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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
