Question

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

• A. $119.8c{m}^2;11.42$ cm
• B. $136.74c{m}^2;19.58$ cm
• C. $127cm;5.85$ cm
• D. None of these

Answer 1

The correct option is D None of these

Given : $AB=12$ cm, $BC=10$ cm
$BD$ is the diagonal of parallelogram $ABCD$ and $BD=16$ cm
In $△ABD$ and $△DCB$

$AB=CD$ ......... [opposite sides of a parallelogram]

$AD=BC$ ......... [opposite sides of a parallelogram]

$BD=BD$ ......... [common side]
$\therefore △ABD\cong △DCB$ ...... [By SSS Congruence]

Area of parallelogram $ABCD=$ Area of $△ABD+$Area of $△DCB$
$=2A\left(△ABD\right)$
$=2\sqrt{s(s-12)(s-10)(s-16)}$

Here, $s=\frac{\text{Perimeter of ABD}}{2}=\frac{12+10+16}{2}$
$=19$
$\therefore$ Area of parallelogram $ABCD=2\sqrt{19(19-12)(19-10)(19-16)}$
$=2\sqrt{3591}$
$=2\left(59.92\right)$
$=119.8$ sq. cm
Let $h$ be the distance of the two shorter sides.
We know, Area of parallelogram $=\text{base}\times \text{height}$
$=BC\times h$
$\therefore h=\frac{A\left(\square ABCD\right)}{BC}$
$⟹h=\frac{119.8}{10}$
$⟹h=11.98$ cm

Hence, option D is correct.