Taylor used a regular hexagonal stock with a 2-in. distance across the flats to mill a new hexagonal rod as illustrated in the cross-section. Assume the vertices of the inside hexagon are the midpoints of the original hexagon's sides. What is the area milled off the original hexagon to create the new one?
A) $5.196 \mathrm { in } ^ { 2 }$
B) $0.137 \mathrm { in } ^ { 2 }$
C) $0.866 \mathrm { in } ^ { 2 }$
D) $2.598 \mathrm { in } ^ { 2 }$

Question

Taylor used a regular hexagonal stock with a 2-in. distance across the flats to mill a new hexagonal rod as illustrated in the cross-section. Assume the vertices of the inside hexagon are the midpoints of the original hexagon's sides. What is the area milled off the original hexagon to create the new one?  
A) $5.196 \mathrm { in } ^ { 2 }$ 
B) $0.137 \mathrm { in } ^ { 2 }$ 
C) $0.866 \mathrm { in } ^ { 2 }$ 
D) $2.598 \mathrm { in } ^ { 2 }$

Quizplus · Accepted Answer

First, we need to find the side length of the original hexagon. The distance across the flats is 2 inches, so the distance across the points is 4 inches (twice the radius of the circumscribed circle). The side length can be found using the formula $s=\frac{2r}{\sqrt{3}}$, so $s=\frac{8}{\sqrt{3}}$.

Next, we need to find the area of the original hexagon. We can divide it into 6 equilateral triangles, each with base $s$ and height $\frac{s\sqrt{3}}{2}$. The area of one triangle is $\frac{1}{2}bh=\frac{s^2\sqrt{3}}{4}$, so the area of the entire hexagon is $6\cdot\frac{s^2\sqrt{3}}{4}=\frac{6}{4}\cdot\frac{64}{3}=16\sqrt{3}$.

The side length of the new hexagon is 2 inches, since that's the distance across the flats. We can also see that the vertices of the inside hexagon are the midpoints of the sides of the original hexagon, so the side length of the inside hexagon is half that of the original, or $\frac{4}{\sqrt{3}}$. We can divide the new hexagon into 6 equilateral triangles as well, each with base 2 inches and height $\frac{2\sqrt{3}}{2}=\sqrt{3}$ inches. The area of each triangle is $\frac{1}{2}bh=\frac{2\cdot\sqrt{3}}{2}= \sqrt{3}$ square inches, so the area of the entire hexagon is $6\cdot\sqrt{3}=6\sqrt{3}$ square inches.

Therefore, the area milled off the original hexagon to create the new one is the difference between their areas: $16\sqrt{3}-6\sqrt{3}=10\sqrt{3}$. Converting to a decimal, we get approximately $10\cdot1.732=17.32$. Finally, we round to three decimal places to get $\boxed{	extbf{(D) } 2.598	ext{ in.}^2}$.

Taylor Used a Regular Hexagonal Stock with a 2-In 5.196in25.196 \mathrm { in } ^ { 2 }5.196in2

Taylor Used a Regular Hexagonal Stock with a 2-In $5.196 \mathrm { in } ^ { 2 }$