# Finding Area of a Triangular Pyramid

I an trying to implement a program that needs to have a triangular pyramid and calcualte its surface area, and need some conceptual help.

Suppose I have a pyramid with a triangular base, and the base can be any kind of triangle with 3 sides. We have information about the length of 3 sides of the base, as well as the angles. (The way I did this was to initialize the triangle with 2 known lengths and the angle between them, and calculate the rest.)

Now we want to construct a pyramid with that triangle as the base, and we have information about the height of the pyramid. For simplicity, we can assume that the tip of the pyramid is above the centroid of the base triangle. (?)

How do we find the surface area of this pyramid? I understand that the formula is Base Area + 0.5(perimeter of base + slant height), and the part where I'm really stumped is how to obtain the slant height based on the values I have. Is it even possible?

• When you "for simplicity" : be conscious that if it is above another point, you will not get the same fianal result ! Commented Oct 13, 2021 at 10:30
• What does slant height refer to, is it the length of the slant or the given height? Commented Oct 13, 2021 at 10:41
• @AaaLol_dude slant height refers to the height of the triangles that are on the side (i.e not the base triangle, but the height of the "walls" of the pyramid)
– Ymi
Commented Oct 13, 2021 at 10:44
• It's probably possible with coordinate geometry, as the centroid needs to be found. Commented Oct 13, 2021 at 10:46
• I would work with Cartesian coordinates for the vertices. The areas of the triangular sides are then easily computed. Commented Oct 13, 2021 at 11:10

Not only is it possible, but since you have already rendered the area of the base it's easy.

Assume the base has vertices $$A,B,C$$ with opposite side lengths $$a,b,c$$ and known area $$S$$. Add the fourth vertex $$D$$ of the pyramid at height $$h$$ above the centroid $$G$$ of the base triangle.

Construct the perpendicular segment from $$G$$ to side $$BC$$ of the base triangle. This together with the altitude of the pyramid and the slant height of lateral face $$BCD$$ forms a right triangle. One leg of this right triangle is $$h$$. The second leg is the altitude of $$\triangle BCG$$ from side $$BC$$, and with $$G$$ as the centroid of $$\triangle ABC$$ the area of $$\triangle BCG$$ would be $$S/3$$. Therefore the second leg of our right triangle would be $$2S/3a$$. The Pythagorean Theorem then renders the hypotenuse which is the slant height from $$BC$$ to $$D$$ as

$$\sqrt{h^2+\dfrac{4S^2}{9a^2}}.$$

Compute the other slant heights similarly anf get the lateral face areas from there.

• Thanks for the answer! Can you elaborate further on how we got the second leg of the right Triangle? How did we go from S/3 to 2S/3a? I'm not sure about this part.
– Ymi
Commented Oct 13, 2021 at 11:26
• Area of the small triangle, $S/3$, is half the edge $a$ times the corresponding altitude which is our leg. Commented Oct 13, 2021 at 11:35