# The side of a square is equal with the width of a rectangle.The length of the rectangle is 6 units longer than it's width. The sum of their areas is 176 square units. Which is the side of the square?

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Lets assume that the length of the rectangle = L=6+W, and the width= W = the side of the square

We kow that the area of a square = side^2 = w^2

and the area of a rectangle = L*W = (6+W)W

from the problem we know that the sum of both area = 176

so, w^2 + LW = 176

W^2 + W(W+6) = 176

then, W^2 + W^2 +6W = 176

2W^2 +6W-176= 0 divide by 2

==> w^2+3W - 88 = 0

==> (W+11) (W-8)= 0

then W = 8 (ignore W=-11 because the width can not be negative value)

then the width of the square = 8 units

The width of the rectangle be x. Then its length = x+6 units.

So the area of the rectangle = x(x+6) = x^2+6x.

The side of the square being equal to the width of the rectangle, is x. So the area of the square = x^2.

Therefore the agebraically the sum of the areas of the rectangle and the square = x^2+6x+x which is equal to 176. Or

2x^2+6x=176. Or

2x^2+6x-176 = 0. Dviding by 2 , we get:

x^2+3x-88 = 0. Or

x^2+11x-8x-88 = 0. Or

x(x+11) -8(x+11) = 0. Or

(x+11)(x-8) = 0.

So x= 8 is the practical solution.

Therefore the rectangle measure 8 by 8+6 or 8 by 14

We'll consider the side of the square = x. Also, from enunciation, the width of the rectangle is = x, too.

The length of the rectangle is x+6.

Now, we'll calculate the areas of the squares and rectangle:

A of square = x*x = x^2

A of rectangle = x*(x+6) = x^2 + 6x

We'll work over the constraint given by the enunciation:

A square + A rectangle = 176

x^2 + x^2 + 6x = 176

2x^2 + 6x - 176 = 0

We'll divide by 2:

x^2 + 3x - 88 = 0

x1 = [-3+sqrt(9+352)]/2

x1 = (-3+19)/2

x1 = 16/2

x1 = 8

x2 = (-3-19)/2

x2 = -22/2

x2 = -11 Impossible, because a length is positive always!

**So, the side of the square is x=8.**