Question

A chord 10 cm long is drawn in a circle whose radius is 5√2 cm. Find the area of both
segments

Answer 1

Given radius = r = `5sqrt(2)` cm = OA = OB

Length of chord AB = 10cm

In ΔOAB, OA = OB =` 5sqrt(2)` 𝑐𝑚 AB = 10cm

`OA^2 + OB^2 = (5sqrt(2))^2+ (5sqrt(2))^2= 5 0 + 50 = 100 = (AB)^2`

Pythagoras theorem is satisfied OAB is right triangle

𝜃 = angle subtended by chord = ∠AOB = 90°

Area of segment (minor) = shaded region

= area of sector - area of `triangle`OAB

=`theta/360× pir^2 −1/2`× 𝑂𝐴 × 𝑂𝐵

=`90/360×22/7(5sqrt(2))^2−1/2× 5sqrt(2) × 5sqrt(2)`

=`275/7− 25 −100/7 cm^2`

Area of major segment = (area of circle) – (area of minor segment)

= `pir^2`2 −`100/7`

=`22/7× (5sqrt(2))^2−100/7`

=`1100/7−100/7`

=`1000/7 cm^2`