The area of a trapezium is 91 cm

Question:

The area of a trapezium is 91 cm2 and its height is 7 cm. If one of the parallel sides is longer than the other by 8 cm, find the two parallel sides.

Solution:

Given:

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

Height $=7 \mathrm{~cm}$

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

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

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

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

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

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

We can rewrite it as follows:

$7 \times[8+2 \mathrm{x}]=182$

$\Rightarrow[8+2 x]=\frac{182}{7}=26$

$\Rightarrow 8+2 x=26$

$\Rightarrow 2 x=26-8=18$

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

$\therefore$ Length of the shorter side of the trapezium $=9 \mathrm{~cm}$

And, length of the longer side $=8+\mathrm{x}=8+9=17 \mathrm{~cm}$

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