A triangular pyramid is a polyhedron with a triangular base bounded by three lateral faces meeting at a common point, known as the apex.
Triangular Pyramid
The lateral faces are triangular.
Tents and combination puzzles are some real-life example of a triangular pyramid shape.
How many faces, vertices, and edges does a triangular pyramid have?
A triangular pyramid has 4 faces, 4 vertices, and 6 edges. Since all the 4 faces are triangular, a triangular pyramid is also called a tetrahedron.
A net for a triangular pyramid can illustrate its shape from a 2-D view. This net can be folded along the dotted lines to form a triangular pyramid as shown in the diagram below.
Triangular Pyramid Net
Types
Based on the regularity of the base, a triangular pyramid can be – (1) Regular triangular pyramid, (2) Irregular triangular pyramid.
Regular
A regular triangular pyramid is a pyramid that has its base in shape of an equilateral triangle. So it is also called an equilateral triangular pyramid.
Regular Triangular Pyramid
In contrast, when the base of a triangular pyramid is an irregular, it is an irregular triangular pyramid.
Based on the position of its apex, a a triangular pyramid can be (1) Right triangular pyramid, (2) Oblique triangular pyramid.
Right
A right triangular pyramid is a pyramid that has its apex right above its base center. So, an imaginary line drawn from the apex intersects the base at its center at a right angle. This line is its height.
Right Triangular Pyramid
In contrast, when the apex is away from the base center, the pyramid is an oblique triangular pyramid.
Like all other polyhedrons, we can calculate the surface area and the volume of a triangular pyramid.
Formulas
Volume
The formula is:
Volume (V) = ${\dfrac{1}{3}Bh}$, here B = base area, h = height
Let us solve some examples involving the above formula.
Find the volume of a regular triangular pyramid with a base area of 97 cm2 and a height of 26 cm.
Solution:
As we know, Volume (V) = ${\dfrac{1}{3}Bh}$,here B = 97 cm2, h = 26 cm ∴ V = ${\dfrac{1}{3}\times 97\times 26}$ = 840.6 cm3
Find the volume of a right triangular pyramid with a base of 21 cm, a base height of 8 cm, and a height of 17 cm.
Solution:
As we know, Volume (V) = ${\dfrac{1}{3}Bh}$ B = ${\dfrac {1}{2}bH}$,here b = base, H = base height plugging the value of B, we get, Volume (V) = ${\dfrac{1}{6}bHh}$, here b = 21 cm, H = 8 cm, h = 17 cm ∴ V = ${\dfrac{1}{6}\times 21\times 8\times 17}$ = 476 cm3
Finding the volume of a regular triangular pyramid when the BASE and HEIGHT are known
Find the volume of a regular triangular pyramid with a base of 7 cm, and a height of 16 cm.
Solution:
We will use an alternative formula here. Volume (V) = ${ \dfrac{\sqrt{3}}{12}b^{2}h }$, here b = 7 cm, h = 16 cm ∴ V = ${\dfrac{\sqrt{3}}{12}\times 7^{2}\times 16}$ = 21.21762 × 1/3 × 16 = 113.16 cm3
Surface Area
Surface Area (SA) = ${B+\dfrac{1}{2}Ps}$, here B = base area, P = base perimeter, s = slant height,
Also ${\dfrac{1}{2}Ps}$ = lateral surface area (LSA)
∴ SA = B + LSA
Let us solve some examples to understand the above concept better.
Find the surface area of a regular triangular pyramid with a base area of 62.35 cm2, a base perimeter of 36 cm, and a slant height of 7 cm.
Solution:
As we know, Total Surface Area (TSA) = ${B+\dfrac{1}{2}Ps}$, here B = 62.35 cm2, P = 36 cm, s = 7 cm ∴ TSA = ${62.35+\dfrac{1}{2}\times 36\times 7}$ = 188.35 cm2
Find the lateral and total surface area of a triangular pyramid with a base of 8 cm, a base height of 4.6 cm, and a slant height of 9 cm.
Solution:
As we know, Lateral Surface Area (LSA) = ${\dfrac{1}{2}Ps}$ P = 3 × b, here b = 8 cm ∴ LSA = ${\dfrac{3}{2}bs}$, here b = 8 cm, s = 9 cm ∴ LSA = ${\dfrac{3}{2}\times 8\times 9}$ = 108 cm2 Total Surface Area (TSA) = B + LSA Now, B = ${\dfrac{1}{2}bH}$, here b = 8 cm, H = 4.6 cm ∴ TSA = ${\dfrac{1}{2}\times bH+LSA}$, here b = 8 cm, H = 4.6 cm, LSA = 108 cm2 ∴ TSA = ${\dfrac{1}{2}\times 8\times 4.6 +108}$ = 126.4 cm2
Finding the surface area of an regular triangular pyramid when the BASE and SLANTHEIGHT are known
Find the surface area of a regular triangular pyramid with a base of 11 mm, and a slant height of 7 mm.
Solution:
Here we will use an alternative formula for a regular (equilateral) triangular pyramid. Total Surface Area (TSA) = ${\dfrac{\sqrt{3}}{4}b^{2}+\dfrac{1}{2}\times 3bs}$, here b = 11 mm,s = 7 mm ∴ TSA = ${\dfrac{\sqrt{3}}{4}\times 11^{2}+\dfrac{1}{2}\times 3\times 11\times 7}$ = 167.89 mm2
