two circles each of radius 7cm intersect each other the distance between the centres is 7 root 2 cm find the area common to both the circles
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If you draw this, there is a 7 sqrt(2) distance between the centers and the radius goes to the two points of intersection. Take the upper point. That is 7 on both sides with 7 sqrt(2) for the bottom, an isosceles triangle. If the radius is the hypotenuse, then we have sides of (7/2) sqrt(2) on both circles to give a distance between of 7 sqrt(2). But 7/sqrt(2)=7 sqrt(2)/2 is a 45-45-90 right triangle. Therefore the distance between the middle or widest part and the point of intersection is 7/2* sqrt(2). That makes the area of the whole triangle (1/2)(7 sqrt(2))*(7/2) sqrt(2)=49/2 cm^2.
The combined area of both sectors including the intersection area is (1/8) of each circle, since the arc is 45 degrees. That makes it 49*pi/8 *2 or 49*pi/4 cm^2 for the combined area.
Numerically, this is 38.48 cm^2, whereas the area of the whole triangle, which has overlap, is 24.5 cm^2. This makes the overlap 13.48 cm^2 for half of the the intersection area.
Double it, and the total area is 26.96 cm^2.
The combined area of both sectors including the intersection area is (1/8) of each circle, since the arc is 45 degrees. That makes it 49*pi/8 *2 or 49*pi/4 cm^2 for the combined area.
Numerically, this is 38.48 cm^2, whereas the area of the whole triangle, which has overlap, is 24.5 cm^2. This makes the overlap 13.48 cm^2 for half of the the intersection area.
Double it, and the total area is 26.96 cm^2.
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