E is the midpoint of diagonal BD of Parallelogram ABCD If the point E is joined to a point F - Maths - Areas of Parallelograms and Triangles

Question

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

Lalit Mehra · Accepted Answer

ABCD is a parallelogram E is the mid point of BD E is joined to F, where FD $= \frac{1}{3} DA$ Draw BP âŠ¥ AD and EQ âŠ¥ AD Since BP âŠ¥ AD and EQ âŠ¥ AD, âˆ´ BP || EQ In Î”BPD, E is the mid point of BP and EQ || BP âˆ´ Q is the mid point of PD (Converse of mid point theorem) $EQ = \frac{1}{2} BP (Mid point theorem) Area of Δ ABD = \frac{1}{2} \times AD × BP Area of Δ EFD = \frac{1}{2} \times DF × EQ \frac{Area of Δ EFD}{Area of Δ ABD} = \frac{\frac{1}{2} \times DF × EQ}{\frac{1}{2} \times AD × BP} = \frac{\frac{1}{2} \times \frac{1}{3} AD × \frac{1}{2} BP}{\frac{1}{2} \times AD × BP} = \frac{1}{6} Area of Δ EFD = \frac{1}{6} Area of Δ ABD$ Area of quadrilateral ABEF = Area of Î”ABD â€“ Area of Î”EFD $= Area of Δ ABD - \frac{1}{6} Area of Δ ABD = \frac{5}{6} Area of Δ ABD$ $\frac{Area of Δ EFD}{Area of quadrilateral ABEF} = \frac{\frac{1}{6} Area of Δ ABD}{\frac{5}{6} Area of Δ ABD} = \frac{1}{5}$ Thus, the ratio of area of Î”DFE to area of quadrilateral ABEF is 1 : 5

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

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