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Question
In parallelogram ABCD, P is a point on side AB and Q is a point on side BC.
Prove that:
(i) ΔCPD and ΔAQD are equal in the area.
(ii) Area (ΔAQD) = Area (ΔAPD) + Area (ΔCPB)
Solution
Given ABCD is a parallelogram. P and Q are any points on the sides AB and BC respectively, join diagonals AC and BD.
proof:
(i) since triangles with the same base and between the same set of parallel lines have equal areas
area ( CPD ) = area( BCD ) …… (1)
again, diagonals of the parallelogram bisect area in two equal parts
area ( BCD ) = ( 1/2 ) area of parallelogram ABCD …… (2)
from (1) and (2)
area( CPD ) = 1/2 area( ABCD ) …… (3)
similarly area ( AQD ) = area( ABD ) = 1/2 area( ABCD )…… (4)
from (3) and (4)
area( CPD ) = area( AQD ),
hence proved.
(ii) We know that area of triangles on the same base and between same parallel lines are equal
So Area of AQD= Area of ACD= Area of PDC = Area of BDC = Area of ABC=Area of APD + Area of BPC
Hence Proved
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