Outside point

The square ABCD and the point E lying outside the given square are given. What is the area of the square when the distance | AE | = 2, | DE | = 5 a | BE | = 4?

Correct result:

S =  9.1973

Solution:

AE=2 DE=5 BE=4  a,x,y>0  1. AE2=y2+x2 BE2=(ay)2+x2 DE2=y2+(x+a)2  2. AE2=y2+x2 BE2=(a+y)2+x2 DE2=y2+(x+a)2  3. AE2=y2+x2 BE2=(a+y)2+x2 DE2=y2+(xa)2  2: a=3.032713.0327 x=1.945891.9459 y=AE2x2=221.945920.4621  S=a2=3.032729.1973=9.1973   Verifying Solution:   l1=y2+x2=0.46212+1.94592=2 l2=(a+y)2+x2=(3.0327+0.4621)2+1.945924 l3=y2+(x+a)2=0.46212+(1.9459+3.0327)25



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