A square has a side 5 centimeters shorter than the side of a second square. The area of the larger square is four times the area of the smaller...

Let the side of the first square be
x.

The side of the second square is
y.

Given that the side of square 1 is 5 cm shorter than
square 2.

==> x = y-5
...........(1)

Also, the area of the larger square ( square
2) is 4 times the area of square 1.

==> y^2 = 4*x^2
.........(2)

We will solve by
substitution.

We will substitue (1) into
(2).

==> y^2 =
4(y-5)^2

==> y^2 = 4(y^2-10y +
25)

==> y^2 = 4y^2 - 40y
+100

==> 3y^2 -40y +100 =
0

==> (3y -10)(y-10)=
0

==> y1= 10 ==> x1= 10-5 =
5

==> y2= 10/3 ==> x2= 10/3 -5 = -5/3 (
impossible).

Then, the sides of the squares
are 5 and 10.