Question

$ABCD$ is an isosceles trapezium. If $\mathrm{AB}\parallel \mathrm{CD}$, $\mathrm{AB}=20\mathrm{cm}$, $\mathrm{CD}=10\mathrm{cm}$ and each of the non-parallel sides is $13\mathrm{cm}$, then find the area of trapezium.

Answer 1

Step $1:$ Drawing the diagram:

$ABCD$ is an isosceles trapezium.

From $D$ and $C$, draw a perpendicular to side $AB$, such that it intersects at $E$ and $F$, we get,

$\mathrm{AE}=5\mathrm{cm},\mathrm{EF}=10\mathrm{cm},\mathrm{FB}=5\mathrm{cm}$

$DE$ is the height of trapezium.

Step $2:$ Finding the height of trapezium:

In, $∆\mathrm{ADE}$, right angled at $E$,

Using Pythagoras theorem, we get

$\begin{array}{rcl}{\mathrm{AD}}^2& =& {\mathrm{DE}}^2+{\mathrm{AE}}^2\\ {13}^2& =& {\mathrm{DE}}^2+5^2\\ 169& =& {\mathrm{DE}}^2+25\\ {\mathrm{DE}}^2& =& 169-25\\ {\mathrm{DE}}^2& =& 144\\ \mathrm{DE}& =& \sqrt{144}\\ \mathrm{DE}& =& 12\mathrm{cm}\end{array}$

Step $3:$ Finding the area of trapezium:

Therefore, Area of trapezium is

$$\begin{gathered} =\frac{1}{2}\times (\mathrm{sum}\mathrm{of}\mathrm{lengths}\mathrm{of}\mathrm{parallel}\mathrm{sides})\times \mathrm{height}\mathrm{of}\mathrm{trapezium} \\ =\frac{1}{2}\times (\mathrm{AB}+\mathrm{CD})\times \mathrm{DE} \\ =\frac{1}{2}\times (20+10)\times 12 \\ =\frac{1}{2}\times 30\times 12 \\ =180{\mathrm{cm}}^2 \end{gathered}$$

Hence, the area of trapezium is $180{\mathrm{cm}}^2$.