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

 Video Solution
Triangles
Ex 6.4 | Question 2

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)
\(\Rightarrow \; \Delta AOB \sim \Delta COD\)

(AA criterion)
\(\begin{align}\Rightarrow \frac{{{\rm{Area \,of\,}}{{\Delta AOB}}}}{ \text{Area} \,of \Delta COD} & = \frac{{{{(AB)}^2}}}{{{{(CD)}^2}}} \\ & (\text{Theorem} \, 6.6) \end{align}\)

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

\[\begin{align} \Rightarrow \; &\text{Area of } \Delta AOB :  \text{Area of } \Delta COD \\=\;& 4:1 \end{align}\]

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