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CAT 2024 Slot 3 QA Question & Solution

GeometryMedium

Question

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° at the center. Then, the area, in square meters, of the smaller region is

Options

$20\left(\cfrac{4 \pi}{3} + \sqrt{3}\right)$
$25\left(\cfrac{4 \pi}{3} + \sqrt{3}\right)$
$20\left(\cfrac{4 \pi}{3} - \sqrt{3}\right)$
$25\left(\cfrac{4 \pi}{3} - \sqrt{3}\right)$

Solution

This is the situation that is described in the question above, Angle AOB is 120 degrees and the chord AB is of length 10cm. 

circle image - geometry diagram

Using Cosine rule we can find the length of AO and AB
$\cos\left(AOB\right)=\frac{\left(r^2+r^2-300\right)}{2r^2}$
$-\frac{1}{2}=\frac{\left(2r^2-300\right)}{2r^2}$
$3r^2=300$
$r=10$

Area of the Sector AOB is $\frac{120}{360}\times\ \pi\ \times\ r^2$
$\frac{1}{3}\times\ \pi\ \times\ \frac{100}{1}$ which is $\frac{100\pi}{3}$

Area of triangle AOB is $\frac{1}{2}\times\ r^2\times\ \sin\left(120\right)$

$\frac{1}{2}\times\ \frac{100}{1}\times\ \frac{\sqrt{3}}{2}$

Area of triangle AOB is $\frac{75}{\sqrt{3}}$

The smaller region will be, $\frac{100\pi\ }{3}-\frac{75}{\sqrt{3}}$
Taking 25 common we will get, 
$25\left(\cfrac{4 \pi}{3} - \sqrt{3}\right)$