Question

An isosceles trapezium has an area of $36c{m}^2$, the parallel sides are $12cm$ and $6cm$ respectively. The perimeter of the trapezium is

• A. $28cm$
• B. $32cm$
• C. $24cm$
• D. $20cm$

Answer 1

The correct option is A

$28cm$

The area of a trapezium is given by:
$\frac{1}{2}\times \text{Sum of parallel sides}\times \text{height}$
$\Rightarrow 36=\frac{1}{2}\times (12+6)\times h$
$\Rightarrow 36=\frac{18h}{2}$

$\Rightarrow h=\frac{36\times 2}{18}=4cm$
So, now in the trapezium:
$h=4cm$
$a=b=6cm$

In $△AFB$ and $△DEC$,

$AB=CD$ [As the trapezium is isosceles]
$\therefore AB+a+CD=12cm$

$\Rightarrow 2AB+a=12cm$
$\Rightarrow 2AB=12-6$
$\Rightarrow AB=\frac{6}{2}=3cm=CD$
Applying Pythagoras Theorem in $△ABF$,

$A{B}^2+h^2=d^2$​
$\Rightarrow d^2=3^2+4^2=25$​
$\Rightarrow d=5cm$
As the trapezium is isosceles, the slant sides of the trapezium are equal in length
$\Rightarrow d=c=5cm$
$\therefore$ the perimemter of the trapezium
$=(2\times AB)+a+c+b+d$
$=(2\times 3)+6+5+6+5=6+22=28cm$