Area of right-angled triangle inscribed in circle

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

Accepted Answer

D. Since one side of the triangle is a diameter of circle $C$, the triangle is a right-angled triangle, and the hypotenuse is $2r$. Let the other two sides be $x = ak$ and $y = bk$. By Pythagoras theorem: $$x^2 + y^2 = (2r)^2 \implies (a^2 + b^2)k^2 = 4r^2 \implies k^2 = \frac{4r^2}{a^2 + b^2}$$ Area of the right-angled triangle $= \frac{1}{2} x y = \frac{1}{2} (ak)(bk) = \frac{1}{2} ab k^2$. Substitute $k^2$: $$	ext{Area} = \frac{1}{2} ab \left(\frac{4r^2}{a^2 + b^2}ight) = \frac{2 a b r^2}{a^2 + b^2}$$

Answer

$\frac{a b r^2}{2(a^2 + b^2)}$

Answer

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

Answer

$\frac{4 a b r^2}{2(a^2 + b^2)}$

Answer

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

Area of right-angled triangle inscribed in circle

Explanation

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