# Ex.6.4 Q2 Triangles Solution - NCERT Maths Class 10

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## Question

Diagonals of a trapezium $$ABCD$$ with $$AB || DC$$ intersect each other at the point $$O.$$ If $$AB = 2 \,CD,$$ find the ratio of the areas of triangles $$AOB$$ and $$COD.$$

Diagram

## Text Solution

Reasoning:

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

$$AA$$ criterion.

Steps:

In trapezium $$ABCD,$$  $$AB\parallel CD\,$$ and $$AB = 2\,CD$$

Diagonals $$AC, BD$$ intersect at ‘$$O$$

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

$$\angle AOB = \angle COD$$ (vertically opposite angles)
$$\angle ABO = \angle CDO$$ (alternate interior angles)
$$\Rightarrow \qquad \Delta AOB \sim \Delta COD$$ (AA criterion)
\begin{align}\Rightarrow \qquad \frac{{{\rm{Area \,of\,}}{{\Delta AOB}}}}{Area \,of \Delta COD} = \frac{{{{(AB)}^2}}}{{{{(CD)}^2}}} \end{align} (Theorem 6.6)

\begin{align}=\frac{{{{(2CD)}^2}}}{{{{(CD)}^2}}}&=\frac{{4C{D^2}}}{{C{D^2}}} = \frac{{4}}{1}\end{align}

$$\Rightarrow$$ Area of $$\Delta AOB$$ : Area of $$\Delta COD$$$$= 4:1$$

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