Question

A chord of a circle subtends an angle 𝜃 at the centre of circle. The area of the minor segment cut off by the chord is one eighth of the area of circle. Prove that 8 sin`theta/2 "cos"theta/2+pi =(pitheta)/45`

Answer 1

Let radius of circle = r

Area of circle = 𝜋𝑟²

AB is a chord, OA, OB are joined drop OM ⊥ AB. This OM bisects AB as well as ∠AOB.

∠AOM = ∠MOB =`1/2(0) =theta/2`                        AB = 2AM

In ΔAOM, ∠AMO = 90°

`"Sin"theta/2=(AM)/(AD)⇒ AM = R."sin"theta/2`         AB = 2R sin`theta/2`

`"Cos"theta/2=(OM)/(AD)⇒ OM = R"cos"theta/2`

Area of segment cut off by AB = (area of sector) – (area of triangles)

=`theta/360× pir^2 −1/2`× 𝐴𝐵 × 𝑂𝑀

`= r^2 [(pitheta)/360^@−1/2. 2"rsin"theta/2. R" cos"theta/2]`

`= R^2 [(pitheta)/360^@− "sin"theta/2. "cos"theta/2]`

Area of segment =`1/2`(𝑎𝑟𝑒𝑎 𝑜𝑓 𝑐𝑖𝑟𝑐𝑙𝑒)

`r^2 [(pitheta)/360− "sin"theta/2." cos"theta/2] =1/8pir^2`

`(8pitheta)/360^@− 8 "sin"theta/2. "cos"theta/2= pi`

`8 "sin"theta/2. "cos"theta/2+ pi =(pitheta)/45`