CBSE Class 10 Mathematics Standard 2025: question 29

Hints

Open one hint at a time and try again before the next.

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. Let chord AB subtend $\angle AOB=90^{\circ }$ at centre O, with $r=10$ cm.

   Area of minor segment = Area of sector OAB − Area of $△OAB$.

   Area of sector $=\frac{\theta }{360^{\circ }}\times \pi r^2=\frac{90}{360}\times 3.14\times 10\times 10$

   ½ mark

2. $=78.5{\text{ cm}}^2$

   ½ mark

3. $△OAB$ is right-angled at O, so its area $=\frac{1}{2}\times OA\times OB=\frac{1}{2}\times 10\times 10$

   ½ mark

4. $=50{\text{ cm}}^2$

   ½ mark

5. Area of minor segment $=78.5-50$

   ½ mark

6. $=28.5{\text{ cm}}^2$

   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