Correct answer is (b) 15.28, 20.28
Explanation: Let the sides of two squares be x and y.
Their areas will be x^2 and y^2 respectively.
Sum of their areas is 625 m^2
x^2+y^2=625 (1)
Their perimeters will be 4x and 4y
Difference in their perimeters is 20 m
4x-4y=20
x-y=5
x=5+y (2)
Substituting (2) in (1) we get,
(y+5)^2+y^2=625
y^2+10y+25+y^2=625
2y^2+10y+25=625
2y^2+10y-620=0
y^2+5y=310
Adding \(\frac {b^2}{4}\) on both sides, where b=5
y^2 + 5y + \(\frac {5^2}{4}=\frac {5^2}{4}\) + 310
y^2 + 5y + \(\frac {25}{4}=\frac {25}{4}\) + 310
y^2 + 5y + \(\frac {25}{4}=\frac {1265}{4}\)
\((y+\frac {5}{2})\)^2 = \((\frac {\sqrt {1265}}{2})\)^2
y+\(\frac {5}{2}\) = ±\(\frac {\sqrt {1265}}{2}\)=±\(\frac {35.56}{2}\)
y = \(\frac {35.56}{2}-\frac {5}{2}=\frac {30.56}{2}\) = 15.28 and y = \(\frac {-35.56}{2}-\frac {5}{2}=\frac {-40.56}{2}\) = -20.28
Since, length cannot be negative hence, y=15.28
Now, x=5+y=5+15.28=20.28