Pyramids : Surface Area and Volume

Overview

**Surface Area and Volume of pyramid**:
Total surface area of the pyramid

$=$ area of the base-surface$+$ sum of the area of the side-faces
Volume of the pyramid

$=\frac{1}{3}\times \phantom{\rule{1ex}{0ex}}\text{area of the base-face}\phantom{\rule{1ex}{0ex}}\times \phantom{\rule{1ex}{0ex}}\text{height}$

pyramid

The shape shown in the figure is "a pyramid". Pyramid is a 3D solid shape with a 2D-polygon base at the bottom with triangular faces on the sides converging to a single point on the top . The base can be any 2D-polygon, for example, square, rectangle, triangle, etc, and the pyramids are square-pyramid, rectangular-pyramid, triangular-pyramid respectively.

We consider only right-pyramids with its axis at right angle to the base-surface. The oblique-pyramids have the angle between its axis and the base is not a right-angle. A right pyramid is shown in orange. And an oblique pyramid is shown in blue.

surface area

The surface area of the square-pyramid of side $a$ and height $h$ is

$\phantom{\rule{1ex}{0ex}}\text{area of square}\phantom{\rule{1ex}{0ex}}$ $+4\times \phantom{\rule{1ex}{0ex}}\text{area of triangles}$ or

$\phantom{\rule{1ex}{0ex}}\text{area of square}\phantom{\rule{1ex}{0ex}}$ $+\frac{1}{2}\phantom{\rule{1ex}{0ex}}\text{perimeter of base-surface}\times \left(slantheight\right)$

Note: A generic pyramid consists of

base-surface a 2D-polygon, (in this problem a square)

a set of triangular faces, (in this problem $5$ triangles of base $a$ and height $s$).

$s$ is the slant height computed as $\sqrt{{a}^{2}/4+{h}^{2}}$.

A pyramid of height $h$ is shown in orange. The triangular faces are shown in blue. Total surface area of the pyramid

$=$ area of the base-surface $+$ sum of the area of the triangular faces

volume of square pyramid

The volume of the square-pyramid of side $l$ and height $h$ is $\frac{1}{3}\times \phantom{\rule{1ex}{0ex}}\text{area of base-surface}\phantom{\rule{1ex}{0ex}}$ $\times \phantom{\rule{1ex}{0ex}}\text{height}$

Note: As per the Cavalieri's principle in 3D, the pyramid is equivalently represented by the oblique-pyramid in blue.

The modified-pyramid fits into a cuboid. Each of the pyramids in blue, green, and purple, equal in volume, fit into the cuboid, and spans the entire cuboid. So, each fills exactly one-third of the cuboid as shown.

A square-pyramid of height $h$ is shown in orange. As per the Cavalieri's principle in 3D, the pyramid is equivalently represented by the oblique-pyramid in blue.

The modified-pyramid fits into a cuboid. Each of the pyramids, shown on the lower half of the figure, in blue, green, and purple, are

• equal in volume,

• fit into the cuboid, and

• spans the entire cuboid.
So, each fills exactly one-third of the cuboid as shown.

Volume of the pyramid

$=$ volume of the modified pyramid with identical cross sectional area along vertical axis

$=\frac{1}{3}\times \phantom{\rule{1ex}{0ex}}\text{area of the base}\phantom{\rule{1ex}{0ex}}\times \phantom{\rule{1ex}{0ex}}\text{height}$

volume of any pyramid

The volume of the pentagonal-pyramid shown in the figure is $\frac{1}{3}\times \phantom{\rule{1ex}{0ex}}\text{area of base-surface}\phantom{\rule{1ex}{0ex}}$ $\times \phantom{\rule{1ex}{0ex}}\text{height}$

A pentagonal-pyramid is shown in orange. As per the Cavalieri's principle in 3D, the pyramid is equivalently represented by the oblique-pyramid in blue.

The modified-pyramid has

• the area at the bottom surface equal to the pentagonal pyramid

• the height equal to the pentagonal pyramid.

• the cross-sectional area at any point along vertical axis equals to the same of pentagonal pyramid

Volume of the pyramid

$=$ volume of the modified pyramid with identical cross sectional area along vertical axis

$=\frac{1}{3}\times \phantom{\rule{1ex}{0ex}}\text{area of the base}\phantom{\rule{1ex}{0ex}}\times \phantom{\rule{1ex}{0ex}}\text{height}$

What is the volume of a pyramid with base area $30c{m}^{2}$ and height $2$ cm? Volume can be computed without specifying what type of pyramid is that. Volume $=\frac{1}{3}$ base-area $\times$ height $=20c{m}^{3}$

summary

**Surface Area and Volume of pyramid**:
Total surface area of the pyramid

$=$ area of the base-surface$+$ sum of the area of the side-faces
Volume of the pyramid

$=\frac{1}{3}\times \phantom{\rule{1ex}{0ex}}\text{area of the base-face}\phantom{\rule{1ex}{0ex}}\times \phantom{\rule{1ex}{0ex}}\text{height}$

Outline

The outline of material to learn *Mensuration : Length, Area, and Volume* is as follows.

Note 1: * click here for the detailed overview of Mensuration High *

Note 2: * click here for basics of mensuration, which is essential to understand this. *

• ** Basics of measurement**

→ __Summary of Measurement Basics__

→ __Measurement by superimposition__

→ __Measurement by calculation__

→ __Measurement by equivalence__

→ __Measurement by infinitesimal pieces__

→ __Cavalieri's Principle (2D)__

→ __Cavalieri's Principle (3D)__

• **Perimeter & Area of 2D shapes**

→ __Circumference of Circles__

→ __Area of Circles__

• **Surface area & Volume of 3D shapes**

→ __Prisms : Surface Area & Volume__

→ __Pyramids : Surface Area & Volume__

→ __Cone : Surface Area & Volume__

→ __Sphere : Surface Area & Volume__

• **Part Shapes **

→ __Understanding part Shapes__

→ __Circle : Sector and Segment__

→ __Frustum of a Cone__