+0

Polygons Area

0
114
1
+973

Fido's leash is tied to a stake at the center of his yard, which is in the shape of a regular hexagon. His leash is exactly long enough to reach the midpoint of each side of his yard. If the fraction of the area of Fido's yard that he is able to reach while on his leash is expressed in simplest radical form as $$\frac{\sqrt{a}}{b}\pi$$, what is the value of the product $$ab$$?

Dec 2, 2018

#1
+4471
+1

$$\text{Fido can reach a circle of radius }r\\ A_c = \pi r^2 \\ \text{the apothem of the hexagon is also }r \\ \text{a side will be such that }\tan(30^\circ) = \dfrac{\frac s 2}{r}\\ s = \dfrac{2r}{\sqrt{3}}\\ A_h = \dfrac 1 2 s a n = \dfrac 1 2 \cdot \dfrac{2r}{\sqrt{3}}\cdot r\cdot 6 = 2\sqrt{3}r^2\\ \dfrac{A_c}{A_h} = \dfrac{\pi r^2}{2\sqrt{3}r^2} = \dfrac{\pi}{2\sqrt{3}}= \dfrac{\sqrt{3}}{6} \pi\\ a = 3,~b=6,~ab=18$$

.
Dec 2, 2018

$$\text{Fido can reach a circle of radius }r\\ A_c = \pi r^2 \\ \text{the apothem of the hexagon is also }r \\ \text{a side will be such that }\tan(30^\circ) = \dfrac{\frac s 2}{r}\\ s = \dfrac{2r}{\sqrt{3}}\\ A_h = \dfrac 1 2 s a n = \dfrac 1 2 \cdot \dfrac{2r}{\sqrt{3}}\cdot r\cdot 6 = 2\sqrt{3}r^2\\ \dfrac{A_c}{A_h} = \dfrac{\pi r^2}{2\sqrt{3}r^2} = \dfrac{\pi}{2\sqrt{3}}= \dfrac{\sqrt{3}}{6} \pi\\ a = 3,~b=6,~ab=18$$