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CAT 2023 Slot 2 — QA Question 17

Basics of TrianglesEasy

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

Answer & solution

  • 2abr2a2+b2

  • B

    4abr2a2+b2

  • C

    abr2a2+b2

  • D

    abr22(a2+b2)

Solution

Easy

A triangle inscribed in a circle with one side as the diameter is automatically right-angled (angle in a semicircle =90∘=90^\circ). So the two other sides are the legs, the diameter is the hypotenuse, and the area is just 12×\tfrac12\times leg ×\times leg. Put everything in terms of rr using the ratio a:ba:b.

diameter = 2r a b
1

Use the semicircle right angle. The angle facing the diameter is 90∘90^\circ, so the triangle is right-angled with legs in ratio a:ba:b and hypotenuse 2r2r. Write the legs as kaka and kbkb:

(ka)2+(kb)2=(2r)2⇒ k2(a2+b2)=4r2⇒ k2=4r2a2+b2\begin{aligned} &(ka)^2+(kb)^2=(2r)^2\\ &\Rightarrow\ k^2(a^2+b^2)=4r^2\\ &\Rightarrow\ k^2=\frac{4r^2}{a^2+b^2} \end{aligned}
2

Area = half the product of the legs. The legs are perpendicular, so:

Area=12 (ka)(kb)=12 k2ab=12⋅4r2a2+b2⋅ab=2abr2a2+b2\begin{aligned} &\text{Area}=\tfrac12\,(ka)(kb)=\tfrac12\,k^2 ab\\ &=\tfrac12\cdot\frac{4r^2}{a^2+b^2}\cdot ab\\ &=\frac{2abr^2}{a^2+b^2} \end{aligned}
Area=2abr2a2+b2\text{Area}=\frac{2abr^{2}}{a^{2}+b^{2}}
CAT 2023 Slot 2 QA Q17: A triangle is drawn with its vertices on the circle C such that one of its sides is a diameter of C and the ot — Solution | TheCATExam