1/2 of the chord is r sin(B/2)
So, the area of the triangle is
2 * (1/2 r sin(B/2)(r cos(B/2))
= r^2 sin(B/2)cos(B/2)
= 1/2 r^2 sinB
The area of the whole sector is 1/2 r^2 B
So, the segment's area is as noted above: the sector minus the triangle.
A chord PQ of a circle with radius r subtends an angle B at the center. Show that the area of the minor segment PQ=1/2.r^2.(B-sinB) and write down the area of the major segment PQ interms of r and B.
1 answer