Question

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: Divide the wall shape

The wall can be divided into a rectangle and a triangle. The rectangle has a base of 18 feet and a height of 9 feet. The triangle has a base of 18 feet and a height of 6 feet.

Step2: Calculate rectangle area

The area formula for a rectangle is $A = lw$. Here, $l = 18$ feet and $w=9$ feet. So the area of the rectangle $A_{rectangle}=18\times9 = 162$ square - feet.

Step3: Calculate triangle area

The area formula for a triangle is $A=\frac{1}{2}bh$. Here, $b = 18$ feet and $h = 6$ feet. So the area of the triangle $A_{triangle}=\frac{1}{2}\times18\times6= 54$ square - feet.

Step4: Calculate total wall area

The total area of the wall $A = A_{rectangle}+A_{triangle}=162 + 54=216$ square - feet. This is the amount of wallpaper needed to cover the wall for part a.

Step5: Calculate length of wallpaper roll

For part b, the width of the wallpaper roll is 2 feet. Let the length of the roll be $L$. The area of the wallpaper roll is $A_{roll}=L\times2$. We know the area of the wall is 216 square - feet. We set $A_{roll}=A$. So $2L=216$, and $L=\frac{216}{2}=108$ feet.

Answer:

a. 216 square feet. The wall is composed of a rectangle and a triangle. The rectangle has area $18\times9 = 162$ square - feet and the triangle has area $\frac{1}{2}\times18\times6 = 54$ square - feet, so the total area is $162 + 54=216$ square - feet.
b. 108 feet. Since the area of the wall is 216 square - feet and the width of the wallpaper roll is 2 feet, using the formula $A = lw$ (where $A = 216$ and $w = 2$), we find $l=\frac{216}{2}=108$ feet.