Question

In right triangle $QRS,QR$ is perpendicular to $RS,QR=12$, and $RS=12\sqrt{3}$. The area of the circle that circumscribes triangle $QRS$ is

• A. $108\pi$
• B. $144\pi$
• C. $288\pi$
• D. $576\pi$
• E. $1728\pi$

Answer 1

The correct option is B $144\pi$

QS will be the diameter of circle. Then in right angled $△$QRS, using Pythajoras theorem

$QS=\sqrt{{QR}^2+{RS}^2}=\sqrt{{12}^2+{\left(12\sqrt{3}\right)}^2}=\sqrt{576}$

$QS=24$

$\therefore$ Radius of circle $=\frac{24}{2}=12$

Area of circle $={\pi r}^2=\pi \times {12}^2=144\pi$