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

Given: \( \angle BAC = 90^\circ \) because \( BC \) is the diameter of the circle.

Let \( AB = a \) cm and \( AC = b \) cm.

Since \( \angle BAC = 90^\circ \), triangle \( ABC \) is a right-angled triangle, and we can apply the Pythagoras Theorem:

\[ BC = \sqrt{a^2 + b^2} \]

Since \( BC \) is the diameter of the circle, its length is \( 2r \), where \( r \) is the radius of the circle. So,

\[ 2r = \sqrt{a^2 + b^2} \Rightarrow 4r^2 = a^2 + b^2 \]

Area of triangle \( ABC \) is: \[ \text{Area} = \frac{1}{2} \times a \times b \] 

Now, we multiply and divide this expression by \( a^2 + b^2 \):

\[ \text{Area} = \frac{ab}{2(a^2 + b^2)} \times (a^2 + b^2) \]

Substitute \( a^2 + b^2 = 4r^2 \):

\[ = \frac{ab}{2(a^2 + b^2)} \times 4r^2 = \frac{ab}{(a^2 + b^2)} \times 2r^2 \]

Therefore, \[ \text{Area} = \frac{2abr^2}{a^2 + b^2} \]

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