CBSE Class 10 Mathematics Standard 2024: question 32

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Step-by-step solution

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1. Figure: a circle with centre O, chord AB, $\angle AOB=60^{\circ }$ and $r=21$ cm. The minor segment lies between chord AB and arc ALB.

   (i) Arc length $=\frac{\theta }{360^{\circ }}\times 2\pi r=\frac{60}{360}\times 2\times \frac{22}{7}\times 21$

   1½ marks

2. (i) $=\frac{1}{6}\times 132=22$ cm.

   ½ mark

3. (ii) Area of sector OALB $=\frac{60}{360}\times \frac{22}{7}\times 21\times 21=\frac{1}{6}\times 1386=231$ cm².

   1 mark

4. (ii) In $\Delta OAB$, $OA=OB=21$ cm and $\angle AOB=60^{\circ }$. The base angles are equal and add to $120^{\circ }$, so each is $60^{\circ }$ and the triangle is equilateral.

   Area $=\frac{\sqrt{3}}{4}\times 21^2=\frac{441\sqrt{3}}{4}$ cm² $\approx 190.95$ cm².

   1½ marks

5. (ii) Area of minor segment = area of sector − area of triangle $=\left(231-\frac{441\sqrt{3}}{4}\right)$ cm² $\approx 231-190.95=40.05$ cm².

   ½ mark

Answer: (i) 22 cm; (ii) $\left(231-\frac{441\sqrt{3}}{4}\right)$ cm² ≈ 40.05 cm²

Revise the chapter

• Chapter: Areas Related to Circles (notes, formulas, where marks are lost and every past question)
• Topic: Area of a Segment of a Circle