# 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 square. Find the side of each square.

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.

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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

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