Table of Contents

Last modified on August 3rd, 2023

A triangular pyramid is a polyhedron with a triangular base bounded by three lateral faces meeting at a common point, known as the apex.

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.

Based on the regularity of the base, a triangular pyramid can be – **(1) Regular triangular pyramid**, **(2) Irregular triangular pyramid**.

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.

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**.

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.

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.

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 cm ^{2} and a height of 26 cm.**

Solution:

As we know,

Volume (*V*) = ${\dfrac{1}{3}Bh}$,** **here B = 97 cm^{2}, h = 26 cm

∴ *V* = ${\dfrac{1}{3}\times 97\times 26}$

= 840.6 cm^{3}

**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 cm^{3}

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 cm^{3}

**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 cm ^{2}, 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 cm^{2}, P = 36 cm, s = 7 cm

∴ *TSA* = ${62.35+\dfrac{1}{2}\times 36\times 7}$

= 188.35 cm^{2}

**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 cm^{2}

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 cm^{2}

∴ *TSA* = ${\dfrac{1}{2}\times 8\times 4.6 +108}$

= 126.4 cm^{2}

Finding the surface area of an regular triangular pyramid when the **BASE** and **SLANT** **HEIGHT** 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 mm^{2}

Last modified on August 3rd, 2023