Chapter E, Problem 11.7EX

To determine

To find: the ratio of area of $\Delta \text{AOB}$ and $\Delta \text{COD}$ .

Answer to Problem 11.7EX

The ratio of area of $\Delta \text{AOB}$ and $\Delta \text{COD}$ is $9:1$ .

Explanation of Solution

Given information:

  $AB\parallel CD$in the given figure:

  

Calculation:

In $\Delta \text{AOB}$ and $\Delta \text{COD}$ ,

  $\angle OAB=\angle OCD$   … (Alternate angle)

  $\angle OBA=\angle ODC$  … (Alternate angle)

By AA similarity criterion:

  $\Delta AOB\sim \Delta COD$

It is known that, if two triangles are similar then the ratio of areas is equal to the square of ratio of corresponding sides.

  $\begin{array}{l}\frac{ar(\Delta AOB)}{ar\left(\Delta COD\right)}={\left(\frac{AB}{CD}\right)}^2\\ ={\left(\frac{18}{6}\right)}^2\\ =\frac{3^2}{1}\\ =\frac{9}{1}\end{array}$

Therefore, the ratio of area of $\Delta \text{AOB}$ and $\Delta \text{COD}$ is $9:1$ .