# The adjacent sides of a parallelogram $ABCD$ measures $34\ cm$ and $20\ cm$, and the diagonal $AC$ measures $42\ cm$. Find the area of the parallelogram.

Given:

The adjacent sides of a parallelogram $ABCD$ measures $34\ cm$ and $20\ cm$, and the diagonal $AC$ measures $42\ cm$.

To do:

We have to find the area of the parallelogram.

Solution:

In parallelogram $ABCD$,

$AB = 34\ cm, BC = 20\ cm$ and $AC = 42\ cm$

The diagonal of a parallelogram divides into two triangles equal in area.

Area of $\triangle ABC$,

$s=\frac{a+b+c}{2}$

$=\frac{34+20+42}{2}$

$=\frac{96}{2}$

$=48$

Area of the triangle $=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{48(48-34)(48-20)(48-42)}$

$=\sqrt{48 \times 14 \times 28 \times 6}$

$=\sqrt{4 \times 4 \times 3 \times 2 \times 7 \times 7 \times 2 \times 2 \times 2 \times 3}$

$=7 \times 4 \times 3 \times 2 \times 2$

$=336 \mathrm{~cm}^{2}$

Area of parallelogram $\mathrm{ABCD}=2 \times$ area of triangle

$=2 \times 336$

$=672 \mathrm{~cm}^{2}$.

Updated on: 10-Oct-2022

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