What is the total area of the doors and windows in the room?

[drewcifur]

I am having trouble figuring out the answer to the following question, I am not sure how to get to my answer.Calculate the area of each wall, including any doors and windows (The room is rectangular in shape. The floor is 18 ft. 5 in long by 9 ft 6 in wide. All of the walls are 8 ft high. You will not paint the windows nor the doors. There are two doors each of which is 7 ft 3 in. high by 3 ft wide. One window is 4 ft wide by 5 ft. high. There are also two windows that are 2 ft 6 in wide by 5 ft high)

 

[MarkFL]

Let's ignore the doors and windows for now. You have 4 walls, which are rectangular. The walls opposite each other have identical areas. Can you find the total area of the 4 walls?

 

[drewcifur]

Let's ignore the doors and windows for now. You have 4 walls, which are rectangular. The walls opposite each other have identical areas. Can you find the total area of the 4 walls?

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

 

[MarkFL]

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:

\(\displaystyle \text{Area}=\text{base}\times\text{height}\)

or:

\(\displaystyle A=bh\)

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:

\(\displaystyle 18\text{ ft}\,5\text{ in}=\left(18+\frac{5}{12}\right)\text{ ft}=\frac{221}{12}\,\text{ft}\)

And we will have two walls whose bases are:

\(\displaystyle 9\text{ ft}\,6\text{ in}=\left(9+\frac{6}{12}\right)\text{ ft}=\frac{19}{2}\,\text{ft}\)

Now, if we add these all up, then the total area \(A\) of the 4 walls is then:

\(\displaystyle A=2\left(\frac{221}{12}\,\text{ft}\right)\left(8\text{ ft}\right)+2\left(\frac{19}{2}\,\text{ft}\right)\left(8\text{ ft}\right)\)

Let's factor this:

\(\displaystyle A=2(8)\left(\frac{221}{12}+\frac{19}{2}\right)\text{ ft}^2\)

Adding the fractions, and computing the product of the first two factors, we can write:

\(\displaystyle A=16\left(\frac{335}{12}\right)\text{ ft}^2\)

Reduce:

\(\displaystyle A=\frac{4\cdot335}{3}\,\text{ft}^2=\frac{1340}{3}\,\text{ft}^2\)

Before we proceed, does all this make sense so far?

 

[drewcifur]

The formula for the area of a rectangle is:

\(\displaystyle \text{Area}=\text{base}\times\text{height}\)

or:

\(\displaystyle A=bh\)

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:

\(\displaystyle 18\text{ ft}\,5\text{ in}=\left(18+\frac{5}{12}\right)\text{ ft}=\frac{221}{12}\,\text{ft}\)

And we will have two walls whose bases are:

\(\displaystyle 9\text{ ft}\,6\text{ in}=\left(9+\frac{6}{12}\right)\text{ ft}=\frac{19}{2}\,\text{ft}\)

Now, if we add these all up, then the total area \(A\) of the 4 walls is then:

\(\displaystyle A=2\left(\frac{221}{12}\,\text{ft}\right)\left(8\text{ ft}\right)+2\left(\frac{19}{2}\,\text{ft}\right)\left(8\text{ ft}\right)\)

Let's factor this:

\(\displaystyle A=2(8)\left(\frac{221}{12}+\frac{19}{2}\right)\text{ ft}^2\)

Adding the fractions, and computing the product of the first two factors, we can write:

\(\displaystyle A=16\left(\frac{335}{12}\right)\text{ ft}^2\)

Reduce:

\(\displaystyle A=\frac{4\cdot335}{3}\,\text{ft}^2=\frac{1340}{3}\,\text{ft}^2\)

Before we proceed, does all this make sense so far?

yes so far i am understanding a bit more clearly

 

[MarkFL]

yes so far i am understanding a bit more clearly

Okay, now we need to compute the total area of the windows and doors. What do you get for that?

 

[drewcifur]

Okay, now we need to compute the total area of the windows and doors. What do you get for that?

i'm not sure I'm right but I for the area of the doors I got 7ft 6in (7.5) and for the windows I have the first window an area of 20 ft and the two other windows with an area of 12 ft 6 in (12.5)

 

[MarkFL]

i'm not sure I'm right but I for the area of the doors I got 7ft 6in (7.5) and for the windows I have the first window an area of 20 ft and the two other windows with an area of 12 ft 6 in (12.5)

You are finding areas, so you want your units to be \(\displaystyle \text{ft}^2\), because you are multiplying two linear measures in units of \(\displaystyle \text{ft}\).

There are 2 doors of equal measure and so the area \(A_D\) of the doors is:

\(\displaystyle A_D=2(3\text{ ft})\left(\left(7+\frac{1}{4}\right)\text{ft}\right)=?\)