives $\frac{1}{2}=\frac{75}{OB}$

   ½ mark

4. (b) $OB=150$ cm

   ½ mark

5. (c)(a) $Q$ lies on the circle and on $OB$, so $OQ=75$ cm. Then $QB=OB-OQ=150-75=75$ cm, so $Q$ is the mid-point of $OB$.

   1 mark

6. (c)(a) In $△BOA$, the line through the mid-point $Q$ of $OB$ parallel to $OA$ bisects $AB$ [converse of the mid-point theorem]. So $P$ is the mid-point of $AB$, and
   $AP=\frac{AB}{2}=\frac{75\sqrt{3}}{2}$ cm ($\approx 64.95$ cm).

   1 mark

7. (c)(b) As above, $QB=150-75=75$ cm. Since $PQ\parallel OA$, $△BQP\sim △BOA$ [AA similarity: common $\angle B$, corresponding angles equal].

   ½ mark

8. (c)(b) $\frac{QB}{OB}=\frac{PQ}{OA}$, which gives $\frac{75}{150}=\frac{PQ}{75}$, i.e. $\frac{1}{2}=\frac{PQ}{75}$

   1 mark

9. (c)(b) $PQ=\frac{75}{2}=37.5$ cm

   ½ mark

Answer: (a) $AB=75\sqrt{3}$ cm; (b) $OB=150$ cm; (c)(a) $AP=\frac{75\sqrt{3}}{2}$ cm; (c)(b) $PQ=\frac{75}{2}=37.5$ cm

Revise the chapter

• Chapter: Circles (notes, formulas, where marks are lost and every past question)
• Topic: Tangent Perpendicular to Radius Theorem