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  • Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.

Q: 1 Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.

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We have $\| g m \mathrm{ABCD}$ and a rectangle $A B E F$ both lie on the same base $A B$ such that, $\boldsymbol{a r}(\| g m \mathrm{ABCD})=\boldsymbol{a r}(\mathrm{ABEF})$ for rectangle, $\mathrm{ABEF}$
and for $\| g m, \mathrm{ABCD}$
$\Rightarrow C D=E F$
$\Rightarrow A B+C D=A B+E F$
Since $\triangle B E C$ and $\triangle A F D$ are right-angled triangles
Therefore, $A D>A F$ and $B C>B E$
$\Rightarrow(B C+A D)>(A F+B E)$
Adding equation (i) and (ii), we get
$(A B+C D)+(B C+A D)>(A B+B E)+(A F+B E)$
$Rightarrow(A B+B C+C D+D A)>(A B+B E+E F+F A)$
Hence proved, that the perimeter of ||gm $ABCD$ is greater than the perimeter of the rectangle $ABEF$.

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manish painkra

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