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A Triangle in a Square (Posted on 2019-08-21) Difficulty: 3 of 5
ABCD is a square with points E and F on sides AB and BC respectively. Given DF=4, EF=3, and DE=5 find the area of the square.

No Solution Yet Submitted by Brian Smith    
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solution | Comment 3 of 7 |
Call the side of the square S and angle FDC a.

Then angle DFC=90-a, and since angle DFE=90, angle BFE=a and angle BEF=90-a.  So triangles DCF and FBE are similar and S/4=BF/3.

Also BF=S-FC=S-sqrt(16-S^2).

Substitute and solve to find S^2=256/17.


  Posted by xdog on 2019-08-21 13:29:30
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