question_answer

In trapezium ABCD, AB || CD and AB = 2CD. Its diagonals intersect at O. If the area of \[\Delta AOB=84\,c{{m}^{2}}\], then the area of \[\Delta COD\]is equal to [SSC CGL Tier II, 2015]

A) \[72\text{ }c{{m}^{2}}\]

B) \[42\,c{{m}^{2}}\]

C) \[26\,c{{m}^{2}}\]

D) \[21\,c{{m}^{2}}\]

Correct Answer: D

Solution :

[d]                                                 \[\angle \,AOB=\angle COD\]                         \[\angle \,CAB=\angle \,DCA\]                         \[\angle \,DBA=\angle \,CDB\]             \[\Delta \,AOB\] is similar to \[\Delta \,COD\,\frac{AB}{CD}\] \[\Rightarrow \]\[\frac{\text{Area}\,\text{of}\,\Delta \,AOB}{\text{Area}\,\text{of}\,\Delta COD}={{\left( \frac{2}{1} \right)}^{2}}\] Area of \[\Delta \,COD=84\times \frac{1}{4}=21\,c{{m}^{2}}\]