If $P$ is any point in the interior of a parallelogram $ABCD$, then prove that area of the triangle $APB$ is less than half the area of parallelogram.

Given:

$P$ is any point in the interior of a parallelogram $ABCD$.

To do:

We have to prove that area of the triangle $APB$ is less than half the area of parallelogram.

Solution:

Join $AP$ and $BP$.
Draw $DN \perp AB$ and $PM \perp AM$.

$\operatorname{area}(parallelogram \mathrm{ABCD})=\mathrm{AB} \times \mathrm{DN}$..........(i)

$\operatorname{area}(\Delta \mathrm{APB})=\frac{1}{2} \mathrm{AB} \times \mathrm{PM}$...............(ii)

From (i) and (ii), we get,

$\mathrm{DN}>\mathrm{PM}$ or $\mathrm{PM}

$\mathrm{AB} \times \mathrm{PM}

$\frac{1}{2} \mathrm{AB} \times \mathrm{PM}

$\operatorname{ar}(\Delta \mathrm{PAB})

Hence proved.

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Updated on: 10-Oct-2022

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