**Question:**

The area of an equilateral triangle is $49 \sqrt{3} \mathrm{~cm}^{2}$. Taking each angular point as centre, circles are drawn with radius equal to half the length of the side of the triangle. Find the area of the triangle not included in the circles. [Take π = 1.73]

**Solution:**

Area of equilateral triangle $=49 \sqrt{3}$

$\Rightarrow \frac{\sqrt{3}}{4} \times(\text { Side })^{2}=49 \sqrt{3}$

$\Rightarrow(\text { Side })^{2}=7^{2} \times 2^{2}$

$\Rightarrow$ Side $=14 \mathrm{~cm}$

Radius of the circle = Half of the side of the triangle = 7 cm

Area of triangle not included in the circle = Area of triangle − Area of 3 sectors having central angle 60∘

$=49 \sqrt{3}-3 \times \frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 7 \times 7$

$=84.77-77$

$=7.77 \mathrm{~cm}^{2}$

Hence, the required area is 7.77 cm2