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

The Correct Option is A

\(∠BAC\) = 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): \(\frac{2abr^2}{a^2+b^2}\)