Find the area of a trapezium whose parallel sides are 11 ma and 25 m long,

Solution:

Draw $D E \| B C$ and $D L$ perpendicular to $A B$.

The opposite sides of quadrilateral DEBC are parallel. Hence, DEBC is a parallelogram.
∴ DE = BC = 13 m
Also

$A E=(A B-E B)=(A B-D C)=(25-11)=14 \mathrm{~m}$

For $\Delta D A E$

Let:
AE = a =14 m
DE = b = 13 m
DA = c =15 m

Thus, we have:

$s=\frac{a+b+c}{2}$

$s=\frac{14+13+15}{2}=21 \mathrm{~m}$

Area of $\Delta D A E=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{21 \times(21-14)} \times(21-13) \times(21-15)$

$=\sqrt{21 \times 7 \times 8 \times 6}$

$=\sqrt{7056}$

 

$=84 \mathrm{~m}^{2}$

Area of $\Delta D A E=\frac{1}{2} \times A E \times D L$

$\Rightarrow 84=\frac{1}{2} \times 14 \times D L$

$\Rightarrow \frac{84 \times 2}{14}=D L$

 

$\Rightarrow D L=12 \mathrm{~m}$

Area of trapezium $=\frac{1}{2} \times($ Sum of parallel sides $) \times($ Distance between them $)$

$=\frac{1}{2} \times(11+25) \times 12$

$=\frac{1}{2} \times 36 \times 12$

 

$=216 \mathrm{~m}^{2}$