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?
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 = 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.