Question

A field is in the shape of a trapezium whose parallel sides are $25$ m and $10$ m. The non-parallel sides are $14$ m and $13$ m. And the area of the field in $m^2$ is:

Answer 1

Area of trapezium

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

Draw, $BF\parallel AD$, also

$AB\parallel DF$

Since, opposite sides are parallel, therefore ABFD is a parallelogram. Hence,

$BF=AD=13m$ and $DF=AB=10m,FC=DC-DF=25-10=15m$

$ar△BFC=\sqrt{s(s-a)(s-b)(s-c)},s=\frac{a+b+c}{2}=\frac{15+13+14}{2}=21m$

$\therefore ar△BFC=\sqrt{21(21-15)(21-13)(21-14)}=\sqrt{3\times 7\times 6\times 8\times 7}$

$=\sqrt{3\times 7\times 1\times 2\times 2\times 2\times 2\times 7}=2\times 2\times 3\times 7=84㎡$

Also, $ar△BFC=\frac{}{}\times FC\times BE$

$\Rightarrow 84=\frac{1}{2}\times 15\times BE\Rightarrow BE=\frac{84\times 2}{15}=11.2m$

$\therefore$Area of trapezium ABCD$=\frac{1}{2}(10+25)\times 11.2=\frac{1}{2}\times 35\times 11.2$

$=196㎡$