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

Hints

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Step-by-step solution

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1. The unshaded region is a square of side $a$ in the centre. On each of its four sides stands a semicircle of diameter $a$.
   $$\text{Area}=a^2+4\times \left(\text{area of a semicircle of diameter }a\right)$$

   ½ mark

2. Across the square, the unshaded region leaves a 3 cm gap on each side. So its full width is $14-3-3=8$ cm.

   ½ mark

3. This width is made of a radius, the side of the square and another radius:
   $$\frac{a}{2}+a+\frac{a}{2}=8\Rightarrow 2a=8\Rightarrow a=4\text{ cm}$$
   So the radius of each semicircle is $2$ cm.

   ½ mark

4. $$\text{Area}=4^2+4\times \frac{1}{2}\pi (2{)}^2=16+8\pi {\text{cm}}^2$$
   With $\pi =\frac{22}{7}$, this is $16+\frac{176}{7}\approx 41.14{\text{cm}}^2$.

   ½ mark

Answer: $(16+8\pi )$ cm² ≈ 41.14 cm²

Revise the chapter

• Chapter: Areas Related to Circles (notes, formulas, where marks are lost and every past question)