AB = 80 m, BC = 60 m
DF = 50 m, GE = 50 m
HE = 100 m
The area of the region ACDF is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 3750 m2
The region ACDF (shaded) is the triangle ACE with the small triangle FDE cut off its top, so its area is the difference of the two areas. Each triangle has a base along a horizontal line and its height along EH, which is perpendicular to both FD and AC.
AC from the right triangle ABC: \(AC=\sqrt{80^2+60^2}=\sqrt{10000}=100\) m
Triangle ACE: \(\frac{1}{2}\times AC\times EH=\frac{1}{2}\times100\times100=5000\) m2
Triangle FDE: \(\frac{1}{2}\times DF\times GE=\frac{1}{2}\times50\times50=1250\) m2
ACDF \(=5000-1250=3750\) m2