plss solve this my two second question for slc..please Its urgent ..I will mark u brainlist
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Angle PAQ = 90 ( ANGLE IN SEMI CIRCLE) . PR parallel to AQ so RPQ = PQA = 30. now in triangle PAQ, QPA + PAQ + AQP =180 ( ASP). SO , QPA = 60., QPA = OAP = 60( BECAUSE there are radius , so angle opp. to equal sides) so OAP = 60
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if PR||AQ and RPQ=30°then OAP is 60°
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1) Diagonal of square =root2×side
4root3=root2×side
4root3/root2=side
2root2×root3=side
2root6=side
now area of triangleABC=1/2×BC×AB=1/2×2root6×2root6=12cm^2
Now area of triangleABC=area of triangleBCE ( coz they both lie between same parallel line)
so area of triangleBCE=12cm^2
hope this will be helpful to u :-)
4root3=root2×side
4root3/root2=side
2root2×root3=side
2root6=side
now area of triangleABC=1/2×BC×AB=1/2×2root6×2root6=12cm^2
Now area of triangleABC=area of triangleBCE ( coz they both lie between same parallel line)
so area of triangleBCE=12cm^2
hope this will be helpful to u :-)
why Diagonal of square is equal to root2 n which side and is root2side formula
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