Question

In parallelogram ABCD, points E and F are the midpoints of side AB and BC, respectively. A F meets DE at G and BD at H. What is the area of quadrilateral BHGE if the area of ABCD is 60?

Answer 1

Answer:

The area of the quadrilateral B-H-G-E is 7

Step-by-step explanation:

The area of the parallelogram = 60  

It can be shown that ΔG-D-F = 3/10 × Area of the parallelogram S = 3/10×S

Because the diagonal B-D shares A-F into the ratio 1:2, the ΔH-D-A = 2/3×ΔD-B_A  where ΔD-B_A = 1/2·S, therefore, ΔD-B_A = 2/3×1/2×S = 1/3×S

ΔB-F-H is similar to ΔD-B_A scaled to 1/2,

Therefore;

The area of ΔB-F-H = 1/4×Area of ΔD-B_A = 1/4×1/3×S = 1/12×S,

ΔF-C-D = 1/4×S

ΔE-A-D = 1/4×S

The area of the quadrilateral B_H_G_E found as follows;

Area B_H_G_E  = S - (ΔF-C-D + ΔG-D-F +ΔE-A-D +ΔB-F-H)

Area B_H_G_E  = S - (1/4×S + 1/10×S +1/4×S +1/12×S) = S - 53/60×S = 7/60×S

Whereby we are given that the area, S of the parallelogram AB-CD = 60 we have;

Area B_H G_E  = 7/60×60 = 7