Question

Two adjacent sides of a parallelogram are $10cm$ and $12cm$. If one diagonal of it is $16cm$ long; find area of the parallelogram. Also, find distance between its shorter sides.

• A. $119.8c{m}^2;11.98cm$
• B. $119.9c{m}^2;11.20cm$
• C. $117.6c{m}^2;10.28cm$
• D. $120.8c{m}^2;15.98cm$

Answer 1

The correct option is A $119.8c{m}^2;11.98cm$

Given, $AB=12cm$, $BC=10cm$
$BD$ is the diagonal of parallelogram $ABCD$.
Also $BD=16cm$
We know, $△ABD\cong △DCB$ $[S-S-S\text{Congruence}]$
Also, area of parallelogram $ABCD$ $=$ Area of $△ABD$ $+$ Area of $△DCB$
$=$ $2\times$ Area of $△ABD$
$=$ $2\sqrt{s(s-12)(s-10)(s-16)}$
$s=\frac{12+10+16}{2}$
$=$ $19$
Therefore, area of parallelogram $ABCD=$ $2\sqrt{19(19-12)(19-10)(19-16)}$
$=$ $2\sqrt{3591}$
$=$ $2\left(59.92\right)$
$=$ $119.8c{m}^2$
Also, $h$ is the distance of the two shorter sides.
We know, area of parallelogram $=$ $\text{base}\times \text{height}$
$=$ $BC\times h$
Therefore, $h=\frac{\text{Area of parallelogram}ABCD}{BC}$
$\Rightarrow h=\frac{119.8}{10}$
$\Rightarrow h=11.98cm$