Class 9MathematicsAreas of Parallelograms and TrianglesTextbook - Areas of Parallelograms and Triangles
Medium$P$ and $Q$ are any two points lying on the sides $DC$ and $AD$ respectively of a parallelogram $ABCD$. Show that $\text{ar}(APB) = \text{ar}(BQC)$.
(N/A) Given: $ABCD$ is a parallelogram.
Since $ABCD$ is a parallelogram,we have $AB \parallel CD$ and $BC \parallel AD$.
Now,$\Delta APB$ and parallelogram $ABCD$ are on the same base $AB$ and between the same parallels $AB$ and $CD$.
Therefore,$\text{ar}(\Delta APB) = \frac{1}{2} \text{ar}(\text{parallelogram } ABCD) \quad \dots(1)$
Also,$\Delta BQC$ and parallelogram $ABCD$ are on the same base $BC$ and between the same parallels $BC$ and $AD$.
Therefore,$\text{ar}(\Delta BQC) = \frac{1}{2} \text{ar}(\text{parallelogram } ABCD) \quad \dots(2)$
From $(1)$ and $(2)$,we have:
$\text{ar}(\Delta APB) = \text{ar}(\Delta BQC)$
Since $ABCD$ is a parallelogram,we have $AB \parallel CD$ and $BC \parallel AD$.
Now,$\Delta APB$ and parallelogram $ABCD$ are on the same base $AB$ and between the same parallels $AB$ and $CD$.
Therefore,$\text{ar}(\Delta APB) = \frac{1}{2} \text{ar}(\text{parallelogram } ABCD) \quad \dots(1)$
Also,$\Delta BQC$ and parallelogram $ABCD$ are on the same base $BC$ and between the same parallels $BC$ and $AD$.
Therefore,$\text{ar}(\Delta BQC) = \frac{1}{2} \text{ar}(\text{parallelogram } ABCD) \quad \dots(2)$
From $(1)$ and $(2)$,we have:
$\text{ar}(\Delta APB) = \text{ar}(\Delta BQC)$
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