Question

how many square yards of cement are needed to create the walkway around the rectangular pool?
176 square yards
196 square yards
208 square yards
280 square yards

Explanation:

Answer:

To find the area of the walkway, we calculate the area of the larger shape (including the pool and walkway) and subtract the area of the pool.

Step 1: Area of the Pool

The pool is a rectangle with length \( 12 \) yds and width \( 6 \) yds.
\[
\text{Area of pool} = \text{length} \times \text{width} = 12 \times 6 = 72 \text{ square yards}
\]

Step 2: Area of the Larger Shape (Pool + Walkway)

The larger shape can be decomposed into:

• Two rectangles (top and bottom) with dimensions \( 12 \) yds (length) and \( 4 \) yds (width).
• Two rectangles (left and right) with dimensions \( 6 + 4 + 4 = 14 \) yds (length) and \( 4 \) yds (width) – wait, no, actually, looking at the diagram, the vertical sides of the walkway: the height of the vertical rectangles (left and right) is \( 6 + 4 + 4 = 14 \)? Wait, no, let’s re-examine the diagram.

Wait, the larger shape’s horizontal length (top and bottom) is \( 12 \) yds, and the vertical height (left and right) is \( 6 + 4 + 4 = 14 \)? No, actually, the diagram shows:

• The top and bottom horizontal sections: length \( 12 \) yds, width \( 4 \) yds (two of these: top and bottom).
• The left and right vertical sections: height \( 6 + 4 + 4 = 14 \)? Wait, no, the pool’s height is \( 6 \) yds, and the walkway extends \( 4 \) yds above and below? Wait, no, the diagram has:
• The pool is \( 12 \) yds (length) and \( 6 \) yds (width).
• The walkway has:
• Top and bottom: each is a rectangle with length \( 12 \) yds and width \( 4 \) yds (so two rectangles: \( 2 \times 12 \times 4 \)).
• Left and right: each is a rectangle with height \( 6 + 4 + 4 = 14 \)? No, wait, the vertical sides (left and right) of the walkway: the height is \( 6 + 4 + 4 = 14 \)? Wait, no, the pool’s height is \( 6 \) yds, and the walkway extends \( 4 \) yds above and \( 4 \) yds below? No, the diagram shows the vertical segments (left and right) have length \( 6 \) yds? Wait, no, the diagram labels:
• The left and right sides of the walkway (the vertical parts) are \( 6 \) yds in length? Wait, no, let’s look again.

Wait, the correct way is to calculate the area of the larger octagon-like shape by breaking it into rectangles and squares. Alternatively, notice that the larger shape’s total length (horizontal) is \( 12 + 4 + 4 = 20 \) yds? No, the top and bottom are \( 12 \) yds, and the left and right have \( 4 \) yds extensions. Wait, maybe a better approach:

The larger shape can be considered as:

• A central rectangle (pool) plus the walkway. But actually, the walkway consists of:
• Two rectangles (top and bottom) with dimensions \( 12 \) yds (length) and \( 4 \) yds (width): \( 2 \times 12 \times 4 = 96 \)
• Two rectangles (left and right) with dimensions \( 6 \) yds (height) and \( 4 \) yds (width): \( 2 \times 6 \times 4 = 48 \)
• Four squares (the corners) with dimensions \( 4 \) yds (side): \( 4 \times 4 \times 4 = 64 \)

Wait, let’s check:

• Top and bottom: each is \( 12 \times 4 \), so \( 2 \times 12 \times 4 = 96 \)
• Left and right: each is \( 6 \times 4 \), so \( 2 \times 6 \times 4 = 48 \)
• Corners: four squares, each \( 4 \times 4 \), so \( 4 \times 16 = 64 \)

Total area of walkway: \( 96 + 48 + 64 = 208 \)? Wait, no, that can’t be. Wait, no—actually, the total area of the larger shape (including pool) is:

Wait, the larger shape’s horizontal length (top and bottom) is \( 12 \) yds, and the vertical height (left and right) is \( 6 + 4 + 4 = 14 \)? No, let’s use the formula for the area of the walkway: total area (outer) minus pool area.

Wait, the outer shape:

• The top and bottom are rectangles with length \( 12 \) yds and width \( 4 \) yds (two of them: \( 2 \times 12 \times 4 = 96 \))
• The left and right are rectangles with height \( 6 + 4 + 4 = 14 \) yds? No, the pool’s height is \( 6 \) yds, and the walkway extends \( 4 \) yds above and \( 4 \) yds below? Wait, no, the diagram shows the vertical segments (left and right) of the walkway are \( 6 \) yds in length? Wait, the diagram labels:
• The left side of the walkway (the vertical part) is \( 6 \) yds, and the horizontal extension (from the pool) is \( 4 \) yds.

Wait, maybe the outer shape is a rectangle with length \( 12 + 4 + 4 = 20 \) yds and width \( 6 + 4 + 4 = 14 \) yds? No, that would make the outer area \( 20 \times 14 = 280 \), and the pool area is \( 12 \times 6 = 72 \), so walkway area is \( 280 - 72 = 208 \). Wait, but let’s check:

Wait, if the outer length is \( 12 + 4 + 4 = 20 \) yds (since the walkway extends \( 4 \) yds on both sides of the pool’s length), and the outer width is \( 6 + 4 + 4 = 14 \) yds (walkway extends \( 4 \) yds on both sides of the pool’s width), then outer area is \( 20 \times 14 = 280 \). Pool area is \( 12 \times 6 = 72 \). Then walkway area is \( 280 - 72 = 208 \).

Yes, that matches one of the options: 208 square yards.

So the answer is 208 square yards (option C).