drewcifur said:
the formula it says to use to find the area of one wall is L x W x H, since i am only given the height of the wall (8ft) i figure
I should use the dimensions of the floor for L and W but I am thrown off by having to multiply foot and inch together rather than just, for example, (18 ft length 9 ft width) I have tried to convert to decimal and solve that way but my answers do not look right
The formula for the area of a rectangle is:
$$\text{Area}=\text{base}\times\text{height}$$
or:
$$A=bh$$
The height of each rectangle is 8 ft, as we are told all walls are 8 ft tall. The bases of the rectangular walls we get from the given dimensions of the floor, as you surmised.
And so we will have two walls whose bases are:
$$18\text{ ft}\,5\text{ in}=\left(18+\frac{5}{12}\right)\text{ ft}=\frac{221}{12}\,\text{ft}$$
And we will have two walls whose bases are:
$$9\text{ ft}\,6\text{ in}=\left(9+\frac{6}{12}\right)\text{ ft}=\frac{19}{2}\,\text{ft}$$
Now, if we add these all up, then the total area \(A\) of the 4 walls is then:
$$A=2\left(\frac{221}{12}\,\text{ft}\right)\left(8\text{ ft}\right)+2\left(\frac{19}{2}\,\text{ft}\right)\left(8\text{ ft}\right)$$
Let's factor this:
$$A=2(8)\left(\frac{221}{12}+\frac{19}{2}\right)\text{ ft}^2$$
Adding the fractions, and computing the product of the first two factors, we can write:
$$A=16\left(\frac{335}{12}\right)\text{ ft}^2$$
Reduce:
$$A=\frac{4\cdot335}{3}\,\text{ft}^2=\frac{1340}{3}\,\text{ft}^2$$
Before we proceed, does all this make sense so far?