71@Aditya. Yes I have done that. But that is troublesome (especially if you don't have time)
@Nishant Sir, Yes Sir! I was wondering for the same too. I'll try to scratch my head hard,to see if something comes.
6 Answers
262i think it would be better to use integration .
construct two eqns. of circles with radius 2 such that they pass through each other's centers.
eg . x2+y2=4 and (x-2)2+y2=4.
find their point of intersection.
integrate each one with proper limits.
62A far far far better method is pure geometry..
A bit of trigo
Hint: Area of 2 secants :)
71
11
It can be easily done after this by finding the area of the two secants.
2(\frac{\pi r^{2}}{3} +\frac{1}{2}r^{2}sin120')
r=2
1The points of intersection at (-sqrt(3), 1) and (sqrt(3), 1)
the length of the common chord is therefore 2*sqrt(3)
using Pythagoras theorem, the perpendicular distance from the center to the chord = 1
area of triangle ADC = 1/2 * 2*sqrt(3) * 1 = sqrt(3)
area of sector ADC = 1/6 * pi * r^2 = 0.67 *pi
area of the shaded area = 2* (0.67*pi - sqrt(3)) = 0.74 units (using symmetry)