Pyramid

Solid objects, as shown below are the pyramids.

As we see above, the pyramid is solid with a polygonal base and triangular faces with a common vertex. A line through the vertex to the centre of the base is called the height of the pyramid. Height perpendicular to the base is called right pyramid otherwise, pyramid is an oblique pyramid. A pyramid is regular if it's all lateral faces are a congruent isosceles triangle.

A pyramid whose base is an equilateral triangle is a tetrahedron. In tetrahedron, all the faces are congruent equilateral triangles.

A perpendicular line segment drawn from the vertex to any side of its base is called the slant height for the face consisting that side.

A pyramid is a three-dimensional solid figure in which the base is a polygon of any number of sides, and other faces are triangles that meet at a common point.

\( \therefore \text {Area of triangular face} = \frac {1} {2} base side \times slant \: height\)

The surface area of the pyramid is the total surface area of its all triangular faces together with the base.

Volume of a pyramid

Let's take a cubical container of side 'a' units. Take a pyramid of a square base with
a side of length 'a' units and height is same to that of the previous cube. Fill up water in cube by a pyramid.

Cube is filled up when the water is poured three times by the pyramid. By the
above experiment, we can say that the volume of the pyramid is one-third of the
volume of cube whose base and height are the same as that of pyramid. That is, if
V be the volume of the pyramid then, \( V = \frac {1} {3} a^3 \)

\(\boxed { \therefore V= \frac {1} {3} \times volume \: of \: the \: cube } \)

It can be written as, \( V= \frac {1} {3} a^2 \times a \). Hence, \( V= \frac {1} {3} \times base \: area \times height \)

Alternatively,

Take a cube of side '2a' units. Draw the space diagonal as shown in the figure.

There are six equal pyramids inside the cube, each has a square base of a side 2a units and height is half of the above cube. One of them is shown to the right of the diagram.

Let V be the volume of each pyramid. The total volume of such six pyramids is same as that of the cube. That is,
\begin{align*} 6V &= (2a)^3 \\ or, 6V &= (2a)^2 2a \\ or, V &= \frac {1} {6} (2a)^2 . 2a \\ \therefore V &= \frac {1} {3} (2a)^2 . a \\ \end{align*}

This means volume of each pyramid is equal to the one-third of product of its base area and height.

\(\therefore V =\frac {1} {3} \times base \: area \times height \)