Answer :
Sure, let’s go through the calculations step-by-step to determine how many square feet of wallpaper a woman will need to cover the walls and ceiling of a laundry room, accounting for openings.
1. Dimensions of the Room:
- The room's dimensions are 10 feet by 10 feet, with a height of 10 feet.
2. Calculate the Surface Area of the Walls:
- Front and Back Walls:
The room has two walls each measuring 10 feet in width and 10 feet in height.
[tex]\[ \text{Area of one wall} = 10 \, \text{ft (width)} \times 10 \, \text{ft (height)} = 100 \, \text{ft}^2 \][/tex]
Since there are two such walls:
[tex]\[ \text{Total area of front and back walls} = 2 \times 100 \, \text{ft}^2 = 200 \, \text{ft}^2 \][/tex]
- Side Walls:
The other two walls measure 10 feet in length and 10 feet in height.
[tex]\[ \text{Area of one wall} = 10 \, \text{ft (length)} \times 10 \, \text{ft (height)} = 100 \, \text{ft}^2 \][/tex]
Since there are two such walls:
[tex]\[ \text{Total area of side walls} = 2 \times 100 \, \text{ft}^2 = 200 \, \text{ft}^2 \][/tex]
3. Calculate the Surface Area of the Ceiling:
- The ceiling has an area equal to the length times the width of the room.
[tex]\[ \text{Ceiling area} = 10 \, \text{ft (length)} \times 10 \, \text{ft (width)} = 100 \, \text{ft}^2 \][/tex]
4. Total Surface Area to be Covered:
Add all the surface areas of the walls and the ceiling.
[tex]\[ \text{Total surface area} = \text{Area of front and back walls} + \text{Area of side walls} + \text{Ceiling area} \][/tex]
[tex]\[ \text{Total surface area} = 200 \, \text{ft}^2 + 200 \, \text{ft}^2 + 100 \, \text{ft}^2 = 500 \, \text{ft}^2 \][/tex]
5. Account for Openings:
Subtract the area of the openings from the total surface area.
[tex]\[ \text{Area of openings} = 93 \, \text{ft}^2 \][/tex]
[tex]\[ \text{Wallpaper needed} = \text{Total surface area} - \text{Area of openings} \][/tex]
[tex]\[ \text{Wallpaper needed} = 500 \, \text{ft}^2 - 93 \, \text{ft}^2 = 407 \, \text{ft}^2 \][/tex]
Therefore, the woman will need [tex]\( 407 \, \text{ft}^2 \)[/tex] of wallpaper to cover the walls and ceiling, accounting for the 93 square feet of openings in the room.
1. Dimensions of the Room:
- The room's dimensions are 10 feet by 10 feet, with a height of 10 feet.
2. Calculate the Surface Area of the Walls:
- Front and Back Walls:
The room has two walls each measuring 10 feet in width and 10 feet in height.
[tex]\[ \text{Area of one wall} = 10 \, \text{ft (width)} \times 10 \, \text{ft (height)} = 100 \, \text{ft}^2 \][/tex]
Since there are two such walls:
[tex]\[ \text{Total area of front and back walls} = 2 \times 100 \, \text{ft}^2 = 200 \, \text{ft}^2 \][/tex]
- Side Walls:
The other two walls measure 10 feet in length and 10 feet in height.
[tex]\[ \text{Area of one wall} = 10 \, \text{ft (length)} \times 10 \, \text{ft (height)} = 100 \, \text{ft}^2 \][/tex]
Since there are two such walls:
[tex]\[ \text{Total area of side walls} = 2 \times 100 \, \text{ft}^2 = 200 \, \text{ft}^2 \][/tex]
3. Calculate the Surface Area of the Ceiling:
- The ceiling has an area equal to the length times the width of the room.
[tex]\[ \text{Ceiling area} = 10 \, \text{ft (length)} \times 10 \, \text{ft (width)} = 100 \, \text{ft}^2 \][/tex]
4. Total Surface Area to be Covered:
Add all the surface areas of the walls and the ceiling.
[tex]\[ \text{Total surface area} = \text{Area of front and back walls} + \text{Area of side walls} + \text{Ceiling area} \][/tex]
[tex]\[ \text{Total surface area} = 200 \, \text{ft}^2 + 200 \, \text{ft}^2 + 100 \, \text{ft}^2 = 500 \, \text{ft}^2 \][/tex]
5. Account for Openings:
Subtract the area of the openings from the total surface area.
[tex]\[ \text{Area of openings} = 93 \, \text{ft}^2 \][/tex]
[tex]\[ \text{Wallpaper needed} = \text{Total surface area} - \text{Area of openings} \][/tex]
[tex]\[ \text{Wallpaper needed} = 500 \, \text{ft}^2 - 93 \, \text{ft}^2 = 407 \, \text{ft}^2 \][/tex]
Therefore, the woman will need [tex]\( 407 \, \text{ft}^2 \)[/tex] of wallpaper to cover the walls and ceiling, accounting for the 93 square feet of openings in the room.