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

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.

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 * w`

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

`a = l * w\\9 = (2w + 4) * 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 = √(5.5)\\w = √(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(√(5.5) - 1) + 4\\l = 2 * √(5.5) - 2 + 4\\l = √(4) * √(5.5) + 2\\l = √(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 * 6.690 \approx 8.998\\`

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