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

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Step-by-step solution

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1. Each animal grazes a sector of radius $r=14$ m whose central angle is the angle of the triangle at its corner. The sides are 35 m, 40 m and 50 m, each longer than $2\times 14=28$ m, so the three sectors do not overlap. Let the angles at A, B, C be ${\theta }_1,{\theta }_2,{\theta }_3$.

2. Total grazed area $=\frac{{\theta }_1}{360^{\circ }}\pi r^2+\frac{{\theta }_2}{360^{\circ }}\pi r^2+\frac{{\theta }_3}{360^{\circ }}\pi r^2=\frac{\pi r^2}{360^{\circ }}({\theta }_1+{\theta }_2+{\theta }_3)$. Since ${\theta }_1+{\theta }_2+{\theta }_3=180^{\circ }$ [angle sum property of a triangle], this is $\frac{\pi r^2}{360^{\circ }}\times 180^{\circ }=\frac{\pi r^2}{2}$.

   1 mark

3. $=\frac{1}{2}\times \frac{22}{7}\times 14\times 14=308$ m${}^2$. So the three animals can graze 308 m${}^2$ of the field.

   1 mark

Answer: 308 m²

Revise the chapter

• Chapter: Areas Related to Circles (notes, formulas, where marks are lost and every past question)
• Topic: Area of a Sector of a Circle