Question

Find the area of the major segment $APB$, in figure, of a circle of radius $35cm$ and $\angle AOB={90}^0$.

Answer 1

radius of circle = $35cm$
see attachment,
you can say that , major segment is the sum of right angled triangle and $3/4$ th part of circle { as you know, centre of circle is ${360}^o$ , means area of complete circle is $\pi r^2$ , means for ${360}^o$ , area of circle $=\pi r^2$, so, for ${270}^o$ , area of circular part = ${270}^o/{360}^o\pi r^2=3/4\pi r^2$
area of major segment = area of triangle $+3/4$ area of circle
$=1/2height\times base+3/4\pi r^2$
$=1/2\times r\times r+3/4\pi r^2$
$=r^2/4(2+3\pi )$
$=r^2/4\times (2+66/7)=35\times 35/4\times 80/7$
$=35\times 5/4\times 80$
$=35\times 5\times 20$
$=3500c{m}^2$
Hence, area of major segment is $3500c{m}^2$