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

• A. $\frac{2abr^2}{a^2+b^2}$
• B. $\frac{abr^2}{a^2+b^2}$
• C. $\frac{abr^2}{2(a^2+b^2)}$
• D. $\frac{4abr^2}{a^2+b^2}$

Answer 1

Answer: (A)

$\frac{2abr^2}{a^2+b^2}$

Answer 2

$∠BAC$ = 90° ($BC$ is the diameter of the circle)
Let $AB = a$ cm, then $AC = b$ cm.
Apply Pythagoras theorem,
$BC =\sqrt {a^2 + b^2}$
$2r =\sqrt {a^2 + b^2}$
$4r^2 =a^2 +b^2$
Area of the triangle = $\frac 12×a×b$

$⇒ \frac {ab}{2(a^2+b^2)}× (a^2 +b^2)$

$⇒ \frac {ab}{2(a^2+b^2)}× 4r^2$

$⇒ \frac {ab}{(a^2+b^2)}× 2r^2$

$⇒ \frac {2abr^2}{(a^2+b^2)}$

So, the correct option is (A): $\frac{2abr^2}{a^2+b^2}$