Question

Find the area of a trapezium whose parallel sides are $11$ m and $25$ m and the non parallel sides are $15$ m and $13$ m.

Answer 1

Trapezium $ABCD$ is divide into parallelogram $AECD$ and triangle $CEB$.

Now in $△CEB$, we have

$EB=25-11=14cm$

Using Heron’s theorem, we will first evaluate the semi-perimeter of $△CEB$ and then evaluate its area.

$S=\frac{BE+EC+BC}{2}$

$\Rightarrow S=\frac{13+14+15}{2}=21cm$

Therefore,

area of $△CEB=\sqrt{S(S-BE)(S-EC)(S-BC)}$

$\Rightarrow$ Area of $△CEB=\sqrt{21(21-13)(21-14)(21-15)}=84{cm}^2$

As we know that,

Area of triangle $=\frac{1}{2}\times$ Base $\times$ height

Therefore,

Area of $△CEB=\frac{1}{2}\times 14\times BF$

$\Rightarrow BF=\frac{84\times 2}{14}=12cm$

Again, as we know that,

Area of trapezium $=$ Sum of parallel sides $\times$ height

Therefore,

Area of trapezium $ABCD=\frac{1}{2}\times (11+25)\times 12=216{cm}^2$

Thus the area of trapezium is $216{cm}^2$.