Question

problem 7
the figure is a diagram of a wall. lengths are given in feet.
a. how many square feet of wallpaper would be needed to cover the wall? explain your reasoning.
b. wallpaper is sold in rolls that are 2 feet wide. what is the minimum length you would need to purchase to cover the wall?
grade 6 unit 1: mid - unit assessment - version a
tennessee standards edition cc by - nc 4.0. download for free at openup.org.

Explanation:

Step1: Decompose the wall shape

The wall can be decomposed into a rectangle and a triangle. The rectangle has length 48 feet and height 6 feet, and the triangle has base 48 feet and height 9 - 6=3 feet.

Step2: Calculate rectangle area

The area formula for a rectangle is $A_{rect}=l\times h$. So, $A_{rect}=48\times6 = 288$ square - feet.

Step3: Calculate triangle area

The area formula for a triangle is $A_{tri}=\frac{1}{2}\times b\times h$. So, $A_{tri}=\frac{1}{2}\times48\times3=72$ square - feet.

Step4: Calculate total wall area

The total area of the wall $A = A_{rect}+A_{tri}=288 + 72=360$ square - feet. This answers part a.

Step5: Calculate length of wallpaper roll

We know the width of the wallpaper roll is $w = 2$ feet. Let the length be $L$. The area of the wallpaper roll is $A_{roll}=L\times w$. We want $A_{roll}\geq A$. Since $A = 360$ square - feet and $w = 2$ feet, from $A_{roll}=L\times w$, we can solve for $L$: $L=\frac{A}{w}=\frac{360}{2}=180$ feet. This answers part b.

Answer:

a. 360 square feet. The wall is composed of a rectangle with area $48\times6 = 288$ square - feet and a triangle with area $\frac{1}{2}\times48\times3 = 72$ square - feet, and $288+72 = 360$ square - feet.
b. 180 feet. Since the area of the wall is 360 square feet and the width of the wallpaper roll is 2 feet, the length $L=\frac{360}{2}=180$ feet.