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.
My solution is basically the same as xdog but I found it enjoyable enough to extend it.
The inner triangle is right with short leg DE and long leg EF.
In general if the right triangle has short leg = a and long leg = b the area of the square is
b^4/(2b^2-2ab+a^2)
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Posted by Jer
on 2019-08-21 20:30:18 |
general solution
