Question

Find the area of trapezium $ABCD$ whose parallel sides are $AB=19cm$, $DC=9cm$, and non parallel sides are $BC=8cm$ and $DA=6cm$.

Answer 1

Given: $ABCD$ is the given trapezium in which $AB$ and $CD$ are parallel sides.

And, $AD$ and $BC$ are non-parallel sides.

Here, $AB=19cm,CD=9cm,BC=8cm,$ and $DA=6cm$

Construction: Draw $CE\parallel DA$ and a perpendicular on $AB$ from the vertex $C$, i.e $CF\perp AB$

In $AECD,$ we have

$AE\parallel CD$ and $CE\parallel DA$. So, $AECD$ is a paralleogram.

$AE=CD=9cm$ [opposite side of parallelogram]

$EB=AB-AE=19-9=10cm$

$DA=CE=6cm$ [opposite side of parallelogram]

In triangle $CEB,$ the sides of the triangle are

$a=6cm,b=8cm,$ and $c=10cm$

Semi-perimeter of the triangle, $s=\frac{a+b+c}{2}=\frac{6+8+10}{2}=12$

USing Heron's formula, Area of triangle $CEB=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{12(12-6)(12-8)(12-10)}$

$=\sqrt{12\times 6\times 4\times 2}$

$=\sqrt{6\times 2\times 6\times 2\times 2\times 2}$

$=24c{m}^2$

Area of triangle $CEB$ is also given by $\frac{1}{2}\times$ base $\times$ height

$\frac{1}{2}\times EB\times CF=24$ [$\because EB=$ base, $CF=$ height]

$\Rightarrow \frac{1}{2}\times 10\times CF=24$

$\Rightarrow CF=4.8cm$

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

$=\frac{1}{2}\times (AB+CD)\times CF$

$=\frac{1}{2}\times (19+9)\times 4.8=67.2c{m}^2$

$\therefore$ Area of trapezium $ABCD=67.2c{m}^2$