Question

charisse is adding a weather protective stain to the exterior of her shed. she measures the walls to be 8 feet long and 6 feet high. there are no windows on the shed, and the roof and the bottom do not need the protective stain. additionally, the front door, which is 3 feet wide and 6 feet high, does not need to be stained. what is the total surface area charisse needs to stain?
a) 66 ft²
b) 144 ft²
c) 174 ft²
d) 192 ft²

Explanation:

Step1: Assume the shed is a rectangular prism. First, find the surface area of the walls (excluding the bottom, roof, and the door). The shed has two pairs of walls: length×height and width×height. Wait, actually, let's clarify the dimensions. Wait, maybe the shed is a rectangular structure with length 8 ft, height 6 ft, and let's assume the width is also 8 ft? Wait, no, maybe the shed has length 8 ft, height 6 ft, and the front/back walls are length×height, and the side walls are width×height. Wait, maybe the shed is a rectangular prism with length \( l = 8 \) ft, height \( h = 6 \) ft, and let's assume the width \( w = 8 \) ft? No, maybe the base is a rectangle with length 8 ft and width, say, let's re - read the problem. The problem says "the walls to be 8 feet long and 6 feet high". Wait, maybe the shed is a rectangular prism with length \( l = 8 \) ft, height \( h = 6 \) ft, and the width (depth) is also 8 ft? No, maybe the shed has four walls: two walls with dimensions 8 ft (length) × 6 ft (height) and two walls with dimensions, let's say, the width is also 8 ft? Wait, no, maybe the shed is a rectangular structure where the length is 8 ft, height is 6 ft, and the front door is on one of the 8 ft×6 ft walls. Wait, the door is 3 ft wide and 6 ft high.

First, calculate the total surface area of the four walls (excluding bottom and roof). The formula for the lateral surface area (walls) of a rectangular prism is \( 2(lh+wh) \). But we need to subtract the area of the door. Wait, let's assume the shed has length \( l = 8 \) ft, width \( w = 8 \) ft? No, maybe the length is 8 ft, and the width is, let's see, maybe the shed is a rectangular prism with length \( l = 8 \) ft, height \( h = 6 \) ft, and width \( w = 8 \) ft? Wait, no, the problem says "the walls to be 8 feet long and 6 feet high". So maybe there are two walls of 8 ft (length) × 6 ft (height) and two walls of, let's say, the other dimension. Wait, maybe the shed is a square - like base? No, let's re - approach.

Wait, the problem says "the walls to be 8 feet long and 6 feet high". Let's assume that the shed has a rectangular base with length \( l = 8 \) ft and width \( w = 8 \) ft (so it's a square base), but no, maybe the length is 8 ft, height is 6 ft, and the front door is on one of the 8 ft×6 ft walls. The door is 3 ft wide and 6 ft high.

First, calculate the area of all four walls. The lateral surface area (walls) of a rectangular prism is \( 2(lh + wh) \). If we assume that the length \( l = 8 \) ft and the width \( w = 8 \) ft (so it's a square - shaped base for the walls), then the lateral surface area is \( 2(8\times6+8\times6)=2(48 + 48)=192 \) square feet. But we need to subtract the area of the door. The door is 3 ft wide and 6 ft high, so the area of the door is \( 3\times6 = 18 \) square feet. Wait, but also, the problem says "the roof and the bottom do not need the protective stain". So we are only dealing with the walls. Wait, no, the problem says "the exterior of her shed. She measures the walls to be 8 feet long and 6 feet high. There are no windows on the shed, and the roof and the bottom do not need the protective stain. Additionally, the front door, which is 3 feet wide and 6 feet high, does not need to be stained. What is the total surface area Charisse needs to stain?"

Wait, maybe the shed has four walls: two walls with length 8 ft and height 6 ft, and two walls with length, let's say, the same as the other side. Wait, maybe the shed is a rectangular prism with length \( l = 8 \) ft, height \( h = 6 \) ft, and width \( w = 8 \) ft (so it's a cube - like in length and width). Then the lateral surface area (walls) is \( 2(lh+wh)=2(8\times6 + 8\times6)=192 \) square feet. Then we subtract the area of the door. The door is 3 ft×6 ft, so area of door is \( 3\times6 = 18 \) square feet. So \( 192-18 = 174 \) square feet. Let's check:

If the shed has two walls of 8 ft (length) × 6 ft (height): area of each is \( 8\times6 = 48 \), two of them is \( 2\times48 = 96 \). Then the other two walls: wait, maybe the width is also 8 ft? No, maybe the length is 8 ft, and the width is, say, 6 ft? Wait, no, let's re - calculate.

Alternative approach: The shed has four walls. Let's assume the shed is a rectangular structure with length \( l = 8 \) ft, height \( h = 6 \) ft, and the front and back walls are \( l\times h \) (8×6) and the left and right walls are \( w\times h \). But we don't know \( w \). Wait, maybe the shed is a square with side 8 ft? No, the problem says "the walls to be 8 feet long and 6 feet high". Maybe "long" here refers to the length of the wall, so each wall is 8 ft long (in the horizontal direction) and 6 ft high. So there are four walls: two walls with dimensions 8 ft (length) × 6 ft (height) and two walls with dimensions, let's say, the other side is also 8 ft? No, this is confusing. Wait, the answer options include 174. Let's see:

Total area of four walls (excluding bottom and roof): Let's assume the shed has length \( l = 8 \) ft, height \( h = 6 \) ft, and width \( w = 8 \) ft. Then lateral surface area is \( 2(lh+wh)=2(8\times6 + 8\times6)=192 \). The door is 3 ft wide and 6 ft high, area of door is \( 3\times6 = 18 \). So \( 192 - 18=174 \), which is option C.

Step2: Calculate the area of the walls, subtract the area of the door.

- Area of four walls (lateral surface area of rectangular prism): Assume the shed has length \( l = 8 \) ft, height \( h = 6 \) ft, and width \( w = 8 \) ft. The formula for lateral surface area is \( 2(lh + wh) \).
- \( lh=8\times6 = 48 \), \( wh = 8\times6=48 \)
- \( 2(48 + 48)=2\times96 = 192 \) square feet.
- Area of the door: The door is 3 ft wide and 6 ft high, so area \( A_{door}=3\times6 = 18 \) square feet.
- Area to be stained: \( 192-18 = 174 \) square feet.

Answer:

C) \( 174\space\text{ft}^2 \)