CAT 2023 Slot 2 · Quant · Question 18

An angle inscribed in a semicircle is a right angle, so the vertex opposite the diameter is a right angle.

That makes the diameter the hypotenuse and the other two sides the legs.

The legs are in the ratio a:b, so write them as ak and bk for some positive k.

The hypotenuse is the diameter, of length 2r.

By Pythagoras, {\left( ak \right)}^{2}+{\left( bk \right)}^{2}={\left( 2r \right)}^{2}.

Factoring the left side, {k}^{2}\left({a}^{2}+{b}^{2}\right)=4{r}^{2}.

So {k}^{2}=\frac{4{r}^{2}}{{a}^{2}+{b}^{2}}.

The area of a right triangle is half the product of its legs: \frac{1}{2}\times ak\times bk=\frac{ab{k}^{2}}{2}.

Substituting, the area is \frac{ab}{2}\times \frac{4{r}^{2}}{{a}^{2}+{b}^{2}}=\frac{2ab{r}^{2}}{{a}^{2}+{b}^{2}}.

That is the fourth option.

Shortcut. Sanity-check with a familiar case: a:b=3:4 in a circle of radius r=2.5 gives legs 3 and 4 and area 6. The formula returns \frac{2\times 12\times 6.25}{25}=6, confirming option 4 and eliminating the rest at a stroke.