Question

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

• A. $\frac{900\sqrt{6}}{7}{m}^2$
• B. $\frac{1100\sqrt{6}}{7}{m}^2$
• C. $\frac{1300\sqrt{6}}{7}{m}^2$
• D. $\frac{1500\sqrt{6}}{7}{m}^2$

Answer 1

The correct option is C

$\frac{1300\sqrt{6}}{7}{m}^2$

Given:
AB = 15m, BC = 20m, CD = 50m, DA = 25m
Drawing BE parallel to DA, we get two shapes: a parallelogram, ABED, and a
triangle, BCE, which combine to form the trapezium, ABCD.
Now, area(trapezium ABCD) = area(parallelogram ABED) + area($\Delta$ BCE)
For $\Delta BCE):$

i. BC = 20m (given)
ii. CE = (CD - DE) = (CD - AB) = (50 - 15)m = 35m
( DE = AB : opposite sides of a parallelogram are congruent)
And iii. EB = 25m (EB = DA : opposite sides of a parallelogram are congruent)
Area of $\Delta$ BCE:
By heron’s formula:
2S = (20+35+25) = 80
S = 40 m
$area\left(\Delta BCE\right)=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{\left(40\right)(40-20)(40-35)(40-25)}=\sqrt{40\times 20\times 5\times 15}=\sqrt{60000}=100\sqrt{6}sq.m$
Now, the height of the $\Delta$ BCE can be found as:
$\left(\frac{1}{2}\right)\left(CE\right)\left(BF\right)=100\sqrt{6}\{areaof\Delta =\left(\frac{1}{2}\right)\times (base)\times (height\left)\right\}\Rightarrow \left(\frac{1}{2}\right)\times 35\times h=100\sqrt{6}\text{(h = BF = height of the triangle)}\Rightarrow h=\left(\frac{100\sqrt{6}}{\frac{35}{2}}\right)\Rightarrow h=\left(\frac{40}{7}\right)\sqrt{6}m$
Also, h ($\Delta$ BCE) = h (parallelogram ABED)
So, area (parallelogram ABED) = $(b\times h)$ = $\left(DE\right)\times \left(BF\right)$
$=(15\times (\frac{40}{7}\left)\sqrt{6}\right)sq.m$
$=\left(\frac{600}{7}\right)\sqrt{6}sq.m$
So , area (trapezium ABCD) = area(parallelogram ABED) + area($\Delta$ BCE)
$=\{\left(\frac{600}{7}\right)\sqrt{6}+100\sqrt{6}\}sq.m=\left\{\left(\frac{600+700}{7}\right)\sqrt{6}\right\}sq.m=\left(\frac{1300}{7}\right)\sqrt{6}sq.m$