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Question
In the following figure, ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then ar (DPC) = `1/2` ar (EFGD).
Options
True
False
Solution
This statement is False.
Explanation:
In the given figure, join PG.
Since, G is the mid-point of CD.
Thus, PG is a median of ΔDPC and it divides the triangle into parts of equal areas.
Then, ar (ΔDPG) = ar (ΔGPC) = `1/2` ar (ΔDPC) ...(i)
Also, we know that, if a parallelogram and a triangle lie on the same base and between the same parallels, then area of triangle is equal to half of the area of parallelogram.
Here, parallelogram EFGD and ΔDPG lie on the same base DG and between the same parallels DG and EF.
So, ar (ΔDPG) = `1/2` ar (EFGD) ...(ii)
From equations (i) and (ii),
`1/2` ar (ΔDPG) = `1/2` ar (EFGD)
⇒ ar (ΔDPC) = ar (EFGD)
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