Answer:
I. Length, L = 11 cm
II. Width, W = 6 cm
Step-by-step explanation:
Let the length of the rectangle be L.
Let the width of the rectangle be W.
Given the following data;
Area of rectangle = 66 cm²
Translating the word problem into an algebraic expression, we have;
L = 2W - 1 .....equation 1
We know that the area of a rectangle is given by the formula;
A = L * W
Substituting
66 = L * W ....equation 2
Substituting eqn 1 into eqn 2, we have;
66 = (2W - 1) * W
66 = 2W² - W
2W² - W - 66 = 0
Note: the standard form of a quadratic equation is ax² + bx + c = 0
a = 2, b = -1 and c = -66
Solving the quadratic equation using the quadratic formula;
The quadratic equation formula is;
[tex] x = \frac {-b \; \pm \sqrt {b^{2} - 4ac}}{2a} [/tex]
Substituting into the equation, we have;
[tex] x = \frac {-(-1) \; \pm \sqrt {1^{2} - 4*2*(-66)}}{2*2} [/tex]
[tex] x = \frac {1 \pm \sqrt {1 - (-528)}}{4} [/tex]
[tex] x = \frac {1 \pm \sqrt {1 + 528}}{4} [/tex]
[tex] x = \frac {1 \pm \sqrt {529}}{4} [/tex]
[tex] x = \frac {1 \pm 23}{4} [/tex]
[tex] x_{1} = \frac {1 + 23}{4} [/tex]
[tex] x_{1} = \frac {24}{4} [/tex]
x1 = 6
We do not need the negative value of x, so we proceed.
Therefore, Width, W = x1 = 6 cm
Next, we find the length L;
From eqn 1;
L = 2W - 1
L = 2(6) - 1
L = 12 - 1
L = 11 cm
