Question

In figure 7.43,A is the centre of the circle.angle ABC =45° and AC=7√2cm. Find the area of segment BXC.​

Answer 1

AB=AC(RADIUS)

therefore,ABC=ACB

In triangle ABC

A+ABC+ACB=180°(angle sum property)

BAC+45°+45°=180°

BAC=180-90°=70

now in sector

ar=theta/360°×πr²

= 70/360×22/7×7√2×7√2=59.88cm²

draw an altitude from A to BC say AE

in triangle ABE

cos45°=base/hypo=BE/AB

1/√2=BE/7√2

BE= 3.5cm

therefore AE=9.2cm by Pythagoras

as BE =EC altitude of an isoceles triangle acts like a median

BC=7cm

also in triangle ABE

ar =½×Base×height

=½×3.5× 9.2=16.1cm²

ar of triangle ABC=2×arABE=2×16.1=32.2cm²

ar of segment BXC=ar of sector-ar of triangle ABC =59.88-32.2

=27.68cm²