Question:
The side of an equilateral triangle is equal to the radius of a circle whose area is 154 cm2. The area of the triangle is
(a) 49 cm2
(b) $\frac{49 \sqrt{3}}{4} \mathrm{~cm}^{2}$
(c) $\frac{7 \sqrt{3}}{4} \mathrm{~cm}^{2}$
(d) $77 \mathrm{~cm}^{2}$
Solution:
(b) $\frac{49 \sqrt{3}}{4} \mathrm{~cm}^{2}$
Area of a circle $=\pi r^{2}$
$\Rightarrow 154=\pi r^{2}$
$\Rightarrow r=\sqrt{\frac{154 \times 7}{22}}$
$=\sqrt{7 \times 7}$
$=7 \mathrm{~cm}$
The radius of the circle is equal to the side of the equilateral triangle.
∴ r = a (Here, a is the side of the equilateral triangle.)
$a=7 \mathrm{~cm}$
$\therefore$ Area of the equilateral triangle $=\frac{\sqrt{3}}{4} a^{2}=\frac{\sqrt{3}}{4} \times 7 \times 7=\frac{49 \sqrt{3}}{4} \mathrm{~cm}^{2}$