SOLUTION: ABCD is a rectangle. BC = 15m and DC = 8 m. BD is the diagonal. F is a point on BD. F is joint to C, such that FC is perpendicular BD. Similarly, E another point on BD and AE jo

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Question 1105661: ABCD is a rectangle. BC = 15m and DC = 8 m. BD is the diagonal. F is a point on BD. F is joint to C, such that FC is perpendicular BD. Similarly, E another point on BD and AE jointed such that AE is perpendicular to BD. What is the distance from point E to F. in m.
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!


By applying the Pythagorean theorem to right ΔBCD,
BD = 17

ΔBCD ∽ ΔCFD because they are right triangles with a common
acute angle, so

FD%2FDC%22%22=%22%22+DC%2FBD
FD%2F8%22%22=%22%22+8%2F17
17%2AFD%22%22=%22%22+8%2A8
17%2AFD%22%22=%22%22+64
FD%22%22=%22%22+64%2F17

ΔCFD ≅ ΔAEB is easy to show.

So FD = EB  by cpct

FD%22%22=%22%22EB%22%22=%22%22+64%2F17

BD = EB + EF + FD
17%22%22=%22%22+64%2F17%2BEF%2B64%2F17
17%22%22=%22%22+128%2F17%2BEF

Multiply though by 17

289%22%22=%22%22+128%2B17%2AEF

161%22%22=%22%22+17%2AEF

161%2F17%22%22=%22%22+EF

Edwin