Question

Question 3
The area of a trapezium is 475 $c{m}^2$ and the height is 19cm. Find the lengths of its two parallel sides, of one side is 4cm greater than the other.

Answer 1

Let the one side of a trapezium, DC = x cm
According to the question,
second side, AB = (x+4) cm
We know that,
Area of a trapezium =$\frac{1}{2}\left(Sumofparallelsides\right)\times Distancebetweenparallelsides$
$[\because areaofatrapezium=\frac{1}{2}(AB+DC)\times h]$
$\Rightarrow 475=\frac{1}{2}(x+x+4)\times 19$
$\Rightarrow 475=\frac{(2x+4)\times 19}{2}$
$\Rightarrow \frac{(2x+4)}{2}=\frac{475}{19}$
$\Rightarrow \frac{2(x+2)}{2}=\frac{475}{19}$
$\Rightarrow x+2=\frac{475}{19}$
$\Rightarrow x+2=25$
$\therefore x=23$
$\therefore$ One side of a trapezium, DC = x cm =23cm
and the second side, AB = (x+4) cm = 23 + 4 = 27 cm
Hence, the parallel sides are 23cm and 27cm.