Ex.6.4 Q2 Triangles Solution - NCERT Maths Class 10

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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 \(COD.\)


Text Solution



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

\(AA\) criterion.


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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