Let the sides of the two squares be x and y.
Their areas will be :
x² and y² respectively.
Now their perimeters will be :
4x and 4y respectively.
We do this since the sides of a square are equal.
From the question we can do the substitution as follows:
Sum of the areas:
x² + y² = 157………..1)
4x + 4y = 68……2)
Divide equation 2 all through by 4 to get :
x + y = 17
We need to solve for x and y.
By substitution we have :
x = 17 – y
Replace this in equation 1 as follows:
(17 – y)² + y ² = 157
289 – 34y + y² + y² = 157
Collecting the like terms together we have :
2y² – 34y + 132 = 0
Solving the quadratic equation:
Divide through by 2 to get :
y² – 17y + 66 = 0
The roots are :
-11 and -6
We expand the equation as follows:
y² – 11y – 6y + 66 = 0
y(y – 11) – 6(y – 11) = 0
(y – 6)(y – 11) = 0
y = 6 or 11
When y is 11 x is (17 – 11) = 6
So the values can either be 6 or 11 for y or vice versa for x.
The sides are thus :
11 meters and 6 meters