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Question
In a parallelogram ABCD, point P lies in DC such that DP: PC = 3:2. If the area of ΔDPB = 30 sq. cm.
find the area of the parallelogram ABCD.
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Solution
The ratio of the area of triangles with the same vertex and bases along the same line is equal to the ratio of their respective bases. So, we have
`"Area of DPB"/"Area of PCB" = "DP"/"PC" = 3/2`
Given: Area of ΔDPB = 30 sq. cm
Let 'x' be the area of the triangle PCB
Therefore, We have,
⇒ `30/x = 3/2`
⇒ x = `30/3 xx 2` = 20 sq.cm.
So area of ΔPCB = 20 sq. cm
Consider the following figure.
From the diagram, it is clear that,
Area( ΔCDB ) = Area( ΔDPB ) + Area( ΔCDB )
= 30 + 20 = 50 sq.cm.
The diagonal of the parallelogram divides it into two triangles ΔADB and ΔCDB of equal area.
Therefore,
Area( parallelogram ABCD ) = 2 x ΔCDB = 2 x 50 = 100 sq.cm.
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