Question

A field is in the shape of a trapezium whose parallel sides are 50m and 15m. The non-parallel sides are 20m and 25 m. Find the area of the trapezium.

Answer 1

Area of $\Delta CBE=\sqrt{s(s-a)(s-b)(s-c)}$
$s=\frac{20+25+35}{2}=40$
So, Area$=\sqrt{40\times (40-20)\times (40-35)\times (40-25)}$
$=\sqrt{40\times 20\times 5\times 15}$
$=\sqrt{40\times 20\times 5\times 15}$
$=\sqrt{800\times 75}=\sqrt{60000}$
$=100\sqrt{6}$
Now, area of triangle $=100\sqrt{6}=\frac{1}{2}\times 35\times h$
$\Rightarrow h=100\sqrt{6}\times \frac{2}{35}$
$\therefore h=\frac{40\sqrt{6}}{7}$
Therefore, the area of trapezium is $A=\frac{1}{2}\times$(sum of parallel sides)(height)
$=\frac{1}{2}\times (50+15)\times \frac{40\sqrt{6}}{7}$
$=\frac{1}{2}\times 65\times \frac{40\sqrt{6}}{7}$
$\therefore A=455m^2$