# 9.2 Measurement

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When the length of one of the sides is multiplied by a constant the effect is to multiply the original volume by that constant, as for the example in [link] .

## Right pyramids, right cones and spheres

A pyramid is a geometric solid that has a polygon base and the base is joined to a point, called the apex. Two examples of pyramids are shown in the left-most and centre figures in [link] . The right-most figure has an apex which is joined to a circular base and this type of geometric solid is called a cone. Cones are similar to pyramids except that their bases are circles instead of polygons.

Surface Area of a Pyramid

The surface area of a pyramid is calculated by adding the area of each face together.

If a cone has a height of $h$ and a base of radius $r$ , show that the surface area is $\pi {r}^{2}+\pi r\sqrt{{r}^{2}+{h}^{2}}$ .

1. The cone has two faces: the base and the walls. The base is a circle of radius $r$ and the walls can be opened out to a sector of a circle.

This curved surface can be cut into many thin triangles with height close to $a$ ( $a$ is called the slant height ). The area of these triangles will add up to $\frac{1}{2}×$ base $×$ height(of a small triangle) which is $\frac{1}{2}×2\pi r×a=\pi ra$

2. $a$ can be calculated by using the Theorem of Pythagoras. Therefore:

$a=\sqrt{{r}^{2}+{h}^{2}}$
3. ${A}_{b}=\pi {r}^{2}$
4. $\begin{array}{ccc}\hfill {A}_{w}& =& \pi ra\hfill \\ & =& \pi r\sqrt{{r}^{2}+{h}^{2}}\hfill \end{array}$
5. $\begin{array}{ccc}\hfill A& =& {A}_{b}+{A}_{w}\hfill \\ & =& \pi {r}^{2}+\pi r\sqrt{{r}^{2}+{h}^{2}}\hfill \end{array}$

Volume of a Pyramid: The volume of a pyramid is found by:

$V=\frac{1}{3}A·h$

where $A$ is the area of the base and $h$ is the height.

A cone is like a pyramid, so the volume of a cone is given by:

$V=\frac{1}{3}\pi {r}^{2}h.$

A square pyramid has volume

$V=\frac{1}{3}{a}^{2}h$

where $a$ is the side length of the square base.

What is the volume of a square pyramid, 3cm high with a side length of 2cm?

1. The volume of a pyramid is

$V=\frac{1}{3}A·h,$

where $A$ is the area of the base and $h$ is the height of the pyramid. For a square base this means

$V=\frac{1}{3}a·a·h$

where $a$ is the length of the side of the square base.

2. $\begin{array}{ccc}& =& \frac{1}{3}·2·2·3\hfill \\ & =& \frac{1}{3}·12\hfill \\ & =& 4c{m}^{3}\hfill \end{array}$

We accept the following formulae for volume and surface area of a sphere (ball).

$\begin{array}{ccc}\hfill \mathrm{Surface area}& =& 4\pi {r}^{2}\hfill \\ \hfill \mathrm{Volume}& =& \frac{4}{3}\pi {r}^{3}\hfill \end{array}$

A triangular pyramid is placed on top of a triangular prism. The prism has an equilateral triangle of side length 20 cm as a base, and has a height of 42 cm. The pyramid has a height of 12 cm.

1. Find the total volume of the object.
2. Find the area of each face of the pyramid.
3. Find the total surface area of the object.

1. We use the formula for the volume of a triangular prism:
$\begin{array}{ccc}V& =& \frac{1}{2}b{h}^{2}\hfill \\ & =& \frac{1}{2}20{42}^{2}\hfill \\ & =& 17640\hfill \end{array}$
2. We use the formula for the volume of a triangular pyramid:
$\begin{array}{ccc}V& =& \frac{1}{6}b{h}^{2}\hfill \\ & =& \frac{1}{6}20{42}^{2}\hfill \\ & =& 5880\hfill \end{array}$
3. We note that we can simply add the volumes of each of the two shapes. So we obtain: $17640+5880=23520$ . This is the answer to part a.
4. We note that we have four triangles that make up the pyramid. So the area of each face is simply:
$\begin{array}{ccc}\mathrm{Area}& =& \frac{1}{2}bh\hfill \\ & =& \frac{1}{2}2042\hfill \\ & =& 420\hfill \end{array}$
This is the answer to part b.
5. The total area of the pyramid is simply $4×420=1680$
6. The surface area of the prism is:
$\begin{array}{ccc}\mathrm{Surface area}& =& b×h+2×H×S+H×b\hfill \\ & =& \mathrm{20}×\mathrm{20}+2×\mathrm{12}×\mathrm{20}+\mathrm{12}×\mathrm{20}\hfill \\ & =& 1120\hfill \end{array}$
7. To find the total surface area, we must subtract the area of one face of the pyramid from the area of the prism. We must also subtract the area of one of the triangular faces of the prism. Doing this gives the total surface area as: $1120-420+1680-420=1960$ This is the answer to part c.

## Surface area and volume

1. Calculate the volumes and surface areas of the following solids: *Hint for (e): find the perpendicular height using Pythagoras.
2. Water covers approximately 71% of the Earth's surface. Taking the radius of the Earth to be 6378 km, what is the total area of land (area not covered by water)?

can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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