CAT 2024 Slot 3 · Quant · Question 4

Let the radius be R and let the chord be AB, subtending {120}^{∘} at the centre O.

Dropping a perpendicular from O to AB splits the isosceles triangle OAB into two right triangles, each with a {60}^{∘} angle at O.

So half the chord is R\sin{60}^{∘}=\frac{R\sqrt{3}}{2}, making the full chord R\sqrt{3}.

Setting R\sqrt{3}=10\sqrt{3} gives R=10.

The chord cuts the circle into a minor segment and a major segment; the smaller region is the minor one, standing on the {120}^{∘} arc.

That minor segment is the {120}^{∘} sector with triangle OAB removed.

The sector area is \frac{120}{360}\times \pi{R}^{2}=\frac{\pi \times 100}{3}=\frac{100\pi}{3}.

The triangle area is \frac{1}{2}{R}^{2}\sin{120}^{∘}=\frac{1}{2}\times 100\times \frac{\sqrt{3}}{2}=25\sqrt{3}.

The smaller region is \frac{100\pi}{3}-25\sqrt{3}.

Taking 25 out as a common factor, this is 25\left( \frac{4\pi}{3}-\sqrt{3}\right), the third option.

Shortcut. A chord subtending {120}^{∘} always equals R\sqrt{3} — the same ratio as the side of an inscribed equilateral triangle — so R=10 is read off instantly. Also, the smaller region must be less than the sector, which eliminates both options carrying a plus sign.