This story problem pertains to both geometry and algebra.
A dog has a dog house shaped like a regular hexagon with a base that measures $2$ yards on each side. The dog is tethered to one of the angles with a $4$ yard rope.
What is the area in square yards that the dog can reach?
I have attempted this problem twice, but each time end up with outrageous answers. I know that I must calculate the area that the dog can walk first, using $4$ as the circumference(?) of the circle. Once I complete this step I get lost. All help would be greatly appreciated. Thanks!
According to the sketch the OP provides, I shall interpret the problem as the leash will trace out a circle of radius $4$ until it hits one side of the doghouse.
The side length of the doghouse is denoted as $\ell = 2$.
For visual clarify, the diagram below shows only the upper half of the "grazing" region.
\begin{align} \text{total area} &= \text{twice of}~\Bigl(\text{one third large disc (green)} + \text{one sixth small disc (purple)} \Bigr) \\ &= 2 \left( \frac13 \pi R^2 + \frac16 \pi r^2\right) \qquad \text{, with}~R = 4 ~~,~~ r = R-\ell = 2 \\ &=2 \pi \left( \frac{16}3 + \frac46 \right) \\ &= 12 \pi \end{align} To see that the discs are a third and a sixth, simply note that the doghouse is hexagonal and the relevant angles are $2\pi/3$ and $\pi/3$, respectively. There's not a third disc because $r - \ell = 0$.