Chapter 12.5, Problem 5ST2

(a)

To determine

To find:The value of the base edge of the smaller pyramid.

Answer to Problem 5ST2

The value of the base edge of the smaller pyramid is $5.34$ .

Explanation of Solution

Given information:

The height of smaller pyramidis $4$ , the height of larger pyramid is $6$ and the base edge of the larger pyramid is $8$ .

Calculation:

The ratio of the corresponding sides of any two similar geometric figures is termed as Scale factor.

Let the edge of the smaller pyramid be $a_1$ and the edge of the larger pyramid be $a_2$ and heightof smaller pyramid be $h_1$ and height of larger pyramid is $h_2$ .

Expression to determine the relation between edge and heights of the similar pyramid is:

  $\frac{a_1}{a_2}=\frac{h_1}{h_2}$

Substitute $4$ for $h_1$ , $6$ for $h_2$ and $8$ for $a_2$ in the above expression.

  $\begin{array}{c}\frac{a_1}{8}=\frac{4}{6}\\ a_1=\frac{4\times 8}{6}\\ =5.34\end{array}$

Therefore, the base edge of the smaller pyramid is $5.34$ .

(b)

To determine

To find:The value of the total areas of the similar pyramid.

Answer to Problem 5ST2

The ratios of total areas of the similar pyramid is $16:36$ .

Explanation of Solution

Given information:

The height of smaller pyramid is $4$ , the height of larger pyramid is $6$ and the base edge of the larger pyramid is $8$ .

Calculation:

The theorem of similar solids states that if the scale factor of two similar solids is $a:b$ , then the ratio of base areas, lateral areas and total areas is $a^2:b^2$ .

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

  $\begin{array}{c}\text{total}\text{area}={\left(\text{scale}\text{factor}\right)}^2\\ \frac{t.a_1}{t.a_2}={\left(\frac{h_1}{h_2}\right)}^2\end{array}$

Substitute $4$ for $h_1$ , and $6$ for $h_2$ in the above expression.

  $\begin{array}{c}\frac{t.a_1}{t.a_2}={\left(\frac{4}{6}\right)}^2\\ =\frac{4\times 4}{6\times 6}\\ =\frac{16}{36}\end{array}$

Therefore, the ratios of the total area of the similar pyramidsis $16:36$ .