This is a problem that I designed two years ago while I was playing with a set square.
Well, as shown below, there are two right-angled triangles ABE and ADC that share the point A where is located the right angle.
Knowing that = 36 = 28 Find the area of the triangle EFC
Round your answer to 3 decimal places
Clarification: meet at F
A,E and C belong to the same line.
A,D and B belong to the same line.
ABE= AEB=
ADC=
(The picture above is a scale drawing)
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Let H be the projection of F on A B . FHB is isosceles thus: F H = H B
H D = F H x tan 30 ∘ = 3 F H
H B - H D = D B = 8 (36-28=8)
So H B - 3 H B = 8 solving for H B we get that it is equal to 3 − 1 8 3 .
The height of the triangle EFC = A H = A B - H B = 36 - 3 − 1 8 3 = 3 − 1 2 8 3 − 3 6
While the base is A C - A E = 2 8 3 − 3 6
The Area of the triangle EFC = 2 1 ( 2 8 3 − 3 6 )* 3 − 1 2 8 3 − 3 6 = 1 0 6 . 6 7 7