Question

The area of a trapezium is $91c{m}^2$ and its height is $7cm$. If one of the parallel sides is longer than the other by $8cm$, find the sum of the length of two parallel sides (in $cm$).

Answer 1

Given:

Area of the trapezium $=91{cm}^2$

Height $=7cm$

Let the length of the smaller side be $x$.

Then, the length of the longer side will be $8$ more than the smaller side, i.e. $8+x$.

Area of trapezium $=\frac{1}{2}\times$(Sum of the parallel sides)$\times$(Height)

$\Rightarrow 91=\frac{1}{2}\times \left[\right(8+x)+x]\times \left(7\right)$

$\Rightarrow 91=\frac{7}{2}\times [8+x+x]$

$\Rightarrow 91\times 2=7\times [8+2x]$

$\Rightarrow 7\times [8+2x]=182$

$\Rightarrow [8+2x]=\frac{182}{7}$

$\Rightarrow 8+2x=26$

$\Rightarrow 2x=26-8$

$\Rightarrow x=\frac{18}{2}=9cm$

$\therefore$ Length of the shorter side of the trapezium $=9cm$

And, length of the longer side $=8+x=8+9=17cm$

Sum of the length of two parallel sides $=17+9=26cm$