Charlyn walks completely around the boundary of a square whose sides are each $5$ km long. From any point on her path she can see exactly $1$ km horizontally in all directions. What is the area of the region consisting of all points Charlyn can see during her walk, expressed in square kilometers and rounded to the nearest whole number? $\textbf{(A) } 24 \qquad$ $\textbf{(B)}\ 27 \qquad$ $\textbf{(C)}\ 39 \qquad$ $\textbf{(D)}\ 40 \qquad$ $\textbf{(E)}\ 42$
Solution
The region Charlyn can see consists of all points within $1$ km of the square's boundary (a $1$-dimensional curve), both inside and outside the square.
Outside the square: expanding the square outward by $1$ gives area $5^2+4(5)(1)+\pi(1)^2$ minus the original square, i.e. $4(5)(1)+\pi=20+\pi$.
Inside the square: the visible strip is the square minus the inner square of side $5-2=3$, giving $25-9=16$.
Total area $=(20+\pi)+16=36+\pi\approx39.14$.
Rounded to the nearest whole number, the answer is (C) $39$.
Maya walks completely around the boundary of a square whose sides are each $6$ km long. From any point on her path she can see exactly $2$ km horizontally in all directions. What is the area of the region consisting of all points Maya can see during her walk, expressed in square kilometers and rounded to the nearest whole number?
