Answer: x = 7
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Explanation:
The largest rectangle (composed of the green and yellow sections combined) has area of 11*12 = 132 cm^2.
The yellow region takes up 112 of those 132 sq cm. This must mean the green region takes up 132-112 = 20 cm^2.
The horizontal portion of the green rectangle is 12-x cm. The vertical portion is 11-x cm. We can form the area of the green rectangle as an algebraic expression like so
area = length*width
area = (11-x)*(12-x)
area = 132 - 11x - 12x + x^2 .... apply the FOIL rule
area = x^2 - 23x + 132
Set this equal to the 20 cm^2 we found earlier.
x^2 - 23x + 132 = 20
x^2 - 23x + 132-20 = 0
x^2 - 23x + 112 = 0
We could factor or we could use the quadratic formula. I'll go with the second option.
We'll plug in a = 1, b = -23, c = 112
[tex]x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(-23)\pm\sqrt{(-23)^2-4(1)(112)}}{2(1)}\\\\x = \frac{23\pm\sqrt{81}}{2}\\\\x = \frac{23\pm9}{2}\\\\x = \frac{23+9}{2}\ \text{ or } \ x = \frac{23-9}{2}\\\\x = \frac{32}{2}\ \text{ or } \ x = \frac{14}{2}\\\\x = 16\ \text{ or } \ x = 7\\\\[/tex]
One of these solutions isn't feasible. Note how if x = 16, then this exceeds both the 11 cm and 12 cm sides. So this x value is not possible.
However, x = 7 is possible.
If x = 7, then the horizontal portion of the green rectangle is 12-x = 12-7 = 5 cm. Also, the vertical portion of the green rectangle would be 11-x = 11-7 = 4 cm. The area then is length*width = 5*4 = 20 cm^2 which matches up with what we got earlier. So the answer is confirmed.
