question_answer

In trapezium ABCD, \[AB\parallel CD\]and AB = 2 CD. Its diagonals intersect at O. If the area of \[\Delta AOB\] is \[84\,c{{m}^{2}},\] then the area of \[\Delta COD\]is

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

B) \[72\,\,c{{m}^{2}}\]

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

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

Correct Answer: D

Solution :

Let CD be x.

\[\therefore \]      AB = 2x

Now,     \[\frac{\text{Area}\,\text{of}\,\Delta AOB}{\text{Area}\,\text{of}\,\Delta COB}=\frac{A{{B}^{2}}}{C{{D}^{2}}}\]

\[\Rightarrow \]   \[\frac{84}{\text{Area}\,\text{of}\,\Delta COD}=\frac{4{{x}^{2}}}{{{x}^{2}}}\]

\[\therefore \]      Area of \[\Delta COD=\frac{84}{4}=21\,c{{m}^{2}}\]