# ISI2017-DCG-2

78 views

The area of the shaded region in the following figure (all the arcs are circular) is

1. $\pi$
2. $2 \pi$
3. $3 \pi$
4. $\frac{9}{8} \pi$

recategorized

It's like semicircle of radius 1 unit is cut from the semicircle of radius 2 units and joined upside down to the semicircle of 2 units. So, the shaded area will be area of the semicircle with 2 units.

Hence, Shaded Area   $= \frac{1}{2}\cdot\pi \cdot (2^2) = 2 \pi$

yes the area of the shaded portion will be 2 π  as the area of the big semicircle is 2 π  and π/2 has been cut in the RHS of the big semi circle and appended in the LHS side below. Hence the area reamins the same which is equal to 2 π.and the option  B is correct.

## Related questions

1
56 views
The area (in square unit) of the portion enclosed by the curve $\sqrt{2x}+ \sqrt{2y} = 2 \sqrt{3}$ and the axes of reference is $2$ $4$ $6$ $8$
If $a,b,c$ are the sides of a triangle such that $a:b:c=1: \sqrt{3}:2$, then $A:B:C$ (where $A,B,C$ are the angles opposite to the sides of $a,b,c$ respectively) is $3:2:1$ $3:1:2$ $1:2:3$ $1:3:2$
If $a,b,c$ are the sides of $\Delta ABC$, then $\tan \frac{B-C}{2} \tan \frac{A}{2}$ is equal to $\frac{b+c}{b-c}$ $\frac{b-c}{b+c}$ $\frac{c-b}{c+b}$ none of these
The angle between the tangents drawn from the point $(-1, 7)$ to the circle $x^2+y^2=25$ is $\tan^{-1} (\frac{1}{2})$ $\tan^{-1} (\frac{2}{3})$ $\frac{\pi}{2}$ $\frac{\pi}{3}$