Question

Length of the fence of a trapezium shaped field $ABCD$ is $120m$. If $BC=48m,CD=17mandAD=40m$, find the area of this field. Side $AB$ is perpendicular to the parallel sides $ADandBC$.

• A. $660m^2$
• B. $6.6m^2$
• C. $660c{m}^2$
• D. None of these

Answer 1

The correct option is A $660m^2$

$\Rightarrow$Let $ABCD$ be the given trapezium in which $BC=48m,CD=17mandAD=40m$

Through $D$, draw $DL\perp BC$.

Now, $BL=AD=40m$

and $LC=BC-BL=(48-40)m=8m$

Applying Pythagoras theorem in right $\Delta DLC$, we have

$\Rightarrow$$D{L}^2=D{C}^2-L{C}^2={17}^2-8^2=289-64=225$

$\Rightarrow DL=\sqrt{225}=15m$

$\Rightarrow$Now, Area of the trapezium $ABCD$

$=\frac{1}{2}\times (BC+AD)\times DL=\frac{1}{2}\times (48+40)\times 15m^2$

$=(44\times 15)m^2=660m^2$.