Chapter 12.5, Problem 10CE

To determine

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

Answer to Problem 10CE

The two similar cones scale factor is $2:3$ , ratio of base circumferences is $2:3$ , ratio of slant heights is $2:3$ , ratio of total areas is $4:9$ and their volumes ratio is $8:27$ .

Explanation of Solution

Given information:

The ratio of lateral areas of two similar cones is $4:9$ .

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 determine the scale factor 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 $4$ for $l.a_1$ and $9$ for $l.a_2$ in the above expression.

  $\frac{4}{9}={\left(\frac{s_1}{s_2}\right)}^2$

To find the ratio of scale factor, apply the square root at both sides in the above expression.

  $\begin{array}{c}\sqrt{\frac{4}{9}}=\sqrt{{\left(\frac{s_1}{s_2}\right)}^2}\\ \frac{s_1}{s_2}=\frac{2}{3}\end{array}$

Therefore, the ratios of scale factor of two similar cones is $2:3$ .Thus the value for $s_1$ is $2$ and the value for $s_2$ is $3$ .

Expression to find the slant height from the above theorem is:

  $\frac{l_1}{l_2}=\frac{s_1}{s_2}$

Substitute $2$ for $s_1$ and $3$ for $s_2$ in the above expression.

  $\frac{l_1}{l_2}=\frac{2}{3}$

Therefore, the ratios of slant height of two similar cones is $2:3$ .

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

  $\frac{c_1}{c_2}=\frac{s_1}{s_2}$

Substitute $2$ for $s_1$ and $3$ for $s_2$ in the above expression.

  $\frac{c_1}{c_2}=\frac{2}{3}$

Therefore, the ratio of base circumferences of two similar cones is $2:3$ .

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 $2$ for $s_1$ and $3$ for $s_2$ in the above expression.

  $\begin{array}{c}\frac{t.a_1}{t.a_2}={\left(\frac{2}{3}\right)}^2\\ =\frac{2\times 2}{3\times 3}\\ =\frac{4}{9}\end{array}$

Therefore, the ratios of total areas of two similar cones is $4:9$ .

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 $2$ for $s_1$ and $3$ for $s_2$ in the above expression.

  $\begin{array}{c}\frac{V_1}{V_2}={\left(\frac{2}{3}\right)}^3\\ =\frac{2\times 2\times 2}{3\times 3\times 3}\\ =\frac{8}{27}\end{array}$

So, the ratios of volumes of two similar cones is $8:27$ .

Therefore,the two similar cones scale factor is $2:3$ , ratio of base circumferences is $2:3$ , ratio of slant heights is $2:3$ , ratio of total areas is $4:9$ and their volumes ratio is $8:27$ .