7.3 Right pyramids, right cones and spheres (EMBHZ)

A pyramid is a geometric solid that has a polygon as its base and sides that converge at a point called the apex. In other words the sides are not perpendicular to the base.

The triangular pyramid and square pyramid take their names from the shape of their base. We call a pyramid a “right pyramid” if the line between the apex and the centre of the base is perpendicular to the base. Cones are similar to pyramids except that their bases are circles instead of polygons. Spheres are solids that are perfectly round and look the same from any direction.

Surface area of pyramids, cones and spheres (EMBJ2)

Square pyramid		\(\begin{array}{r@{\;}c@{\;}l} \text{Surface area} &=& \text{area of base} \;+ \\ && \text{area of triangular sides} \\ &=& b^2 + \text{4}\left(\frac{1}{2}b{h}_{s}\right) \\ &=& b\left(b+2{h}_{s}\right) \end{array}\)
Triangular pyramid		\(\begin{array}{r@{\;}c@{\;}l} \text{Surface area} &= & \text{area of base} \;+ \\ && \text{area of triangular sides} \\ &=& \left(\frac{1}{2}b\times {h}_{b}\right)+3\left(\frac{1}{2}b\times {h}_{s}\right) \\ &=& \frac{1}{2}b\left({h}_{b}+3{h}_{s}\right) \end{array}\)
Right cone		\(\begin{array}{r@{\;}c@{\;}l} \text{Surface area} &=& \text{area of base} \;+ \\ && \text{area of walls} \\ &=& \pi {r}^{2}+\frac{1}{2}\times 2\pi rh \\ &=& \pi r\left(r+h\right) \end{array}\)
Sphere		\(\text{Surface area} = \text{4} \pi r^2\)

Volume of pyramids, cones and spheres (EMBJ3)

Square pyramid		\(\begin{array}{r@{\;}c@{\;}l} \text{Volume} &=& \frac{1}{3}\times \text{area of base} \;\times \\ && \text{height of pyramid} \\ &=& \frac{1}{3}\times {b}^{2}\times H \end{array}\)
Triangular pyramid		\(\begin{array}{r@{\;}c@{\;}l} \text{Volume} &=& \frac{1}{3}\times \text{area of base} \;\times \\ && \text{height of pyramid} \\ &=& \frac{1}{3}\times \frac{1}{2}bh\times H \end{array}\)
Right cone		\(\begin{array}{r@{\;}c@{\;}l} \text{Volume} &=& \frac{1}{3}\times \text{area of base} \;\times \\ && \text{height of cone} \\ &=& \frac{1}{3}\times \pi {r}^{2}\times H \end{array}\)
Sphere		\(\text{Volume}=\frac{4}{3}\pi {r}^{3}\)

Video: 235K

Worked example 4: Finding surface area and volume

The Southern African Large Telescope (SALT) is housed in a cylindrical building with a domed roof in the shape of a hemisphere. The height of the building wall is \(\text{17}\) \(\text{m}\) and the diameter is \(\text{26}\) \(\text{m}\).

1. Calculate the total surface area of the building.

2. Calculate the total volume of the building.

Calculate the total surface area

\begin{align*} \text{Total surface area} &= \text{area of the dome} + \text{area of the cylinder} \\ \text{Surface area} &= \left[\frac{1}{2}(4 \pi r^2)\right] + \left[2 \pi r \times h\right] \\ &= \frac{1}{2}(4 \pi)(13)^2 + 2 \pi (\text{13})(\text{17}) \\ &= \text{2 450}\text{ m$^{2}$} \end{align*}

Calculate the total volume

\begin{align*} \text{Total volume} &= \text{volume of the dome} + \text{volume of the cylinder} \\ \text{Volume} &= \left[\frac{1}{2} \times \left( \frac{4}{3} \pi r^3 \right)\right] + \left[\pi r^2 h\right] \\ &= \frac{2}{3} \pi (\text{13})^3 + \pi (\text{11})^2 (\text{13}) \\ &= \text{9 543}\text{ m$^{3}$} \end{align*}

Test yourself now

High marks in maths are the key to your success and future plans. Test yourself and learn more on Siyavula Practice.

Sign up and test yourself

Finding surface area and volume

Exercise 7.4

An ice-cream cone has a diameter of \(\text{52,4}\) \(\text{mm}\) and a total height of \(\text{146}\) \(\text{mm}\).

Calculate the surface area of the ice-cream and the cone.

\begin{align*} \text{Radius }&=\frac{\text{52,4}}{2} \\ &=\text{26,2}\text{ mm} \\ \text{Height of cone }&= 146 - \text{26,2} \\ &=\text{119,8}\text{ mm} \\ &\approx \text{120}\text{ mm} \\ \text{Surface area cone }&=\pi \times \text{26,2} \times (\text{26,2}+\text{119,8})\\ &= \text{12,0}\text{ mm$^{2}$} \\ &= \text{120}\text{ cm$^{2}$} \end{align*}

Calculate the total volume of the ice-cream and the cone.

\begin{align*} \text{Volume}&=\text{volume(cone)}+\text{volume}(\frac{1}{2}\text{sphere}) \\ &=\frac{1}{3}\pi r^2H+\frac{1}{2}\left(\frac{4}{3}\pi r^3\right) \\ &=\frac{1}{3}\pi(\text{26,2})^2\times120+\frac{2}{3}\pi(\text{26,2})^3 \\ &=\text{86 260,59}\ldots +\text{37 667,12}\ldots \\ &=\text{123 928}\text{ mm$^{3}$} \\ &=\text{124}\text{ cm$^{3}$} \end{align*}

How many ice-cream cones can be made from a \(\text{5}\) \(\text{ℓ}\) tub of ice-cream (assume the cone is completely filled with ice-cream)?

\begin{align*} \text{1 000}\text{ cm$^{3}$}&=\text{1}\text{ ℓ} \\ \therefore \text{5}\text{ ℓ}&= \text{5 000}\text{ cm$^{3}$}\\ \therefore \frac{\text{5 000}}{124} &\approx 40 \text{ cones} \end{align*}

Consider the net of the cone given below. \(R\) is the length from the tip of the cone to its perimeter, \(P\).

Determine the value of \(R\).

\begin{align*} R &= \text{146} - \text{26,2} \\ &= \text{119,8}\text{ mm} \\ &\approx \text{120}\text{ mm} \end{align*}

Calculate the length of arc \(P\).

\begin{align*} P &=\text{circumference of cone} \\ & = 2\pi(\text{26,2}) \\ &\approx \text{165}\text{ mm} \end{align*}

Determine the length of arc \(M\).

\begin{align*} M&=2\pi(120)-165\\ &=\text{589}\text{ mm} \end{align*}