CBSE Class 10 Mathematics Basic 2024: question 35

Hints

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Step-by-step solution

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1. Here $r=14$ cm and $\theta =90^{\circ }$. Area of minor sector $=\frac{\theta }{360^{\circ }}\pi r^2=\frac{90}{360}\times \frac{22}{7}\times 14\times 14=154{\text{cm}}^2$.

   1 mark

2. The chord and the two radii form a right triangle at the centre, since $\theta =90^{\circ }$. Its area $=\frac{1}{2}{r}^2=\frac{1}{2}\times 14\times 14=98{\text{cm}}^2$.

   1 mark

3. Area of minor segment $=$ area of sector $-$ area of triangle $=154-98=56{\text{cm}}^2$.

   1½ marks

4. Area of major segment $=$ area of circle $-$ area of minor segment $=\frac{22}{7}\times 14\times 14-56=616-56=560{\text{cm}}^2$.

   1½ marks

Answer: Minor segment = 56 cm²; major segment = 560 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