Chapter 12.5, Problem 9CE

To determine

To find:The ratios of base circumferences, scale factor, lateral areas, total areas and volumes of two similar cones.

Answer to Problem 9CE

The two similar cones scale factoris $1:6$ , ratio of base circumferences is $1:6$ , ratio of lateral areas is $1:36$ , ratio of total areas is $1:36$ and their volumes ratio is $1:216$ .

Explanation of Solution

Given information:

The ratio of slant heights of two similar cones is $1:6$ .

Calculation:

The ratio of the corresponding sides of any two similar geometric figures is termed as Scale factor.The theorem of similar solids states that if the scale factor of two similar solids is $a:b$ , then

(i) The ratio of base areas, lateral areas and total areas is $a^2:b^2$ .

(ii) The ratio of base circumferences is $a:b$ .

(iii) The ratio of their volumes is $a^3:b^3$ .

Let the two slant heights of cones be $l_1$ and $l_2$ , scale factor for both the cones be $s_1$ and $s_2$ and the base circumferences be $c_1$ and $c_2$ .

Expression to find the scale factor from the above theorem is:

  $\frac{l_1}{l_2}=\frac{s_1}{s_2}$

Substitute $1$ for $l_1$ and $6$ for $l_2$ in the above expression.

  $\frac{1}{6}=\frac{s_1}{s_2}$

Therefore, the ratio of scale factor of two similar cones is $1:6$ .

Expression to find the base circumferences from the above theorem is:

  $\frac{c_1}{c_2}=\frac{s_1}{s_2}$

Substitute $1$ for $s_1$ and $6$ for $s_2$ in the above expression.

  $\frac{c_1}{c_2}=\frac{1}{6}$

Therefore, the ratio of base circumferences of two similar cones is $1:6$ .

Expression to determine the lateral area of two similar cones from the similar solids theorem is:

  $\begin{array}{c}\text{lateral}\text{area}={\left(\text{scale}\text{factor}\right)}^2\\ \frac{l.a_1}{l.a_2}={\left(\frac{s_1}{s_2}\right)}^2\end{array}$

Substitute $1$ for $s_1$ and $6$ for $s_2$ in the above expression.

  $\begin{array}{c}\frac{l.a_1}{l.a_2}={\left(\frac{1}{6}\right)}^2\\ =\frac{1\times 1}{6\times 6}\\ =\frac{1}{36}\end{array}$

Therefore, the ratios of lateral areas of two similar conesis $1:36$ .

Expression to determine the total area of two similar cones from the above theorem is:

  $\begin{array}{c}\text{total}\text{area}={\left(\text{scale}\text{factor}\right)}^2\\ \frac{t.a_1}{t.a_2}={\left(\frac{s_1}{s_2}\right)}^2\end{array}$

Substitute $1$ for $s_1$ and $6$ for $s_2$ in the above expression.

  $\begin{array}{c}\frac{t.a_1}{t.a_2}={\left(\frac{1}{6}\right)}^2\\ =\frac{1\times 1}{6\times 6}\\ =\frac{1}{36}\end{array}$

Therefore, the ratios of total areas of two similar conesis $1:36$ .

Expression to determine the volumes of two similar cones from the above theorem is:

  $\begin{array}{c}\text{Volumes}={\left(\text{scale}\text{factor}\right)}^3\\ \frac{V_1}{V_2}={\left(\frac{s_1}{s_2}\right)}^3\end{array}$

Substitute $1$ for $s_1$ and $6$ for $s_2$ in the above expression.

  $\begin{array}{c}\frac{V_1}{V_2}={\left(\frac{1}{6}\right)}^3\\ =\frac{1\times 1\times 1}{6\times 6\times 6}\\ =\frac{1}{216}\end{array}$

So, the ratios of volumes of two similar conesis $1:216$ .

Therefore,the two similar cones scale factor ratio is $1:6$ , ratio of base circumferences is $1:6$ , ratio of lateral areas is $1:36$ , ratio of total areas is $1:36$ and their volumes ratio is $1:216$ .