Question

In the given figure, A is the center of the circle. ∠ABC = 45^° and AC = `7 sqrt 2` cm. Find the area of segment BXC.

Answer 1

The radius (r) of the circle = AC = `7 sqrt 2` cm.

m(arc BXC) = ∠BAC = θ = 45°.

A(segment BXC) = `r^2[(pi theta)/360 - (sin theta)/2]`

= `(7 sqrt 2)^2 xx [(3.14 xx 45)/360 - (sin 45^\circ)/2]`

= `49 xx 2 [3.14/8 - (1/sqrt2)/2]`    ...`[\because sin 45^\circ = 1/sqrt2]`

= `98 xx [3.14/8 - 1/(2 sqrt 2)]`

= `98 xx [3.14/8 - (1 xx 2 sqrt 2)/(2 sqrt 2 xx 2 sqrt 2)]`

= `98 xx [3.14/8 - (2 sqrt 2)/8]`

= `98 xx [3.14/8 - (2 xx 1.41)/8]`

= `98 xx [(3.14 - 2.82)/8]`

= `98 xx [0.32/8]`

= `31.36/8`

= 3.92 cm²

A(segment BXC) = 3.92 cm².