How this instrument works
This combined page solves a triangular prism with a right-triangle cross-section completely, returning both the enclosed volume and the total exterior surface area from the same three inputs: the cross-section's two legs (p and q) and the prism's own length. The volume follows the usual cross-section-area-times-length rule, V = ½pq × length. The surface area adds the two triangular end caps to the three rectangular side faces running the prism's length: total = 2×(triangle area) + (triangle's own perimeter × length), since each rectangular side face's width matches one of the triangle's three edges exactly.
This goes further than this site's Triangular Prism Volume page, which returns only the enclosed volume from a triangle's base and height. Here, the right-triangle cross-section is specified by its two legs directly (with the hypotenuse computed automatically), and the FULL exterior is solved alongside the volume — useful whenever both a capacity figure and a material or coating estimate are needed for the same solid.
The hypotenuse never needs entering separately: given the two legs p and q, it's found internally via the Pythagorean theorem, √(p²+q²), and folded directly into the perimeter used for the lateral surface area — one less measurement to track by hand.
- Enter the right-triangle cross-section's first leg into the leg p field.
- Enter the second leg into the leg q field.
- Enter the prism's length into the Prism length field.
- Read Volume and Total surface area: the sheet computes both simultaneously, finding the hypotenuse internally.
Worked example — legs 3 and 4, length 10
A triangular prism has a right-triangle cross-section with legs 3 and 4 (hypotenuse 5, found internally), extruded 10 units long. Its volume is ½×3×4×10=60, and its total surface area is 2×6 + 12×10 = 12+120 = 132 — both the enclosed space and the complete exterior, including both triangular ends and all three rectangular sides, solved from the same three inputs.
A larger cross-section with legs 6 and 8 (hypotenuse 10), extruded only 5 units, has a volume of 120 and a surface area of 168 — a shorter, fatter prism reaching a larger volume than the golden example through a bigger cross-section rather than a longer extrusion.
Questions
How is this different from the Triangular Prism Volume page?
That page returns only the volume, from a general triangle's base and height. This page assumes a RIGHT-triangle cross-section specifically (given by its two legs) and returns BOTH the volume and the complete surface area together.
Why does this page need a right-triangle cross-section specifically?
So the hypotenuse — needed for the surface area's perimeter calculation but not otherwise given — can be found internally via the Pythagorean theorem, without requiring it as a separate, potentially inconsistent third input.
What is included in the total surface area?
All five faces: the two triangular end caps (identical, one at each end) plus the three rectangular side faces running the prism's length, each one's width matching one of the triangle's three edges — two legs and the internally computed hypotenuse.
What if one leg is zero?
The cross-section's area and the prism's volume both collapse to zero, but the surface area doesn't vanish entirely — a lateral surface based on the remaining perimeter still exists, even though the solid itself has flattened.
Does the prism's volume depend on its surface area?
No — the two are computed independently from the same three inputs. A given surface area doesn't fix a unique volume, and vice versa, since different length-versus-cross-section combinations can produce the same one while differing in the other.