Question

The length of a rectangle is 8 cm longer than its width. find the dimensions of the rectangle if its area is 108cm

Answer 1

Answer:

$4+2\sqrt{31}\text{ by } -4+2\sqrt{31}$

Or about 15.136 centimeters by 7.136 centimeters.

Step-by-step explanation:

Recall that the area of a rectangle is given by:

$\displaystyle A = w\ell$

Where w is the width and l is the length.

We are given that the length is 8 centimeters longer than the width. In other words:

$\ell = w+8$

And we are also given that the total area is 108 square centimeters.

Thus, substitute:

$(108)=w(w+8)$

Solve for w. Distribute:

$w^2+8w=108$

Subtract 108 from both sides:

$w^2+8w-108=0$

Since the equation is not factorable, we can use the quadratic formula:

$\displaystyle x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In this case, a = 1, b = 8, and c = -108. Substitute and evaluate:

$\displaystyle \begin{aligned} w&= \frac{-(8)\pm\sqrt{(8)^2-4(1)(-108)}}{2(1)} \\ \\ &=\frac{-8\pm\sqrt{496}}{2}\\ \\ &=\frac{-8\pm4\sqrt{31}}{2} \\ \\ &=-4 \pm 2\sqrt{31} \end{aligned}$

So, our two solutions are:

$w=-4+2\sqrt{31} \approx 7.136 \text{ or } w=-4-2\sqrt{31}\approx -15.136$

Since width cannot be negative, we can eliminate the second solution.

And since the length is eight centimeters longer than the width, the length is:

$\ell =(-4+2\sqrt{31})+8=4+2\sqrt{31}\approx 15.136$

So, the dimensions of the rectangle are about 15.136 cm by 7.136 cm.