CAT 2024 Quant was dominated by Algebra followed by Arithmetic. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.
Question 4 A circular plot of land is divided into two regions by a chord of length \(10 \sqrt{3}\) meters such that the chord subtends an angle of \(120^{\circ}\) at the center. Then, the area, in square meters, of the smaller region is
\[
O A=O B=\text { Radii; } \angle A O B=120^{\circ}
\]
So,
\[
\angle OAB=\angle OBA=30^{\circ}
\]
Consider \(O M \perp A B\);
In right triangle \(\triangle OMB\),
\[
\begin{array}{l}
\cos 30^{\circ}=\frac{M B}{O B} \\
\frac{\sqrt{3}}{2}=\frac{5 \sqrt{3}}{O B} \\
\Rightarrow O B=10
\end{array}
\]

\[
\begin{array}{l}
\sin 30^{\circ}=\frac{O M}{O B} \\
\frac{1}{2}=\frac{O M}{10} \\
\Rightarrow O M=5
\end{array}
\]
Area of the smaller region \(=\) Area of sector - Area of \(\triangle OMB\)
\[
\begin{array}{l}
=\frac{120}{360} \times \pi \times 10^2-\frac{1}{2} \times 10 \sqrt{3} \times 5 \\
=25\left(\frac{4 \pi}{3}-\sqrt{3}\right)
\end{array}
\]
The question is "A circular plot of land is divided into two regions by a chord of length \(10 \sqrt{3}\) meters such that the chord subtends an angle of \(120^{\circ}\) at the center. Then, the area, in square meters, of the smaller region is"
Choice C is the correct answer.
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