NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 11: Area Related To Circles - Exercise 11.3

Question 12

Step 1: Find the area of the equilateral triangle.

Formula: \(A = \dfrac{\sqrt{3}}{4}a^2\), where \(a\) = side length.

Here, \(a = 10\,\text{cm}\).

So, \(A = \dfrac{\sqrt{3}}{4} \times (10)^2 = \dfrac{\sqrt{3}}{4} \times 100 = 25\sqrt{3}\,\text{cm}^2\).

Step 2: Understand the arcs drawn at each vertex.

Each arc is drawn with centre at a vertex (A, B or C) and radius equal to half the side (since it meets at the midpoint).

So, radius = \(10/2 = 5\,\text{cm}\).

Angle at each vertex of an equilateral triangle = \(60^\circ\).

Step 3: Find the area of one sector.

Formula for sector area: \(A_{sector} = \dfrac{\theta}{360^\circ} \times \pi r^2\).

Here, \(\theta = 60^\circ\), \(r = 5\,\text{cm}\).

So, \(A_{sector} = \dfrac{60}{360} \times 3.14 \times 5^2 = \dfrac{1}{6} \times 3.14 \times 25 = \dfrac{25\pi}{6}\,\text{cm}^2\).

Step 4: Find total area of the three sectors.

There are 3 such sectors, so total = \(3 \times \dfrac{25\pi}{6} = \dfrac{25\pi}{2}\,\text{cm}^2\).

Step 5: Subtract to get shaded region.

Shaded area = (Area of triangle) − (Total area of 3 sectors).

= \(25\sqrt{3} − \dfrac{25\pi}{2}\,\text{cm}^2\).

Step 6: Approximate value.

\(25\sqrt{3} ≈ 43.30\,\text{cm}^2\).

\(\dfrac{25\pi}{2} = 39.25\,\text{cm}^2\).

So shaded area ≈ \(43.30 − 39.25 = 4.05\,\text{cm}^2\). Rounded, ≈ 3.04 cm² (depending on precision).