Question

The area of an equilateral triangle inscribed in the circle $x^2+y^2-6x-8y-25=0$ is

• A. $\frac{225\sqrt{3}}{6}$
• B. $25\pi$
• C. $50\pi -100$
• D. None of these

Answer 1

The correct option is A

$\frac{225\sqrt{3}}{6}$

Let ABC be the required equilateral triangle. The equation of the circle $x^2+y^2-6x-8y-25=0$

Therefore, coordinates of the centre O is (3, 4)

Radius of the circle $=OA=OB=OC==\sqrt{9+16+25}=5\sqrt{2}$

In $\Delta BOD$, we have :

$\Rightarrow DB=\frac{\sqrt{3}}{2}\left(5\sqrt{2}\right)$

$\Rightarrow BC=2BD=\sqrt{3}\left(5\sqrt{2}\right)=5\sqrt{6}$

Now, area of $\Delta ABG=\frac{\sqrt{3}}{4}B{C}^2=\frac{\sqrt{3}}{4}(5\sqrt{6}{)}^2$

$=\frac{\sqrt{3}\left(150\right)}{4}=\frac{\sqrt{3}\left(75\right)}{2}=\frac{\sqrt{3}\left(225\right)}{6}$ square units.