they are told that they will teach in one large room with a perimeter of 228 ft and the length is 6 ft greater than twice its width. what are the dimensions of

Question

Two teachers are team-teaching for the first time together at Null Elementary school. They are planning on shopping together to purchase decorations for their new room. they are told that they will teach in one large room with a perimeter of 228 ft and the length is 6 ft greater than twice its width. what are the dimensions of their classroom?

Answer 1

Hi Brook,

 

Let's call the perimeter of the classroom P for short.

Likewise, we'll call the width W and the length L.

 

Step 1. First we write an equation for the perimeter of the classroom as follows: P = 2W + 2L = 228

(Why is this the equation? The classroom is in the shape of a rectangle with two short sides (2W) and two long sides (2L). We know that p is also equal to 228 because it says so in the problem).

 

Step 2. Next, we write an equation that describes L in terms of W. The problem says that "the length is 6ft greater than twice its width." Translating that phrase into an equation we have: L = 6 + 2W

 

Step 3. Substitute our equation from step 2 into our equation in step 1.

We get the following: P= 2W + 2(6+2W) = 228

Simplifying this equation gives you: P=2W + 12 +4W = 228

Simplify further... P=6W + 12 = 228

 

Step 4. Solve the equation in Step 3 to get W.

6W = 228 -12

6W = 216

W = 216/6

W = 36

 

Step 5. We can use our answer from step 4 and our equation from step 2 to solve for L.

L = 6 + 2W

L= 6 + 2(36)

L= 78

 

You've got your answers: width=36 and length =78