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...

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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.**

Let the side of the second square be s, the size of the first is 5 less than the second or s - 5.

The area of the larger square is s^2. This is four times the area of the smaller square which is (s - 5)^5

=> s^2 = 4*(s - 5)^2

=> s^2 = 4*(s^2 + 25 - 10s)

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

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

=> 3s^2 - 30s - 10s + 100 = 0

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

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

=> s = 10/3 and s = 10

We ignore 10/3 , as 10/3 - 5 is negative.

**The sides of the two squares are 10 cm and 5 cm**