Question

Diagonals of trapezium $ABCD$ intersect at $O$ and $AB\parallel DC$.
If $AB=3CD$. Find ratio of areas of $△AOB$ & $△COD$

• A. $1:9$
• B. $9:1$
• C. $3:1$
• D. $1:3$

Answer 1

The correct option is B $9:1$

In $△AOB$ & $COD$
$\angle AOB=\angle COD$ ...Given

$\angle OAB=\angle OCD$ ...Given

$\angle OBA=\angle ODC$ ...Given

$△AOB\sim △COD$ ...by AAA test of similarity

By theorem on ratio of areas of similar triangles,

$\frac{A\left(△AOB\right)}{A\left(△COD\right)}={\left(\frac{AB}{CD}\right)}^2$

$\frac{A\left(△AOB\right)}{A\left(△COD\right)}={\left(\frac{3}{1}\right)}^2=\frac{9}{1}$