# E is the midpoint of diagonal BD of Parallelogram ABCD.If the point E is joined to a point F on DA such that DF= 1/3 of DA ,then ratio of th area of triangle DFE to area of quadrilateral ABEF is a) 1 : 3 b) 1 : 4 c) 1 : 5 d)2 : 5 ABCD is a parallelogram. E is the mid point of BD. E is joined to F, where FD .

∴ BP || EQ

In ΔBPD,

E is the mid point of BP and EQ || BP.

∴ Q is the mid point of PD.  (Converse of mid point theorem) Area of quadrilateral ABEF = Area of ΔABD – Area of ΔEFD  Thus, the ratio of area of ΔDFE to area of quadrilateral ABEF is 1 : 5.

• 33

assume that AD is the base of the parallelogram. Since E is the midpoint of BD, the height of DFE would be half the height of ABD and we know that the base of DFE is 1/3 DA. Thus the area of DFE is 1/3 x 1/2 = 1/6 the area of ABD. Thus ABEF is 1 - 1/6 = 5/6 the area of ABD. So DFE/ABEF = (1/6) / (5/6) = 1/5.

Ans: c) 1:5

• 5

that AD is the base of the parallelogram. Since E is the midpoint of BD, the height of DFE would be half the height of ABD and we know that the base of DFE is 1/3 DA. Thus the area of DFE is 1/3 x 1/2 = 1/6 the area of ABD. Thus ABEF is 1 - 1/6 = 5/6 the area of ABD. So DFE/ABEF = (1/6) / (5/6) = 1/5.

Ans: c) 1:5

poated by abhishek

• -1
What are you looking for?