Question

Find the difference of the areas of the two segments of a circle formed by a chord of length 5cm subtending an angle of 90degree at the centre

Answer 1

Let r be the radius of the circle and AB be the chord, which subtend angle of 90 at centre O. AB = 5 cm in the right angled triangle OAB, using Pythagoras theorem:
$O{A}^2+O{B}^2=A{B}^2$
$r^2+r^2=52$
$2r^2=25$
$r=\frac{5}{\sqrt{2}}cm$
Therefore area of the circle =
$\pi r^2=\frac{22}{7}\times \frac{25}{2}=39.2857c{m}^2$
The area of the minor segment (shaded area) = area of the sector OAB - area of the triangle OAB area of the sector OAB =
$\pi r^2\times \frac{90}{360}=39.2857\times \frac{1}{4}=9.8214c{m}^2$
Area of triangle OAB =
$\frac{1}{2}\times OA\times OB=\frac{1}{2}{r}^2=\frac{1}{2}\times \frac{25}{2}=\frac{25}{4}=6.25c{m}^2$
The area of minor segment = 9.8214 - 6.25 = 3.5714 sq cm ……(1)
The area of the major segment = area of the circle - area of the minor segment
= 39.2857 - 3.5714
= 35.7143 sq cm ….(2)
Difference of the areas of two segments = 35.7143 - 3.5714
= 32.1429 $c{m}^2$