TIFR CSE 2024 | Part A | Question: 1

Question

If $\text{O}$ is the center of the circle, what is the area of the shaded portion in square cm?

• A. $4 \pi-4 \sqrt{2}$
• B. $\frac{7}{2} \pi-4 \sqrt{2}$
• C. $\frac{7}{2} \pi-4 \sqrt{3}$
• D. $\frac{8}{3} \pi-4 \sqrt{3}$
• E. $\frac{8}{3} \pi-4 \sqrt{2}$

   

Answer 1

We can make this into an equilateral triangle by joining O and other end of the chord. Since the angle between side of an equilateral triangle is $60^{\circ}$, we can fit $6$ triangle side by side (as a hexagon), so there will be $6$ shaded regions and $6$ triangle within the circle.

Let us assume the area of shaded region be $A$.

Then, Area of circle = $6\cdot A + 6\cdot$Area of equilateral triangles

$\implies6\cdot A =$  Area of circle – $6 \cdot$Area of equilateral triangles

$\implies 6\cdot A = \pi\cdot (4)^2 - 6\cdot \dfrac{\sqrt{3}}{4}\cdot (4)^2$

$\implies6\cdot A = \pi\cdot 16 - 24\cdot \dfrac{\sqrt{3}}{4}$

$\implies A = \dfrac{\pi\cdot 16}{6} - \dfrac{24\cdot \sqrt{3}}{6}$

$\implies A = \dfrac{8\pi}{3} - 4\sqrt{3}$

 

Or we can directly calculate the area of that sector by taking the proportion of area of circle as, $\dfrac{60}{360}\cdot \pi \cdot 4^2$

Aread of sector$= \dfrac{8\pi}{3}$

Area of equilatera triangle $=\dfrac{\sqrt{3}}{4}4^2 = 4\sqrt{3}$

Area of shaded region $=\dfrac{8\pi}{3} - 4\sqrt{3}$