CBSE Class 10 Mathematics Standard sample paper 2025-26: question 24(B)

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. Area of minor segment = area of sector − area of $△$ formed by the two radii and the chord. With $\theta =90^{\circ }$ the triangle is right-angled at the centre:
   $=\frac{90^{\circ }}{360^{\circ }}\pi (5{)}^2-\frac{1}{2}\times 5\times 5=\left(\frac{25\pi }{4}-\frac{25}{2}\right)$ cm${}^2$.

   1 mark

2. Area of major segment = area of circle − area of minor segment
   $=25\pi -\left(\frac{25\pi }{4}-\frac{25}{2}\right)=\left(\frac{75\pi }{4}+\frac{25}{2}\right)$ cm${}^2$.

   1 mark

Answer: $\left(\frac{75\pi }{4}+\frac{25}{2}\right)$ 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