Question

a rectangular yard measuring 36 ft by 67 ft is bordered (and surrounded) by a fence. inside, a walk that is 2 ft wide goes all the way along the fence. find the area of this walk. be sure to include the correct unit in your answer.

Explanation:

Step1: Determine the dimensions of the walk - area

The walk is along the fence, so we need to consider the area of the walk. But actually, maybe we first find the area of the larger rectangle (yard + walk) and subtract the area of the yard? Wait, no, the walk is inside along the fence. Wait, the yard is 36 ft by 67 ft, and the walk is 2 ft wide along the fence. Wait, maybe the walk is a border inside? Wait, no, the problem says "inside, a walk that is 2 ft wide goes all the way along the fence". So the walk is along the inner side of the fence. Wait, maybe we need to find the area of the walk. Let's think: the walk is along the length and width. So the area of the walk can be calculated as the sum of the areas of the two rectangles along the length, two rectangles along the width, and the four squares at the corners? Wait, no, maybe a better way: the area of the walk is equal to the perimeter of the yard multiplied by the width of the walk, plus the area of the four corner squares (but since the width is 2 ft, the corner squares are 2x2, and there are four of them, but actually, when we do perimeter times width, we are already including the corners? Wait, no. Wait, the formula for the area of a border (walk) along a rectangle: if the rectangle has length \( l \) and width \( w \), and the border (walk) has width \( d \), then the area of the border is \( 2d(l + w)+4d^{2}\)? Wait, no, that's if it's a border outside. But here it's inside? Wait, no, the problem says "inside, a walk that is 2 ft wide goes all the way along the fence". So the fence is around the yard, and the walk is inside the yard, along the fence. So the yard is 36 ft (width) by 67 ft (length). The walk is 2 ft wide, so along the length (67 ft) sides, the walk has a width of 2 ft, and along the width (36 ft) sides, also 2 ft. Wait, maybe the walk is a rectangle that is along the length and width. So the area of the walk can be calculated as: the two rectangles along the length: each has length 67 ft and width 2 ft, so two of them: \( 2\times(67\times2) \). The two rectangles along the width: but we have to subtract the parts that are already counted in the length rectangles. Wait, the width of the yard is 36 ft, but the walk along the width: the length of these rectangles is \( 36 - 2\times2 \) ft (because the two length - side walks have already taken 2 ft from each end of the width). Wait, no, maybe I made a mistake. Let's re - read the problem: "A rectangular yard measuring 36 ft by 67 ft is bordered (and surrounded) by a fence. Inside, a walk that is 2 ft wide goes all the way along the fence. Find the area of this walk."

So the yard is 36 ft (let's say width \( w = 36 \)) and 67 ft (length \( l = 67 \)). The walk is inside, along the fence, so it's a border inside the yard. So the area of the walk can be found by calculating the area of the region that is 2 ft wide along the length and width of the yard.

The area of the walk is equal to the sum of the areas of the two rectangles along the length of the yard, the two rectangles along the width of the yard, and the four square corners (but the square corners are \( 2\times2 \) each, and there are four of them). Wait, but actually, when we calculate the area of the walk, we can use the formula:

Area of walk \(= 2\times l\times d+2\times(w - 2d)\times d\)

where \( l = 67 \) ft (length of yard), \( w = 36 \) ft (width of yard), and \( d = 2 \) ft (width of walk).

Let's calculate:

First, the two rectangles along the length: each has area \( l\times d \), so two of them: \( 2\times67\times2=268 \) square feet.

Then, the two rectangles along the width: the length of these rectangles is \( w - 2d \) (because we have already accounted for the 2 ft on both ends from the length - side walks). So \( w-2d=36 - 2\times2 = 32 \) ft. Each of these rectangles has area \( 32\times2 \), and there are two of them: \( 2\times32\times2 = 128 \) square feet.

Then, the total area of the walk is \( 268+128 = 396 \) square feet.

Wait, another way: the area of the walk is equal to the perimeter of the inner rectangle multiplied by the width of the walk plus the area of the four corner squares? No, the inner rectangle (the part of the yard not including the walk) would have length \( l - 2d=67 - 2\times2 = 63 \) ft and width \( w - 2d = 36 - 2\times2=32 \) ft. Then the area of the yard is \( l\times w=67\times36 = 2412 \) square feet. The area of the inner part (yard without walk) is \( (l - 2d)\times(w - 2d)=63\times32 = 2016 \) square feet. Then the area of the walk is \( 2412-2016 = 396 \) square feet. Yes, this is a better way.

Step1: Calculate the area of the yard

The area of the rectangular yard is given by the formula \( A_{yard}=l\times w \), where \( l = 67 \) ft and \( w = 36 \) ft.

\( A_{yard}=67\times36=2412 \) square feet.

Step2: Calculate the dimensions of the inner rectangle (yard without walk)

The walk is 2 ft wide, so along the length, we subtract \( 2d \) (where \( d = 2 \) ft) from the length of the yard, and along the width, we subtract \( 2d \) from the width of the yard.

The length of the inner rectangle \( l_{inner}=l - 2d=67-2\times2 = 63 \) ft.

The width of the inner rectangle \( w_{inner}=w - 2d=36 - 2\times2 = 32 \) ft.

Step3: Calculate the area of the inner rectangle

The area of the inner rectangle is \( A_{inner}=l_{inner}\times w_{inner}=63\times32 = 2016 \) square feet.

Step4: Calculate the area of the walk

The area of the walk is the area of the yard minus the area of the inner rectangle.

\( A_{walk}=A_{yard}-A_{inner}=2412 - 2016=396 \) square feet.

Answer:

\( 396\space ft^{2} \)