Question

In the given figure, an equilateral triangle $ABC$ is inscribed in a circle. If $AB=2\sqrt{2}\text{cm}$ then the area of the circle is

• A. $4\pi {\text{cm}}^2$
• B. $4\sqrt{3}\pi {\text{cm}}^2$
• C. $8\sqrt{3}\pi {\text{cm}}^2$
• D. $\sqrt{3}\pi {\text{cm}}^2$

Answer 1

The correct option is A $4\pi {\text{cm}}^2$

Let $O$ be the orthocentre of the triangle $ABC$ as all the altitudes of the triangle are intersecting at $O$.

Since $\Delta ABC$ is an equilateral triangle and in case of an equilateral triangle, the orthocentre and the centroid are coincident.

Thus, $O$ is also the centroid of the $\Delta ABC$.
Also, in case of an equilateral triangle, altitude is the median.

Thus, $AM$ is the median of the triangle $ABC$ such that it divides $BC$ into two equal parts $BM$ and $MC$.
i.e., $BM=MC=\frac{2\sqrt{3}}{2}=\sqrt{3}\text{cm}$

Thus, $\Delta ABM$ is a right-angled triangle.
Therefore, using Pythagoras theorem in $DABM$, we get
$A{B}^2=A{M}^2+B{M}^2$
$\Rightarrow {\left(2\sqrt{3}\right)}^2=A{M}^2+{\left(\sqrt{3}\right)}^2$
$\Rightarrow 12=A{M}^2+3$
$\Rightarrow A{M}^2=9$
$\Rightarrow AM=3\text{cm}$
As, centroid divides the median in the ratio $2:1$.
$\therefore O$ divides $AM$ in the ratio $2:1$.
Thus, $AO=2\text{cm}$ and $OM=1\text{cm}$. $(AM=3\text{cm})$
Now, clearly $AO$ is the radius of the given circle.
$\therefore$ Radius of circle, $r=2\text{cm}$
Area of the given circle $=\pi \times r^2$
$=\pi \times (2{)}^2$
$=4\pi {\text{cm}}^2$

Hence, the correct answer is option a.