Area of right-angled triangle inscribed in circle

CAT 2023 Slot 2 · QA · Easy · Geometry

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A triangle is drawn with its vertices on the circle CC such that one of its sides is a diameter of CC and the other two sides have their lengths in the ratio a:ba:b. If the radius of the circle is rr, then the area of the triangle is

  1. A.

    abr22(a2+b2)\frac{a b r^2}{2(a^2 + b^2)}

  2. B.

    abr2a2+b2\frac{a b r^2}{a^2 + b^2}

  3. C.

    4abr22(a2+b2)\frac{4 a b r^2}{2(a^2 + b^2)}

  4. D.

    2abr2a2+b2\frac{2 a b r^2}{a^2 + b^2}

Answer

D

Explanation

Since one side of the triangle is a diameter of circle CC, the triangle is a right-angled triangle, and the hypotenuse is 2r2r.

Let the other two sides be x=akx = ak and y=bky = bk.

By Pythagoras theorem: x2+y2=(2r)2  ⟹  (a2+b2)k2=4r2  ⟹  k2=4r2a2+b2x^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 =12xy=12(ak)(bk)=12abk2= \frac{1}{2} x y = \frac{1}{2} (ak)(bk) = \frac{1}{2} ab k^2.

Substitute k2k^2: Area=12ab(4r2a2+b2)=2abr2a2+b2\text{Area} = \frac{1}{2} ab \left(\frac{4r^2}{a^2 + b^2}\right) = \frac{2 a b r^2}{a^2 + b^2}

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