Question

The difference between the area of the circumcircle and the area of the incircle of an equilateral triangle is $2156\ \text{cm}^2$. What is the area of the equilateral triangle?

• A. $686\sqrt{3}$
• B. $1000$
• C. $961\sqrt{2}$
• D. $650\sqrt{3}$
• E. None of the above

Hint:

For an equilateral triangle: $R=\dfrac{a}{\sqrt{3}}$, $r=\dfrac{a}{2\sqrt{3}}$, and $[\triangle]=\dfrac{\sqrt{3}}{4}a^2$. In many objective tests, using $\pi=\dfrac{22}{7}$ quickly yields integer-friendly results.

Answer 1

The Correct Option is A

Step 1: Write $R$ and $r$ in terms of side $a$.
For an equilateral triangle of side $a$, \[ R=\frac{a}{\sqrt{3}},\qquad r=\frac{a\sqrt{3}}{6}=\frac{a}{2\sqrt{3}}. \] Step 2: Use the given difference of circle areas.
\[ \text{Difference}=\pi(R^2-r^2)=\pi\!\left(\frac{a^2}{3}-\frac{a^2}{12}\right)=\pi.....\frac{a^2}{4}. \] Given this equals $2156$, and using the standard exam approximation $\pi=\tfrac{22}{7}$, \[ \frac{22}{7}.....\frac{a^2}{4}=2156 \;\Rightarrow\; \frac{11}{14}a^2=2156 \;\Rightarrow\; a^2=2156.....\frac{14}{11}=196..... 14=2744. \] Step 3: Area of the equilateral triangle.
\[ [\triangle]=\frac{\sqrt{3}}{4}\,a^2=\frac{\sqrt{3}}{4}..... 2744=686\sqrt{3}. \] Area of the equilateral triangle $=\boxed{686\sqrt{3}\ \text{cm}^2}$. }}