A dog's leash is $4 \mathrm{~m}$ long and is attached to the corner of a $2 \mathrm{~m} \times 2 \mathrm{~m}$ square doghouse at $C$, as shown. The $\operatorname{dog}$ is attached to the other end of the leash, at $D$. What is the area outside of the doghouse in which the dog can play?
(A) $14 \pi \mathrm{m}^{2}$
(B) $16 \pi \mathrm{m}^{2}$
(D) $15 \pi \mathrm{m}^{2}$
(E) $24 \pi \mathrm{m}^{2}$
(C) $20 \pi \mathrm{m}^{2}$
We begin by labelling some additional points as shown.

With the leash extended the full $4 \mathrm{~m}$, the dog can reach points $A$, $B$ and $D$ where $A C=B C=D C=4 \mathrm{~m}$ and thus the dog is able to play anywhere inside the shaded area shown.

The doghouse $E F G C$ is a square and so $\angle E C G=90^{\circ}$.

Therefore, the shaded figure is $\frac{3}{4}$ of a circle centred at $C$, has radius $C D=4 \mathrm{~m}$, and thus has area $\frac{3}{4} \pi(4 \mathrm{~m})^{2}=12 \pi \mathrm{m}^{2}$.

There is additional area outside the doghouse in which the dog can play, as shaded in the diagram.

Since $A C=4 \mathrm{~m}$ and $E C=2 \mathrm{~m}$, then $A E=2 \mathrm{~m}$. Since $E C+E F=2 \mathrm{~m}+2 \mathrm{~m}=4 \mathrm{~m}$, the length of the leash, then the dog can reach $F$.

Similarly, $B G=2 \mathrm{~m}$ and $G C+G F=2 \mathrm{~m}+2 \mathrm{~m}=4 \mathrm{~m}$, the length of the leash, and so the $\operatorname{dog}$ can also reach $F$ by travelling around sides $G C$ and $G F$ of the doghouse.

Since $\angle A E F=\angle F G B=90^{\circ}$, then each of these two shaded figures is $\frac{1}{4}$ of a circle (centred at $E$ and $G$ respectively), has radius $2 \mathrm{~m}$,

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-060.jpg?height=421&width=423&top_left_y=131&top_left_x=1537)
and thus each has area $\frac{1}{4} \pi(2)^{2}=\pi \mathrm{m}^{2}$. (For example, when the dog is above and to the right of $E$, it can stretch to at most $2 \mathrm{~m}$ of rope and so can form a quarter circle of radius $2 \mathrm{~m}$ coming from $E$.)

The area outside of the doghouse in which the dog can play is $12 \pi \mathrm{m}^{2}+2 \times \pi \mathrm{m}^{2}=14 \pi \mathrm{m}^{2}$.

ANSwER: (A)