Question

In the trapezium ABCD, AB || DC, AB = 2CD and ar ($△$AOB) = 84 cm${}^2$, find the area of $△$COD

Answer 1

It is given that $ABCD$ is a trapezium and $AB||DC$.

In $△AOB$ and △COD

$\angle ABO=\angle CDO$ (Alternate angles)

$\angle BAO=\angle DCO$ (Alternate angles)

$\angle AOB=\angle COD$ (Vertically opposite angles)

Therefore, △ABC∼△COD

We know that the arc of similar triangles are proportional to squares of their corresponding altitude, therefore with $Ar(△AOB)=84$ cm${}^2$ and $AB=2CD$, we have,

$\frac{Ar(△AOB)}{Ar(△COD)}=\frac{A{B}^2}{C{D}^2}\Rightarrow \frac{84}{Ar(△COD)}=\frac{4C{D}^2}{C{D}^2}\Rightarrow \frac{84}{Ar(△COD)}=4\Rightarrow Ar(△COD)=\frac{84}{4}\Rightarrow Ar(△COD)=21$

Hence, area of $△COD$ is $21$cm${}^2$.