Question

A field is in the shape of an isosceles trapezium with parallel sides measuring $\text{20 cm and 8 cm}$ and its non-parallel sides measuring $\text{10 cm}$. Find its area.

• A. 84 ${cm}^2$
• B. 112 ${cm}^2$
• C. 101 ${cm}^2$
• D. 96 ${cm}^2$

Answer 1

The correct option is B 112 ${cm}^2$

Let ABCD be the trapezium given to us with $\text{AB = 8 cm, CD = 20 cm, AD = 10 cm and BC = 10 cm}$.
We draw a line BQ parallel to AD where Q lies on DC such that, ABQD is a parallelogram.

Hence, $BQ=AD=10cm$ ( opposite sides of parallelogram)
$DQ=AB=8cm$ ( opposite sides of parallelogram)
Also, $\Delta BQC$ is an isosceles triangle
As, $BQ=BC$
Now, drop a perpendicular BL , where L lies on QC
Such that ,
$QL=LC=\frac{QC}{2}$
$=\frac{20-8}{2}$
$=\frac{12}{2}$
$=6cm$ ( BL is the perpendicular bisector of the triangle)
Now, in $△$ BLC,
using Pythagoras theorem:
$(LC{)}^2+(BL{)}^2=(BC{)}^2$
$(6{)}^2+(BL{)}^2=(10{)}^2$
$(BL{)}^2=(10{)}^2-(6{)}^2$
$(BL{)}^2=100-36$
$BL=\sqrt{64}$
$BL=8cm$

Area of trapezium $=\frac{1}{2}\times \left(sumofparallelsides\right)\times height$
$=\frac{1}{2}\times (AB+DC)\times BL$
$=\frac{1}{2}\times (8+20)\times 8$
$=112c{m}^2$