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In figure,ABPC is a quadrant of a circle of radius 14cm and a semi circle is drawn with BC as diameter. Find the area of shaded region. |
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Answer» ong>Step-by-step explanation: in right angle TRIANGLE ABC by using PYTHAGORAS therum , =>AC^2+AB^2 =BC ^2 =>14^2+14^2=BC^2 =>196+196=BC^2 =>BC=√392 =>BC=14√2 diameter of the semi circle =14√2 =>2r=14√2 =>r=14√2=7√2 cm area of the SEGMENT BPOC= area of quadrant - area of triangle =πr^2/4-1/2* base *HEIGHT =1/4*22/7*14*14-1/2*14*14 =154-98 =56 cm sq area of shaded region =area of semi circle -area of segment =πr^2/2-56 cm =1/2*22/6*7√2*7√2-56 =154-56 =98 cm sq |
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