Table of Contents

Last modified on August 3rd, 2023

A right pyramid has its apex aligned right above the center of the base.

A right pyramid is a pyramid whose apex is aligned exactly above the center of the base. Thus, a perpendicular line joining the apex and the base center is the height of a right pyramid.

Unlike a right pyramid, an oblique pyramid does not have its apex aligned away from the center of the base.

Some common right pyramids with different bases are:

**Right Triangular Pyramid**

It has a triangular base with its apex aligned directly above its center.

**Right Square Pyramid**

It has a square base with its apex aligned directly above its center.

**Right Rectangular Pyramid**

It has a rectangular base with its apex aligned directly above its center.

**Right Pentagonal Pyramid**

It has a pentagonal base with its apex aligned directly above its center.

**Right Hexagonal Pyramid**

It has a hexagonal base with its apex aligned directly above its center.

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

The formulas is:

**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 (

∴ *SA* = B + *LSA*

Let us solve some examples to understand the above concept better.

**Find the lateral and the total surface area of a right pyramid with a base perimeter of 36 cm, base area of 81 cm ^{2}, and slant height of 16 cm.**

Solution:

As we know,

Lateral Surface Area (*LSA*) = ${\dfrac{1}{2}Ps}$, here P = 36 cm, s = 16 cm

∴ *LSA * = ${\dfrac{1}{2}\times 36\times 16}$

= 288 cm^{2}

Total Surface Area (*TSA*) = B + *LSA*, here B = 81 cm^{2}, *LSA *= 288 cm^{2}

∴ *TSA * = 81 + 288

= 369 cm^{2}

**Find the** **surface area of a right pyramid with a square base of 6 cm, and a slant height of 9.5 cm.**

Solution:

Here we will use the specific surface area formula for a right square pyramid.

Total Surface Area (*TSA*) = b^{2} + 2bs, here b = 6 cm, s = 9.5 cm

∴ *TSA * = 6^{2} + 2 × 6 × 9.5

= 150 cm^{2}

It is the space it occupies in a 3-dimensional plane. It is expressed in cubic units such as m^{3}, cm^{3}, mm^{3}, and in^{3}.

The general formula is given below:

**Volume ( V) = **${\dfrac{1}{3}Bh}$, here B = base area, h = height

Let us solve an example to understand the above concept better.

**Find the volume of a right pyramid with a base area of 160 cm ^{2} and a height of 16 cm.**

Solution:

As we know,

Volume (*V*) = ${\dfrac{1}{3}Bh}$, here B = 160 cm^{2}, h= 16 cm

∴ *V * = ${\dfrac{1}{3}\times 160\times 16}$

= 853.33 cm^{3}

**Find the volume of a right triangular pyramid with a basearea of 2.3 cm ^{2} and a height of 5 cm.**

Solution:

As we know,

Volume (*V*) = ${\dfrac{1}{3}Bh}$,** **here B = 2.3 cm

∴

= 3.83 cm

**Find the volume of a right square pyramid with a base of 6 cm and a height of 8 cm.**

Solution:

As we know,

Volume (*V*) = ${\dfrac{1}{3}b^{2}h}$, here b = 6 cm, h = 8 cm

∴ *V * = ${\dfrac{1}{3}\times 6^{2}\times 8}$

= 96 cm^{3}

**Find the volume of a right rectangular pyramid with a base length of 10 cm, a base width of 6 cm, and a height of 12 cm.**

Solution:

As we know,

Volume (*V*) = ${\dfrac{1}{3}lwh}$, here l = 10 cm, w = 6 cm, h = 12 cm

∴ *V* = ${\dfrac{1}{3}\times 10\times 6\times 12}$

= 240 cm^{3}

Last modified on August 3rd, 2023

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