Question

The area of an equilateral triangle is $49\sqrt{3}{\mathrm{cm}}^2$. Taking each angular point as centre, circle as 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.

Answer 1

Let the side of the equilateral triangle be $a$ cm.

Thus, we have:

$$\begin{gathered} \frac{\sqrt{3}}{4}{a}^2=49\sqrt{3} \\ \Rightarrow a^2=196 \\ \Rightarrow a=14 \end{gathered}$$

The radius of each circle is 7 cm. The angle at the vertex of each triangle is $60°$.

Area of the sector with angle $60°$ and radius 7 cm:
$$\begin{gathered} \frac{60}{360}\times \frac{22}{7}\times 7\times 7 \\ =\frac{1}{6}\times 22\times 7 \\ =25.67{\mathrm{cm}}^2 \end{gathered}$$

There are three such sectors.
Total area = $25.6666667\times 3=77{\mathrm{cm}}^2$

Area not included in the circles = Area of the triangle $-$ Area of the three sectors
$$\begin{gathered} =\frac{\sqrt{3}}{4}\times 14\times 14-77 \\ =7.77{\mathrm{cm}}^2 \end{gathered}$$