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