Table of Contents

Last modified on September 6th, 2022

A rectangular pyramid is a geometrical solid with a rectangular base bounded by four lateral faces meeting above the base at a common vertex, known as the apex. Each pair of opposite triangular lateral faces is congruent.

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

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

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

Based on the position of its apex, a rectangular pyramid can be classified into 2 types – (1) Right rectangular pyramid, and (2) Oblique rectangular pyramid

A right rectangular pyramid is a pyramid whose apex is aligned right above its base center. 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 away from the base center, it is an oblique rectangular pyramid. Its height is a perpendicular line from the apex to the base.

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

The formula is:

**Volume (V) =** ${\dfrac{1}{3}lwh}$, here l = base length, w = base width, h = height

Let us solve some examples to understand the concept better.

**Find the volume of a rectangular-based pyramid with a base length of 19 cm, a base width of 14 cm, and a height of 20 cm.**

Solution:

As we know,

Volume (V) = ${\dfrac{1}{3}lwh}$, here l = 19 cm, w = 14 cm, h = 20 cm

∴ *V* = ${\dfrac{1}{3}\times 19\times 14\times 20}$

= 1773.33 cm^{3}

**Find the volume of a rectangular pyramid with bases of 15 cm and 11 cm and a height of 23 cm.**

Solution:

As we know,

Volume (V) = ${\dfrac{1}{3}lwh}$, here l = 15 cm, w = 11 cm, h = 23 cm

∴ *V* = ${\dfrac{1}{3}\times 15\times 15\times 23}$

= 1265 cm^{3}

Finding the volume of a truncated rectangular pyramid when **BOTTOM BASE LENGTH**, **BOTTOM BASE WIDTH**, **TOP BASE LENGTH**, **TOP BASE WIDTH**, and **HEIGHT** are known

**Find the volume of a truncated rectangular pyramid given in the figure.**

Solution:

Here, we will use a formula for the truncated rectangular pyramid.

Volume (V) = ${\dfrac{Ab+aB+2\left( ab+AB\right) }{6}\times h}$, here A = 9 cm, B = 8 cm, a = 4.5 cm, b = 4 cm, h = 5 cm

∴ *V* = ${\dfrac{9\times 4+4.5\times 8+2\left( 4.5\times 4+9\times 8\right) }{6}\times 5}$

[9 × 4 + 4.5 × 8 + 2(4.5 × 4 + 9 × 8)] × 5/6

= 210 cm^{3}

The formula is:

**Surface Area (SA) **= ${lw+\dfrac{1}{2}w\sqrt{4h^{2}+l^{2}}+\dfrac{1}{2}l\sqrt{4h^{2}+w^{2}}}$, here l = base length, w = width, h = height

Also, ${\dfrac{1}{2}w\sqrt{4h^{2}+l^{2}}+\dfrac{1}{2}l\sqrt{4h^{2}+w^{2}}}$ = lateral surface area (*LSA*)

∴ *SA* = ** lw** +

Let us solve some examples to understand the concept better.

**Find the lateral and total surface of a rectangular pyramid with a base length of 8 cm, a base width of 5 cm, and a height of 12 cm.**

Solution:

As we know,

Lateral Surface Area (LSA) = ${\dfrac{1}{2}w\sqrt{4h^{2}+l^{2}}+\dfrac{1}{2}l\sqrt{4h^{2}+w^{2}}}$, here l = 8 cm, w = 5 cm, h = 12 cm

∴ *LSA* = ${\dfrac{1}{2}\times 5\sqrt{4\times 12^{2}+8^{2}}+\dfrac{1}{2}\times 8\sqrt{4\times 12^{2}+5^{2}}}$

= 161.31 cm^{2}

Total Surface Area (*TSA*) = lw + *LSA*, here l = 8 cm, w = 5 cm, *LSA* = 161.31 cm^{2}

∴ *TSA* = 8 × 5 + 161.31

= 201.31 cm^{2}

**Find the surface of a right rectangular pyramid with bases of 9 cm, 3 cm, and a height of 7 cm.**

Solution:

As we know,

Surface Area = Total Surface Area (TSA)

∴ Total Surface Area (*TSA*) = ${lw+\dfrac{1}{2}w\sqrt{4h^{2}+l^{2}}+\dfrac{1}{2}l\sqrt{4h^{2}+w^{2}}}$, here l = 9 cm, w = 3 cm, h = 7 cm

∴ *TSA* = ${9\times 3+\dfrac{1}{2}\times 3\sqrt{4\times 7^{2}+9^{2}}+\dfrac{1}{2}\times 9\sqrt{4\times 7^{2}+3^{2}}}$

= 116.4 cm^{2}

Let us learn how to find the surface area of a rectangular pyramid with slant height. We will use the general formula,**Total Surface Area ( TSA) = ${B+\dfrac{1}{2}Ps}$, **here B = base area, P = base perimeter, l = slant height

Finding the surface area of a rectangular pyramid when **BASE LENGTH**, **BASE WIDTH**, and **SLANT** **HEIGHT** are known

**Find the surface of a rectangular pyramid with bases of 10 cm, 6 cm, and a slant height of 13 cm.**

Solution:

Here, we will use the general formula.

Total Surface Area (*TSA*) = ${B+\dfrac{1}{2}Ps}$,** **here B = base area, P = base perimeter, s = slant height*B* = l × w, here l = 10 cm, w = 6 cm,

= 10 × 6

= 60 cm^{2}*P* = 2(l + w), here l = 10 cm, w = 6 cm

= 2 × (10 + 6)

= 32 cm

Plugging the value of B and P in the general formula,

Total Surface Area (*TSA*) = ${B+\dfrac{1}{2}Ps}$, here B = 60 cm^{2}, P = 32 cm, s = 13 cm

∴ *TSA* = ${60+\dfrac{1}{2}\times 32\times 13}$

= 268 cm^{2}

Last modified on September 6th, 2022