A rectangular pool measures $20 \mathrm{~m}$ by $8 \mathrm{~m}$. There is a $1 \mathrm{~m}$ wide walkway around the outside of the pool, as shown by the shaded region. The area of the walkway is
(A) $56 \mathrm{~m}^{2}$
(B) $60 \mathrm{~m}^{2}$
(C) $29 \mathrm{~m}^{2}$
(D) $52 \mathrm{~m}^{2}$
(E) $50 \mathrm{~m}^{2}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-007.jpg?height=198&width=569&top_left_y=2319&top_left_x=1163)
Solution 1

The area of the walkway is equal to the area of the pool subtracted from the combined area of the pool and walkway.

That is, if the area of the walkway is $A_{w}$ and the area of the pool is $A_{p}$, then $A_{w}=\left(A_{w}+A_{p}\right)-A_{p}$. The combined area of the pool and walkway is equal to the area of the rectangle with length $22 \mathrm{~m}$ and width $10 \mathrm{~m}$.

The length, $22 \mathrm{~m}$, is given by the $20 \mathrm{~m}$ pool length plus the $1 \mathrm{~m}$ wide walkway on each end of the pool.

Similarly, the $10 \mathrm{~m}$ width is given by the $8 \mathrm{~m}$ pool width plus the $1 \mathrm{~m}$ wide walkway on each side of the pool.

Thus the area of the walkway is

$$
A_{w}=\left(A_{w}+A_{p}\right)-A_{p}=(22 \mathrm{~m} \times 10 \mathrm{~m})-(20 \mathrm{~m} \times 8 \mathrm{~m})=220 \mathrm{~m}^{2}-160 \mathrm{~m}^{2}=60 \mathrm{~m}^{2}
$$

## Solution 2

We begin by extending each side of the pool $1 \mathrm{~m}$ in each direction. This divides the area of the walkway into four $1 \mathrm{~m}$ by $1 \mathrm{~m}$ squares (in the corners), and two $20 \mathrm{~m}$ by $1 \mathrm{~m}$ rectangles (along the $20 \mathrm{~m}$ sides of the pool), and two $8 \mathrm{~m}$ by $1 \mathrm{~m}$ rectangles (along the $8 \mathrm{~m}$ sides of the pool), as shown.

Thus the area of the walkway is

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-036.jpg?height=231&width=489&top_left_y=1611&top_left_x=1463)

$$
4 \times(1 \mathrm{~m} \times 1 \mathrm{~m})+2 \times(20 \mathrm{~m} \times 1 \mathrm{~m})+2 \times(8 \mathrm{~m} \times 1 \mathrm{~m})=4 \mathrm{~m}^{2}+40 \mathrm{~m}^{2}+16 \mathrm{~m}^{2}=60 \mathrm{~m}^{2}
$$

ANSWER: (B)