The area of an equilateral triangle ABC is \( 17320.5 cm^2\). With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see figure). Find the area of the shaded region. (Use \( \pi \ = \ 3.14 \) and \( \sqrt{3} \ = \ 1.73205 \))

Let each side of the triangle be a cm. Then,

Area = 17320.5 cm²

\( \Rightarrow \ \frac{ \sqrt{3}}{4}a^2 \ = \ 17320.5 \)
[\(\therefore \) Area \(= \ \frac{ \sqrt{3}}{4}(side)^2 \)]

\( \Rightarrow \ a^2 \ = \ \frac{17320.5 \ × \ 4}{ \sqrt{3}} \) \( = \ 40000 \)

\( \Rightarrow \ a \ = \ 200 \)

\(\therefore \) Radius of each circle = \( \frac{a}{2} = 100 cm.\)

Now, required area = Area of \(\triangle \) ABC - 3 × (Area of a sector of angle 60º in a circle of 100 cm)

\( = \ 17320.5 \ - \ 3( \frac{60}{360} \ × \ 3.14 \ × \ 100 \ × \ 100) \)

\( = \ 17320.5 \ - \ 15700 \ = \ 1620.5 \) cm²