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Areas of Parallelograms and Triangles — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsAreas of Parallelograms and Triangles1 Mark
Question
Write True or False and justify your answer: In the figure, $ABCD$ and $EFGD$ are two parallelograms and $G$ is the mid-point of $CD.$ Then, $\text{ar}(\triangle\text{DPC})=\frac{1}{2}\text{ar}(\text{EFGD}).$ 
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Answer

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, $\text{ar}(\triangle\text{DPG})=\text{ar}(\triangle\text{GPC})=\frac{1}{2}\text{ar}(\triangle\text{DPC})$
 Also, we know that, if a parallelogram and a triangle lie on the same base and between the same parallel, then area of triangle is equal to triangle is equal to half of the area of parallelogram.
 Here, parallelogram $FEGD$ and $\triangle\text{DPG}$ lie on the same base $DG$ and between the same parallel $DG$ and $EF.$
So, $\text{ar}(\triangle\text{DPG})=\frac{1}{2}\text{ar}(\text{EFGD})$ From eqs. $(i)$ and $(ii),$
$\frac{1}{2}\text{ar}(\triangle\text{DPC})=\frac{1}{2}\ \text{ar}(\text{EFGD})$
$\Rightarrow\text{ar}(\triangle\text{DPC})=\text{ar}(\text{EFGD})$

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