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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