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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.
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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