Volume of a Pentagonal Prism: Formula and Examples
A pentagonal prism is a solid with two equal size pentagonal faces parallel to each other on opposite ends of the prism, with rectangles or parallelograms connecting the pentagons. If it is a right pentagonal prism, the side faces are rectangles. Otherwise, the five faces connecting the pentagons are parallelograms. It's easy to find the volume of a pentagonal prism if you know the area of the pentagonal face and the height of the prism between the two parallel ends. Here is an explanation of the volume formula with some example calculations worked out.
Pentagonal Prism Volume Equation
If the area of the pentagon face is A and the height of the prism is H, then the volume V is given by the very simple formula
V = A*H
The tricky part in applying this formula is finding the area of the pentagon, and in the case of a slant or oblique prism, finding the height. If a regular pentagon has a side length of S, then the area of the pentagon is (S^2)*sqrt[25 + 10*sqrt(5)]/4 ≈ 1.720477*S^2. If the pentagon is irregular, there is no single formula for its area, but you can find its area by dividing the polygon into three triangles and use one of the many area formulas for triangles to find the total area.
A right pentagonal prism has cross-sections that are regular pentagons. If the side length of the pentagonal faces is 4 cm and the height is 11 cm, what is the volume of the prism in cubic cm?
The area of the pentagon face is 1.720477*(4^2) = 27.527632 square cm. Multiplying this by the height gives us a volume of 27.527632*11 = 302.803952, or roughly 302.8 cubic cm.
An oblique pentagonal prism has a height of 2 feet. The pentagonal faces have dimensions in inches as shown in the image below. What is the volume?
Since the height is given as 2 feet, it doesn't matter if the prism is oblique (slanted) or not, so we have H = 24 inches. Next we need to find the area of the pentagon in square inches. This shape is a rectangle capped by an isosceles triangle; its area is the sum of the rectangular and triangular areas. Using the Pythagorean Theorem, we can see that the height of the triangle is 6. Therefore, the total area of the pentagon is 16*5 + (1/2)16*6 = 128 square inches.
The volume of the pentagonal prism is 24*128 = 3072 cubic inches, or 1.778 cubic feet.
A regular right pentagonal prism has a length of 18.9 cm and a volume of 452.7 cm^3. What is the diagonal length of one of the rectangular faces, rounded to the nearest millimeter?
For this problem we need to first figure the area of the pentagonal faces, then find the side length of the pentagon, and then use the Pythagorean Theorem to find the diagonal length of a rectangular face of the prism.
First we solve 452.7 = 18.9*A, which gives us A = 23.952381. Next, we use the fact that the area of a regular pentagon is 1.720477 times the squared side length. This means the side length of the pentagon is sqrt(23.952381/1.720477) = 3.731212 cm.
If a rectangle has a width of 3.731212 cm and a length of 18.9 cm, then the diagonal length is sqrt(3.731212^2 + 18.9^2) = 19.264785 cm. Rounding to the nearest tenth gives us 19.3 cm.
A regular pentagonal prism has a volume of 278.3105 cubic cm. What should the dimensions of the prism be so that the surface area is minimized?
This is a calculus optimization prism volume problem. Let's say the side length of the pentagonal cross-sections is x and the length of the prism is y. The area of a pentagon with a side length of x is 1.720477x^2. This gives us
278.3105 = (1.720477x^2)y
which tells us that y = 161.763569/x^2. The surface area of a pentagonal prism is the sum of the seven faces: two pentagons and five rectangles. If the surface area is S, then we have
S = 5xy + 3.440954x^2
= 808.817845/x + 3.440954x^2
The derivative of S(x) is
S'(x) = -808.817845/x^2 + 6.881908x
Setting this expression equal to zero and solving for x gives us x ≈ 4.8983 cm. Back-solving for y gives us y ≈ 6.7420 cm.