Chapter 11.7, Problem 16WE

To determine

To Find: The perimeter of similar pentagon whose area is $27c{m}^2$ .

Answer to Problem 16WE

  $P=\frac{75}{4}m$

Explanation of Solution

Given: A pentagon with sides $3cm,4cm,5cm,6cm,7cm$ has area $48c{m}^2$ .

Since two pentagons are equal.

Thus

  $\begin{array}{l}\frac{ar(I)}{ar(II)}=\frac{48}{27}=\frac{16}{9}\\ now\\ \frac{16}{9}=\frac{side{(I)}^2}{side{(II)}^2}\\ \frac{16}{9}={\left(\frac{3}{a}\right)}^2\\ \frac{4}{3}=\frac{3}{a}\\ a=\frac{9}{4}\\ similarly\\ b=3\\ c=\frac{15}{4}\\ d=\frac{18}{4}\\ e=\frac{21}{4}\end{array}$

Thus

Perimeter of second pentagon is

  $\begin{array}{l}P=\frac{12}{4}+\frac{9}{4}+\frac{15}{4}+\frac{18}{4}+\frac{21}{4}\\ P=\frac{12+9+15+18+21}{4}\\ P=\frac{75}{4}m\end{array}$

Hence, $P=\frac{75}{4}m$