CAT 2023 Slot 2 QA Question & Solution
GeometryMedium
Question
A triangle is drawn with its vertices on the circle C such that one of its sides is a diameter of C and the other two sides have their lengths in the ratio a : b. If the radius of the circle is r, then the area of the triangle is
Options
$\frac{abr^{2}}{2(a^{2}+b^{2})}$
$\frac{2abr^{2}}{a^{2}+b^{2}}$
$\frac{4abr^{2}}{a^{2}+b^{2}}$
$\frac{abr^{2}}{a^{2}+b^{2}}$
Solution
Since BC is the diameter of the circle, which implies angle BAC is 90 degrees. Let AB = a cm, which implies AC = b cm. Hence, $BC\ =\ \sqrt{\ a^2+b^2}$, which is diameter of the circle (2r).
Hence, $2r\ =\sqrt{\ a^2+b^2}$
=> $4r^2=a^2+b^2$
The area of the triangle is $\frac{1}{2}\times\ a\times\ b$, which can be written as
=> $\ \frac{\ a\cdot b}{2\left(a^2+b^2\right)}\times\ \left(a^2+b^2\right)$
=> $\frac{\ a\cdot b}{2\left(a^2+b^2\right)}\times\ 4r^2$
=> $\frac{\ a\cdot b}{a^2+b^2}\times\ 2r^2$
The correct option is B