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7.3 Right pyramids, right cones and spheres

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.

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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

f2de2fc3b18b567a75ba79f84eacdf16.png \(\begin{array}{[email protected]{\;}[email protected]{\;}l} \text{Surface area} &=& \text{area of base} \;+ \\ && \text{area of triangular sides} \\ &=& b^2 + \text{4}\left(\frac{1}{2}b{h}_{s}\right) \\ &=& b\left(b+2{h}_{s}\right) \end{array}\)

Triangular pyramid

17defc021a86c6ed1d43fdb30cbb473f.png \(\begin{array}{[email protected]{\;}[email protected]{\;}l} \text{Surface area} &= & \text{area of base} \;+ \\ && \text{area of triangular sides} \\ &=& \left(\frac{1}{2}b\times {h}_{b}\right)+3\left(\frac{1}{2}b\times {h}_{s}\right) \\ &=& \frac{1}{2}b\left({h}_{b}+3{h}_{s}\right) \end{array}\)

Right cone

69c731147a070db541329809cea05d68.png \(\begin{array}{[email protected]{\;}[email protected]{\;}l} \text{Surface area} &=& \text{area of base} \;+ \\ && \text{area of walls} \\ &=& \pi {r}^{2}+\frac{1}{2}\times 2\pi rh \\ &=& \pi r\left(r+h\right) \end{array}\)

Sphere

eb38aa680e7eb59e57afb266b8b3c234.png \(\text{Surface area} = \text{4} \pi r^2\)
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Volume of pyramids, cones and spheres (EMBJ3)

Square pyramid

716c0d985b172126108f862e7345c495.png \(\begin{array}{[email protected]{\;}[email protected]{\;}l} \text{Volume} &=& \frac{1}{3}\times \text{area of base} \;\times \\ && \text{height of pyramid} \\ &=& \frac{1}{3}\times {b}^{2}\times H \end{array}\)

Triangular pyramid

d3da6953c78183614dd7adae82808c63.png \(\begin{array}{[email protected]{\;}[email protected]{\;}l} \text{Volume} &=& \frac{1}{3}\times \text{area of base} \;\times \\ && \text{height of pyramid} \\ &=& \frac{1}{3}\times \frac{1}{2}bh\times H \end{array}\)

Right cone

8550ee917136d64f71cffd5d6b631c7e.png \(\begin{array}{[email protected]{\;}[email protected]{\;}l} \text{Volume} &=& \frac{1}{3}\times \text{area of base} \;\times \\ && \text{height of cone} \\ &=& \frac{1}{3}\times \pi {r}^{2}\times H \end{array}\)

Sphere

cc4dd82776137ad6eaba89c671dbca2a.png \(\text{Volume}=\frac{4}{3}\pi {r}^{3}\)
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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 \(\text{26}\) \(\text{m}\).

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  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*}

Finding surface area and volume

Textbook Exercise 7.4

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

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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} \end{align*}

The surface area of the ice-cream is half a sphere:

\begin{align*} \text{Surface area ice-cream:} &=\dfrac{1}{2}(4 \pi r^{2}) \\ &= \dfrac{1}{2}(4 \times \pi \times (26,2)^{2}) \\ &\approx \text{4313,03 mm}^{2} \end{align*}

The surface area of the cone must not include the surface area of the circular face.

\begin{align*} \text{Surface area cone }&= \pi r(r + \sqrt{h^{2} + r^{2}}) - \pi r^{2} \\ &= \pi \times \text{26,2} \times (\text{26,2}+\sqrt{(119,8)^{2} + (26,2)^{2}}) - \pi \times (26,2)^{2} \\ &\approx \text{10093,76 mm}^{2} \\ \therefore \text{Surface area ice-cream and cone} &= \text{4313,03} + \text{10093,76} \\ &=\text{14406,79 mm}^{2} \\ &\approx \text{144,07}\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\times\text{119,8}+\frac{2}{3}\pi(\text{26,2})^3 \\ &=\text{86 116,82}\ldots +\text{37 667,12}\ldots \\ &=\text{123 783,953}\ldots \text{ mm$^{3}$} \\ &\approx\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, \(P\).

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Determine the value of \(R\).

\(R\) is the slant height.

\begin{align*} R &= \sqrt{r^{2} + h^{2}} \\ &= \sqrt{(\text{26,2})^{2} + (\text{119,8})^{2}} \\ &= \text{122,631}\ldots\text{ mm} \\ &\approx \text{123}\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(123)-165\\ &=\text{608}\text{ mm} \end{align*}