9.2 Measurement  (Page 2/3)

Surface areas

1. Calculate the surface area in each of the following:

   

2. If a litre of paint covers an area of $2m^2$ , how much paint does a painter need to cover:
   1. A rectangular swimming pool with dimensions $4m\times 3m\times 2,5m$ , inside walls and floor only.
   2. The inside walls and floor of a circular reservoir with diameter $4m$ and height $2,5m$

   

Volume

The volume of a right prism is calculated by multiplying the area of the base by the height. So, for a square prism of side length $a$ and height $h$ the volume is $a\times a\times h=a^{2}h$ .

Volume of Prisms

Calculate the area of the base and multiply by the height to get the volume of a prism.

Find the surface area and volume for the a square prism of height $4\mathrm{cm}$ and base length, $3\mathrm{cm}$ .

1. Find the surface area We use the formula for the surface area of a prism:
   $\begin{array}{ccc}\hfill \mathrm{S.A.}& =& 2\left[2\left(L\times b\right)+\left(b\times h\right)\right]\hfill \\ & =& 2\left[2\left(3\times 4\right)+\left(3\times 4\right)\right]\hfill \\ & =& {72\mathrm{cm}}^2\hfill \end{array}$
2. Find the volume To find the volume of the prism, we find the area of the base and multiply it by the height:
   $\begin{array}{ccc}V\hfill & =& l^2\times h\\ & =& \left(3^2\right)\times 4\hfill \\ & =& {36\mathrm{cm}}^3\hfill \end{array}$

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Volume

1. Write down the formula for each of the following volumes:

   

2. Calculate the following volumes:

   

3. A cube is a special prism that has all edges equal. This means that each face is a square. An example of a cube is a die. Show that for a cube with side length $a$ , the surface area is $6a^2$ and the volume is $a^3$ .

   

Now, what happens to the surface area if one dimension is multiplied by a constant? For example, how does the surface area change when the height of a rectangular prism is divided by 2?

The size of a prism is specified by the length of its sides. The prism in the diagram has sides of lengths $L$ , $b$ and $h$ .

1. Consider enlarging all sides of the prism by a constant factor $x$ , where $x>1$ . Calculate the volume and surface area of the enlarged prism as a function of the factor $x$ and the volume of the original volume.
2. In the same way as above now consider the case, where $0<x<1$ . Now calculate the reduction factor in the volume and the surface area.

1. Identify

   The volume of a prism is given by: $V=L\times b\times h$

   The surface area of the prism is given by: $A=2\times (L\times b+L\times h+b\times h)$

2. Rescale

   If all the sides of the prism get rescaled, the new sides will be:

   $$\begin{array}{ccc}\hfill L^{\text{'}}& =& x\times L\hfill \\ \hfill b^{\text{'}}& =& x\times b\hfill \\ \hfill h^{\text{'}}& =& x\times h\hfill \end{array}$$

   The new volume will then be given by:

   $$\begin{array}{ccc}\hfill V^{\text{'}}& =& L^{\text{'}}\times b^{\text{'}}\times h^{\text{'}}\hfill \\ & =& x\times L\times x\times b\times x\times h\hfill \\ & =& x^3\times L\times b\times h\hfill \\ & =& x^3\times V\hfill \end{array}$$

   The new surface area of the prism will be given by:

   $$\begin{array}{ccc}\hfill A^{\text{'}}& =& 2\times (L^{\text{'}}\times b^{\text{'}}+L^{\text{'}}\times h^{\text{'}}+b^{\text{'}}\times h^{\text{'}})\hfill \\ & =& 2\times (x\times L\times x\times b+x\times L\times x\times h+x\times b\times x\times h)\hfill \\ & =& x^2\times 2\times (L\times b+L\times h+b\times h)\hfill \\ & =& x^2\times A\hfill \end{array}$$
3. Interpreting the above results
   1. We found above that the new volume is given by: $V^{\text{'}}=x^3\times V$ Since $x>1$ , the volume of the prism will be increased by a factor of $x^3$ . The surface area of the rescaled prism was given by: $A^{\text{'}}=x^2\times A$ Again, since $x>1$ , the surface area will be increased by a factor of $x^2$ . Surface areas which are two dimensional increase with the square of the factor while volumes, which are three dimensional, increase with the cube of the factor.
   2. The answer here is based on the same ideas as above. In analogy, since here $0<x<1$ , the volume will be reduced by a factor of $x^3$ and the surface area will be decreased by a factor of $x^2$

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