let the side of the square be = x
the square be named as DEFB where DE=EF=FB=DB=x
(point D lies on side AB , E on side AC and F on side BC)
now,
AD=a-x
FC=b-x
by RHS triangle ADE is similar to triangle EFC
therefore AD/DE=EF/FC
a-x/x=x/b-x
ab-ax-bx+x^2=x^2
ab=x(a+b)
x=ab/a+b
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Answers & Comments
AC2 = AB2+BC2
a2+b2
= under root a2+b2
answer is not conform
let the side of the square be = x
the square be named as DEFB where DE=EF=FB=DB=x
(point D lies on side AB , E on side AC and F on side BC)
now,
AD=a-x
FC=b-x
by RHS triangle ADE is similar to triangle EFC
therefore AD/DE=EF/FC
a-x/x=x/b-x
ab-ax-bx+x^2=x^2
ab=x(a+b)
x=ab/a+b