Chapter 11.7, Problem 12PSB

a.

To determine

To calculate:Theratio of side AB andsidePQ of pentagons.

Answer to Problem 12PSB

Theratiois $2:3$ .

Explanation of Solution

Given information:

  $\frac{A(ABCDE)}{A(PQRST)}=\frac{8}{18}$

Formula used:

The below property is used:

Area of polygon is proportional to square of its side.

Calculation:

  

The area ratio is as follows:

  $\begin{array}{l}\frac{A(ABCDE)}{A(PQRST)}=\frac{{(AB)}^2}{{(PQ)}^2}\\ \frac{8}{18}=\frac{{(AB)}^2}{{(PQ)}^2}\\ \frac{{(AB)}^2}{{(PQ)}^2}=\frac{8}{18}=\frac{4}{9}\\ \frac{(AB)}{(PQ)}=\frac{\sqrt{4}}{\sqrt{9}}=\frac{2}{3}\end{array}$

b.

To determine

To find: The ratio of perimeter of pentagon ABCDE and perimeter of pentagon PQRST .

Answer to Problem 12PSB

The ratio is $2:3$ .

Explanation of Solution

Given information:

  $\frac{A(ABCDE)}{A(PQRST)}=\frac{8}{18}$

Formula used:

The below properties are used:

Area of polygon is proportional to square of its side.

The ratio of perimeter of two polygon is equal to ratio of any pairs of corresponding sides.

Calculation:

  

The area ratio is as follows:

  $\begin{array}{l}\frac{A(ABCDE)}{A(PQRST)}=\frac{{(AB)}^2}{{(PQ)}^2}\\ \frac{8}{18}=\frac{{(AB)}^2}{{(PQ)}^2}\\ \frac{{(AB)}^2}{{(PQ)}^2}=\frac{8}{18}=\frac{4}{9}\\ \frac{(AB)}{(PQ)}=\frac{\sqrt{4}}{\sqrt{9}}=\frac{2}{3}\end{array}$

Perimeter is sum of all sides.

The ratio of perimeter of two polygon is equal to ratio of any pairs of corresponding sides.

  $\frac{P(ABCDE)}{P(PQRST)}=\frac{(AB)}{(PQ)}=\frac{2}{3}$