Question

The length of a rectangle is 4 cm more than twice its width, and its area is 9cm^2. Find the width to the nearest hundredth.

Answer 1

Answer:

Note that I slightly misread the question and in the steps below found the answer to the nearest thousandth.  To the nearest hundredth though, the rectangle has a length of 6.69cm, and a width of 1.35 cm.

Step-by-step explanation:

We are told that the rectangle has an area of nine square centimetres, and that it's length is 4 centimetres more than twice its width. We can express those as:

$a = 9cm^2$

and

$l = 2w + 4$

We also know that the area of a rectangle is its length times its width:

$a = l \times w$

We can take that last expression, and plug in the other two, to solve for w

$a = l \times w\\9 = (2w + 4) \times w\\9 = w(2w + 4) \\9 = 2w^2 + 4w\\4.5 = w^2 + 2w\\5.5 = w^2 + 2w + 1\\5.5 = (w + 1)^2\\w + 1 = \sqrt{5.5}\\w = \sqrt{5.5} - 1\\w \approx 2.345 - 1\\w \approx 1.345$

We can then plug that into expression "l = 2w + 4" to find the length:

$l = 2w + 4\\l = 2(\sqrt{5.5} - 1) + 4\\l = 2 * \sqrt{5.5} - 2 + 4\\l = \sqrt{4} * \sqrt{5.5} + 2\\l = \sqrt{22} + 2\\l \approx 4.690 + 2\\l \approx 6.690$

Now let's see if we have that right, we can multiply these and see if we get an area of 9:

$1.345 \times 6.690 \approx 8.998$

Which after accounting for rounding errors is matches the 9 square centimetres.