CBSE Class 10 Mathematics Standard 2023: question 34

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Step-by-step solution

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1. Here $r=14$ cm and $\theta =60^{\circ }$.
   Area of sector $=\frac{\theta }{360^{\circ }}\pi r^2=\frac{60}{360}\times \frac{22}{7}\times 14\times 14=\frac{308}{3}{\text{cm}}^2$

   1 mark

2. The two radii are equal and the angle between them is $60^{\circ }$, so the triangle formed by the chord and the radii is equilateral with side 14 cm.
   Area of triangle $=\frac{\sqrt{3}}{4}\times 14^2=49\sqrt{3}{\text{cm}}^2$ (the same as $\frac{1}{2}\times 14\times 14\times \sin 60^{\circ }$).

   1 mark

3. Area of minor segment $=$ area of sector $-$ area of triangle
   $=\left(\frac{308}{3}-49\sqrt{3}\right){\text{cm}}^2\approx 102.67-84.87\approx 17.8{\text{cm}}^2$ (about $17.9{\text{cm}}^2$ if $\sqrt{3}=1.73$ is used).

   1 mark

4. Area of circle $=\frac{22}{7}\times 14\times 14=616{\text{cm}}^2$
   Area of major segment $=$ area of circle $-$ area of minor segment $=616-\left(\frac{308}{3}-49\sqrt{3}\right)$

   1 mark

5. $=\left(\frac{1540}{3}+49\sqrt{3}\right){\text{cm}}^2\approx 513.33+84.87\approx 598.2{\text{cm}}^2$ (about $598.1{\text{cm}}^2$ if $\sqrt{3}=1.73$ is used).

   1 mark

Answer: Minor segment $=\left(\frac{308}{3}-49\sqrt{3}\right){\text{cm}}^2\approx 17.8{\text{cm}}^2$; Major segment $=\left(\frac{1540}{3}+49\sqrt{3}\right){\text{cm}}^2\approx 598.2{\text{cm}}^2$

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