question_answer

In a trapezium \[ABCD,\]\[\left. AB \right\|CD\] and its diagonals intersect at O. If \[\text{AB}=\text{6 cm}\] and \[\text{DC}=\text{3 cm},\]then find the ratio of the areas of \[\Delta AOB\] and \[\Delta COD\].

A) 4:1

B) 1:2

C) 2:1

D) 1:4

Correct Answer: A

Solution :

[a] In \[\Delta AOB\] and \[\Delta COD,\]

\[\angle AOB=\angle COD\]  (Vertically opposite angles)

\[\angle OAB=\angle OCD\]     (Alternate interior angles)

\[\therefore \,\,\,\Delta AOB\tilde{\ }\Delta COD\](By AA similarity criterion)

\[\therefore \,\,\,\,\frac{ar\,(\Delta AOB)}{ar\,(\Delta COD)}={{\left( \frac{AB}{CD} \right)}^{2}}={{\left( \frac{6}{3} \right)}^{2}}=\frac{4}{1}\]