Question
The given figure shows the parallelograms $\text{ABCD}$ and $\text{APQR}$.Show that these parallelograms are equal in the area.$[$Join $B$ and $R]$
✓
Answer
Join $B$ and $R$ and $P$ and $R.$
We know that the area of the parallelogram is equal to twice the area of the triangle if the triangle and the parallelogram are on the same base and between the parallels.
Consider $\text{ABCD}$ parallelogram:Since the parallelogram $\text{ABCD}$ and the $\triangle ABR$ lie on $AB$ and between the parallels $AB$ and $DC,$ we have
Area$(\square ABCD )=2 \times$ Area$(\triangle ABR )\dots ....(1)$
We know that the area of triangles with the same base and between the same parallel lines are equal.
Since the triangles $\text{ABR}$ and $\text{APR}$ lie on the same base $AR$ and between the parallels $AR$ and $QP,$ we have,
Area $( ΔABR ) =$ Area $( ΔAPR ) \dots....(2)$
From equations $(1)$ and $(2)$, we have,
Area$(\square ABCD )=2 \times$ Area$(\triangle APR ) \dots.....(3)$
Also, the $\triangle APR$ and the parallelograms, $AR$ and $QR,$ lie on the same base $AR$ and between the parallels, $AR$ and $QP,$
Area($\triangle APR )=\frac{1}{2} \times$ Area$(\square ARQP ) \dots....(4)$
Using $(4)$ in equation $(3),$ We have,
Area$(\square ABCD )=2 \times \frac{1}{2} \times$ Area$(\square ARQP )$
Area$(\square ABCD )=$Area$(\square ARQP )$
Hence Proved.
We know that the area of the parallelogram is equal to twice the area of the triangle if the triangle and the parallelogram are on the same base and between the parallels.
Consider $\text{ABCD}$ parallelogram:Since the parallelogram $\text{ABCD}$ and the $\triangle ABR$ lie on $AB$ and between the parallels $AB$ and $DC,$ we have
Area$(\square ABCD )=2 \times$ Area$(\triangle ABR )\dots ....(1)$
We know that the area of triangles with the same base and between the same parallel lines are equal.
Since the triangles $\text{ABR}$ and $\text{APR}$ lie on the same base $AR$ and between the parallels $AR$ and $QP,$ we have,
Area $( ΔABR ) =$ Area $( ΔAPR ) \dots....(2)$
From equations $(1)$ and $(2)$, we have,
Area$(\square ABCD )=2 \times$ Area$(\triangle APR ) \dots.....(3)$
Also, the $\triangle APR$ and the parallelograms, $AR$ and $QR,$ lie on the same base $AR$ and between the parallels, $AR$ and $QP,$
Area($\triangle APR )=\frac{1}{2} \times$ Area$(\square ARQP ) \dots....(4)$
Using $(4)$ in equation $(3),$ We have,
Area$(\square ABCD )=2 \times \frac{1}{2} \times$ Area$(\square ARQP )$
Area$(\square ABCD )=$Area$(\square ARQP )$
Hence Proved.
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