question_answer

ABCD is a trapezium with parallel sides AB = a and DC = b. If E and F are mid- points of non-parallel sides AD and BC respectively, then the ratio of areas of quadrilaterals ABFE and EFCD is

A)                      a : b                            

B)        \[(a+3b):(3a+b)\]          

C)                    \[(3a+b):(a+3b)\]          

D)        \[(2a+b):(3a+b)\]

Correct Answer: C

Solution :

We have, ABCD is a trapezium and let h be the height of ABCD Then, \[\frac{h}{2}\]is the height of trapezium EFBA and EDCF. Now, Area of trapezium EFBA \[=\frac{1}{2}(a+b+EX+YF)\times \frac{h}{2}\] \[=\frac{1}{2}\left( a+b+\frac{a-b}{2} \right)\times \frac{h}{2}\] Similarly, Area of trapezium EFCD \[=\frac{1}{2}(b+b+EX+YF)\times \frac{h}{2}\] \[=\frac{1}{2}\left( 2b\frac{a-b}{2} \right)\times \frac{h}{2}\] \[\therefore \]Required ration \[=\frac{\frac{1}{2}\left( \frac{3a+b}{2} \right)\times \frac{h}{2}}{\frac{1}{2}\left( \frac{3b+a}{2} \right)\times \frac{h}{2}}=\frac{3a+b}{a+3b}\]