Question

Find the area of the trapezium.

• A. $144\sqrt{3}c{m}^2$
• B. $154\sqrt{3}c{m}^2$
• C. $162\sqrt{3}c{m}^2$
• D. None of these

Answer 1

The correct option is A $144\sqrt{3}c{m}^2$

Given : $AB=30cm,BC=12cm,CD=18cm,DA=12cm$

$DM$ and $CN$ are $\perp$ drawn to $AB$

$\square CDMN$ forms a rectangle with $MN=CD$ and $DM=CN$

Now, $AB=AM+MN+NB=2AM+18$ ....... $[\because AM=NB]$

$⟹2AM=30-18=12⟹AM=6$

$\therefore AM=6=NB$ ........ $\left(i\right)$

In $△ADM,\angle AMD={90}^o$
$⟹A{D}^2=A{M}^2+D{M}^2$ ....... [By Pythagoras theorem]
$⟹{12}^2=6^2+D{M}^2$
$⟹D{M}^2=144-36$
$⟹DM=6\sqrt{3}$
$\therefore DM=CN=6\sqrt{3}$ .......... $\left(ii\right)$
Area of trapezium $ABCD=$ Area of $△ADM+$ Area of rectangle $CDMN+$ Area of $△BCN$
$=\frac{1}{2}\times AM\times DM+MN\times NC+\frac{1}{2}\times BN\times CN$
$=\frac{1}{2}\times 6\times 6\sqrt{3}+18\times 6\sqrt{3}+\frac{1}{2}\times 6\times 6\sqrt{3}$
$=144\sqrt{3}c{m}^2$

Hence, area of trapezium is $144\sqrt{3}c{m}^2$.