A hexagonal right prism has a volume of 500 cubic inches. If the base is a regular hexagon with a side 4 inches, what is the altitude of the prism ? Round off your answer to two decimal places.
Hello,
You have to calculate the hexagon area A, because the volume of the prism is altitude*A. As the hexagon is regular, you know that it's composed of 6 identical triangles, isocele in O :
As the angle at O is 360/6 = 60°, we can state that all triangles are equilateral, because each side is already equal.
So each side of a triangle will be 4 inches.
Now let's calculate the length of the altitude from O to [AB], OH, in the triangle ABO, so as to get the area of the hexagon.
The altitude will also be the bisection of the segment ([AB] for instance) -> H is the midpoint of [AB]
So by the Pythagor theorem, OH˛=OA˛-AH˛=4˛-2˛=12 -> $\displaystyle OH=\sqrt{12}=2 \sqrt{3}$
The area of the triangle ABO is $\displaystyle \frac{OH.AB}{2} = \frac{2 \sqrt{3} . 4}{2} = 4 \sqrt{3}$
A, the area of the hexagon will be 6 times ABO's area. So $\displaystyle A=24 \sqrt{3}$ square inches
Hence the altitude of the prism, h, is : $\displaystyle 500=24 \sqrt{3} * h$
-> $\displaystyle h=\frac{500}{24 \sqrt{3}}$
I hope there is no mistake :s