CBSE Class 10 Mathematics Basic sample paper 2023-24: question 25(b)

Hints

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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. Here $r=10$ cm and $\theta =90^{\circ }$.
   Area of minor segment = Area of sector − Area of right triangle $=\frac{\theta }{360^{\circ }}\pi r^2-\frac{1}{2}\times r\times r$

   ½ mark

2. $=\frac{90^{\circ }}{360^{\circ }}\times 3.14\times 10\times 10-\frac{1}{2}\times 10\times 10$

   ½ mark

3. $=\frac{314}{4}-50=78.5-50$

   ½ mark

4. Area of minor segment $=28.5{\text{cm}}^2$

   ½ mark

Answer: 28.5 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