Find the area of the shaded region in fig.12.19,if PQ=24cm,PR=7cm and O is the centre of the circle.
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AngleQPR=90 (Traingle formed in the semi circle is of 90°
IN Traingle QPR
By pythagoras theorem
H^2=P^2+B^2
H^2=(7)^2+(24)^2
H^2=49+576
H^2=625
H=√625
H=25cm
Diameter of circle =25cm
Radius of circle= 25/2
Area of Semi circle=pie(r)^2
=1/2×22/7×25/2×25/2
=245.5
Area of traingle= 1/2×b×h
=1/2×24×7
= 12×7
=84
Area of shaded region = Area of semicircle -Area of traingle
=245.5-84
=161.5 cm
IN Traingle QPR
By pythagoras theorem
H^2=P^2+B^2
H^2=(7)^2+(24)^2
H^2=49+576
H^2=625
H=√625
H=25cm
Diameter of circle =25cm
Radius of circle= 25/2
Area of Semi circle=pie(r)^2
=1/2×22/7×25/2×25/2
=245.5
Area of traingle= 1/2×b×h
=1/2×24×7
= 12×7
=84
Area of shaded region = Area of semicircle -Area of traingle
=245.5-84
=161.5 cm
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