What Is A Pentagonal Prism?
A pentagonal prism is a prism whose two bases are pentagons - five-sided polygons - connected by five flat rectangular faces. Because the two ends are identical and parallel, and the sides are straight rectangles, the shape belongs to the family of prisms, the same family as the rectangular box and the triangular prism.
Count the parts and a pentagonal prism always gives the same three numbers: 7 faces (2 pentagons + 5 rectangles), 15 edges, and 10 vertices (5 corners on each pentagon). A solid with exactly seven faces is also called a heptahedron. If you would like the flat five-sided shape that forms each base on its own, see the pentagon shape.
A pentagonal prism can be a right prism (the rectangles meet the bases at 90 degrees, so it stands straight) or an oblique prism (the bases slide sideways, so it leans). It can also be regular (both pentagons are regular, all five sides equal) or irregular. Most school problems use the right regular pentagonal prism, and that is the shape we measure below.
How the volume and surface-area formulas come from the shape
Every prism follows one rule: fill the base, then push it up through the height. That is why the volume is always the base area multiplied by the height. You are stacking copies of the base until you reach the top.
Volume.
$$V = B \times h$$
Here $B$ is the area of one pentagon base and $h$ is the height of the prism (the distance between the two bases). For a regular pentagon with side length $b$ and apothem $a$ (the distance from the centre to the middle of a side), the base area is
$$B = \frac{5}{2} \times b \times a$$
so the full volume becomes
$$V = \frac{5}{2} \times b \times a \times h$$
Surface area. The surface is the two pentagon bases plus the five rectangles that wrap around the side (the lateral surface).
$$\text{Total Surface Area} = 2B + P \times h$$
$P$ is the perimeter of the base pentagon, so $P \times h$ is the combined area of the five rectangles unrolled into one strip. For a regular pentagonal prism this simplifies to
$$\text{TSA} = 5ab + 5bh$$
where $5ab$ is the two bases written out and $5bh$ is the five identical rectangles. Notice we use one multiplication symbol, $\times$, throughout, and every variable is defined before it appears.
Quantity | Formula | What each symbol means |
|---|---|---|
Base area $B$ | $\frac{5}{2},b,a$ | $b$ = base side, $a$ = apothem |
Volume $V$ | $B \times h$ | $h$ = prism height |
Lateral surface | $P \times h$ | $P$ = base perimeter $= 5b$ |
Total surface area | $2B + P,h = 5ab + 5bh$ | sum of bases and side rectangles |
Examples Of Pentagonal Prism
Work through these in order. They build from a direct count up to solving for a missing dimension, and one of them shows a wrong turn worth walking through.
Example 1
How many faces, edges, and vertices does a pentagonal prism have?
Count each family of parts.
Faces: 2 pentagon bases + 5 rectangular sides $= 7$ faces.
Vertices: each pentagon has 5 corners, and there are 2 pentagons, so $5 \times 2 = 10$ vertices.
Edges: 5 edges around the top pentagon, 5 around the bottom, and 5 vertical edges joining them, so $5 + 5 + 5 = 15$ edges.
A quick check with Euler's formula for solids, $V - E + F = 2$: here $10 - 15 + 7 = 2$. The count is consistent.
Example 2
A student is asked for the volume of a regular pentagonal prism with base side 4 cm, apothem 2.75 cm, and height 10 cm. Watch the tempting shortcut first.
The tempting move is to treat the pentagon base like a square and compute base area as side times side: $4 \times 4 = 16 \text{ cm}^2$, then $16 \times 10 = 160 \text{ cm}^3$.
That answer is wrong, and you can see why: a pentagon is not a square, so multiplying one side by itself does not give its area. A pentagon needs the apothem, because it is really five thin triangles fanning out from the centre.
The correct method uses the base-area formula:
$$B = \frac{5}{2} \times b \times a = \frac{5}{2} \times 4 \times 2.75 = 27.5 \text{ cm}^2$$
Then push the base up through the height:
$$V = B \times h = 27.5 \times 10 = 275 \text{ cm}^3$$
The real base area, $27.5 \text{ cm}^2$, is well above the $16 \text{ cm}^2$ the shortcut produced, so the shortcut was undercounting the whole solid.
Example 3
Find the total surface area of a regular pentagonal prism with base side 6 cm, apothem 4.13 cm, and height 12 cm.
Base area:
$$B = \frac{5}{2} \times 6 \times 4.13 = 61.95 \text{ cm}^2$$
Two bases:
$$2B = 123.9 \text{ cm}^2$$
Lateral surface (perimeter times height, with $P = 5 \times 6 = 30 \text{ cm}$):
$$P \times h = 30 \times 12 = 360 \text{ cm}^2$$
Total surface area:
$$\text{TSA} = 123.9 + 360 = 483.9 \text{ cm}^2$$
Example 4
A pentagonal prism has base area 30 cm² and volume 210 cm³. Find its height.
Start from the volume rule and solve for the unknown.
$$V = B \times h$$
$$210 = 30 \times h$$
$$h = \frac{210}{30} = 7 \text{ cm}$$
Working backwards from volume to a missing dimension is the same rule read in reverse.
Example 5
A gift box is a regular pentagonal prism with base side 5 cm, apothem 3.44 cm, and height 8 cm. How much cardboard (surface area) is needed, and how much space (volume) does it hold?
Base area:
$$B = \frac{5}{2} \times 5 \times 3.44 = 43 \text{ cm}^2$$
Volume:
$$V = 43 \times 8 = 344 \text{ cm}^3$$
Surface area, with $P = 5 \times 5 = 25 \text{ cm}$:
$$\text{TSA} = 2(43) + 25 \times 8 = 86 + 200 = 286 \text{ cm}^2$$
The box needs about $286 \text{ cm}^2$ of cardboard and holds $344 \text{ cm}^3$.
Example 6
Find the lateral surface area only of a pentagonal prism whose base perimeter is 35 cm and height is 9 cm.
Lateral surface skips the two bases and measures only the wrap-around rectangles.
$$\text{Lateral Surface} = P \times h = 35 \times 9 = 315 \text{ cm}^2$$
This is the amount of paper you would need to make a label that goes all the way around the prism without covering the ends.
Why The Pentagonal Prism Matters - "Five Walls Give More Angles To Defend"
The pentagonal prism is not a classroom curiosity. Its five-sided base is chosen on purpose whenever a design needs more than four flat walls but wants every wall to stay straight and every corner to stay rigid.
Fortress design. Star forts and bastion forts built from the 1500s onward used pentagon footprints so defenders had five walls of sightlines and no blind corners - more firing angles than a square could give. The Pentagon building inherited that same footprint for the same efficiency of movement and sightlines.
Packaging and crystals. Some gift boxes, pencils, and nuts are pentagonal prisms because five faces distribute pressure evenly and resist rolling. Certain mineral crystals also grow in prism form.
Where the maths is going. Once you can measure one prism by "fill the base, push it up," you can measure any prism - hexagonal, octagonal, or the general prisms family - and then move to curved solids like the cylinder and the cone, where the same base-times-height idea reappears with a twist.
The reason engineers keep the shape "regular" is stability: a regular pentagon spreads load evenly across all five sides, which is why the formulas above lean on the apothem - the single number that captures how "wide" the pentagon is from its centre.
Mistakes To Watch For With Pentagonal Prisms
Mistake 1: Treating the pentagon base like a square
Where it slips in: the moment a student computes the base area of a pentagonal prism and reaches for side times side.
Don't do this: $B = b \times b$. A pentagon is not a square, and this ignores the apothem entirely. Students meeting the pentagonal prism for the first time usually compute the base as if it were a rectangle and forget the apothem, which throws off both the volume and the surface area.
The correct way: use $B = \frac{5}{2} \times b \times a$, because a regular pentagon is five triangles fanning from the centre, and each triangle needs the apothem as its height.
Mistake 2: Confusing lateral surface area with total surface area
Where it slips in: surface-area problems that ask for "the label around the prism" versus "all the cardboard."
Don't do this: report $P \times h$ when the question wants the total, or report $2B + Ph$ when it only wants the wrap-around.
The correct way: read whether the two pentagon ends are included. Lateral surface is the five rectangles only ($P \times h$); total surface adds the two bases ($2B + Ph$).
Mistake 3: Mixing units within one calculation
Where it slips in: a problem gives the side in centimetres and the height in metres.
Don't do this: multiply 4 cm by 0.1 m and record 0.4 of some undefined unit.
The correct way: convert everything to one unit first. This is exactly the failure that destroyed the Mars Climate Orbiter in 1999, when one team worked in metric units, another in imperial, and the two never reconciled, so a real spacecraft the size of a small prism was lost. Fix the units before you multiply, every time.
Key Takeaways
A pentagonal prism has 7 faces, 15 edges, and 10 vertices - two pentagon bases and five rectangular sides.
Volume $= B \times h$ (base area times height); for a regular prism, $V = \frac{5}{2}bah$.
Total surface area $= 2B + Ph = 5ab + 5bh$; lateral surface is the five rectangles only.
The apothem is essential - the base is a pentagon, not a square.
Real pentagonal prisms include the Pentagon building, some packaging, and certain crystals.
To take pentagonal prisms further with a teacher, explore Bhanzu's geometry tutor sessions, a middle school math tutor for solids and mensuration, or general math classes online.
A Practical Next Step
Practice these problems to solidify your understanding. Work through them in order and check your answers below.
Find the volume of a regular pentagonal prism with base side 3 cm, apothem 2.06 cm, and height 6 cm.
Find the total surface area of a regular pentagonal prism with base side 8 cm, apothem 5.5 cm, and height 10 cm.
A pentagonal prism has volume 450 cm³ and base area 45 cm². Find its height.
Answer to Question 1: $V = \frac{5}{2}(3)(2.06)(6) = 92.7 \text{ cm}^3$. Answer to Question 2: $\text{TSA} = 2\big(\frac{5}{2}(8)(5.5)\big) + (40)(10) = 220 + 400 = 620 \text{ cm}^2$. Answer to Question 3: $h = 450 \div 45 = 10 \text{ cm}$.
Want a live Bhanzu trainer to walk through more pentagonal prism problems? Book a free demo class.
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