Table of Contents

Last modified on August 3rd, 2023

A square pyramid is a pyramid with a square base bounded by four lateral faces meeting at a common point, known as the apex. The lateral faces are triangular.

**How many faces,** **edges and vertices does a square pyramid have?**

A square pyramid has 5 faces, 8 edges, and 5 vertices.

A net for a square pyramid can illustrate its shape from a 2-D view. This net can be folded along the dotted lines to form a square pyramid as shown in the diagram.

Based on the position of its apex, a square pyramid can be classified into 2 types â€“ (1) Right square pyramid, and (2) Oblique square pyramid.

A right square pyramid is a pyramid whose apex is aligned right above its base center. So, an imaginary line drawn from the apex intersects the base at its center at a right angle. A square pyramid is usually right.

In contrast, when the apex is away from the base center, the pyramid is an **oblique square pyramid**. Its height is a perpendicular line from the apex to the base.

A square pyramid has a special type based on its edge lengths.

An equilateral square pyramid has its 3 lateral faces that are equilateral triangles. It means all the edges of are of equal length.

Like all other polyhedrons, we can calculate the surface area and the volume of a square pyramid.

The formula is:

**Volume ( V)**

Let us solve some examples to understand the concept better.

**Find the volume of aÂ squareÂ pyramidÂ with a base of 12 cm, and a height of 6 cm.**

Solution:

As we know,

Volume (*V*) = ${\dfrac{1}{3}b^{2}h}$, here b = 12 cm, h = 6 cm

âˆ´Â *V*Â = ${\dfrac{1}{3}\times 12^{2}\times 6}$

= 288 cm^{3}

**Find the volume of aÂ squareÂ based pyramidÂ with a base of 4 cm, and a height of 14 cm.**

Solution:

As we know,

Volume (*V*) = ${\dfrac{1}{3}b^{2}h}$, here b = 4 cm, h = 14 cm

âˆ´Â *V*Â = ${\dfrac{1}{3}\times 4^{2}\times 14}$,

= 74.66 cm^{3}

The formula is:

**Surface Area (SA) = **${b^{2}+2bs}$, here b = base, s = slant height

Also ${2bs}$ = lateral surface area (*LSA*)

âˆ´ *SA* = ** b^{2} **+

Let us solve some examples to understand the concept better.

**Find the lateral and total surface area of aÂ squareÂ pyramidÂ with a slant height of 8.5 cm and a base of 8 cm.**

Solution:

As we know,

Lateral Surface Area (*LSA*) = ${2bs}$**, **here b = 8 cm, s = 8.5 cm

âˆ´ *LSA*Â = 2Â Ã— 8 Ã— 8.5

= 136 cm^{2}

Total Surface Area (*TSA*) = b^{2} + *LSA*, here b = 8 cm, *LSA*Â = 136 cm^{2}

âˆ´ *TSA*Â = Â 8^{2}+ 136

= 200 cm^{2}

**Find the surface area of aÂ squareÂ based pyramidÂ with a base of 11 cm and a slant height of 15 cm.**

Solution:

As we know,

Total Surface Area (*TSA*) = b^{2} + 2bs, here b = 11 cm, s = 15 cm

âˆ´ *TSA*Â =Â 11^{2}+ 2 Ã— 11 Ã— 15

= 451 cm^{2}

Now, let us learn how to find the slant height and height of a square pyramid with some typical examples.

Finding the slant height of a square pyramid when its **BASE **and **SURFACE AREA** are known.

**Find the slant height of a square pyramid with a base area of 189 cm ^{2} and a base of 9 cm.**

Solution:

As we know,**Surface Area ( SA)** =

âˆ´ s = ${\dfrac{SA-b^{2}}{2b}}$, here

= ${\dfrac{189-9^{2}}{2\times 9}}$

= 6 cm

Finding the height of a square pyramid when its **BASE **and **VOLUME** are known.

**Find the height of a square pyramid with a volume of 256 cm ^{3}, and a base of 8 cm.**

Solution:

As we know,

Volume (*V*) = ${\dfrac{1}{3}b^{2}h}$

âˆ´ h = ${\dfrac{3V}{b^{2}}}$, here *V* = 256 cm^{3}, *b* = 8 cm

= ${\dfrac{3\times 256}{8^{2}}}$

= 12 cm

Last modified on August 3rd, 2023