Class 9MathematicsAreas of Parallelograms and TrianglesTextbook - Areas of Parallelograms and Triangles
DifficultIn the figure,$ABCD$ is a parallelogram and $EFCD$ is a rectangle. Also,$AL \perp DC$. Prove that:
$(i)$ $\text{ar}(ABCD) = \text{ar}(EFCD)$
$(ii)$ $\text{ar}(ABCD) = DC \times AL$
$(i)$ $\text{ar}(ABCD) = \text{ar}(EFCD)$
$(ii)$ $\text{ar}(ABCD) = DC \times AL$
(N/A) $(i)$ Since a rectangle is also a parallelogram,and both $ABCD$ and $EFCD$ are on the same base $DC$ and between the same parallels $DC$ and $EF$,therefore,$\text{ar}(ABCD) = \text{ar}(EFCD)$.
$(ii)$ From the result above,$\text{ar}(ABCD) = \text{ar}(EFCD)$.
Since $EFCD$ is a rectangle,its area is $\text{length} \times \text{breadth} = DC \times FC$.
Thus,$\text{ar}(ABCD) = DC \times FC$ $(1)$.
Since $AL \perp DC$ and $EF \parallel DC$,$AL$ is the height of the parallelogram. In the rectangle $AFCL$ (or by considering the parallel lines),we have $AL = FC$ $(2)$.
Substituting $(2)$ in $(1)$,we get $\text{ar}(ABCD) = DC \times AL$.
$(ii)$ From the result above,$\text{ar}(ABCD) = \text{ar}(EFCD)$.
Since $EFCD$ is a rectangle,its area is $\text{length} \times \text{breadth} = DC \times FC$.
Thus,$\text{ar}(ABCD) = DC \times FC$ $(1)$.
Since $AL \perp DC$ and $EF \parallel DC$,$AL$ is the height of the parallelogram. In the rectangle $AFCL$ (or by considering the parallel lines),we have $AL = FC$ $(2)$.
Substituting $(2)$ in $(1)$,we get $\text{ar}(ABCD) = DC \times AL$.
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