Question

If the distance from the vertex to the centroid of an equilateral triangle is 6 cm, then what is the area of the triangle?

• A. 24 $c{m}^2$
• B. $27\sqrt{3}c{m}^2$
• C. 12 $c{m}^2$
• D. $12\sqrt{3}c{m}^2$

Answer 1

Answer: (B)

The correct option is B $27\sqrt{3}c{m}^2$
Here $OA=6cm$ $\therefore OD=\frac{OA}{2}=3cm$
$AD=6cm+3cm=9cm$
Let each side of $\Delta ABC$ be $x$ cm.
Form right angled $\Delta ADB,$
$A{B}^2-B{D}^2=A{D}^2$
$\Rightarrow x^2-{\left(\frac{x}{2}\right)}^2=9^2\Rightarrow \frac{3x^2}{4}=81$
$x^2=\frac{4\times 81}{3}=4\times 27$
$\Rightarrow x=6\sqrt{3}cm$
Area of equilatcral $\Delta ABC=\frac{\sqrt{3}}{4}\times (6\sqrt{3}{)}^2$
$=\frac{\sqrt{3}}{4}\times (6\sqrt{3}{)}^2$
$=\frac{\sqrt{3}}{4}\times 108c{m}^2=27\sqrt{3}c{m}^2$