Chapter 7.1, Problem 28WE

(a)

To determine

To find: The ratio of the perimeter two rectangle.

Answer to Problem 28WE

The required ratio is $1:1$ .

Explanation of Solution

Given:

The perimeter of the first rectangle is $30\text{cm}$ .

The perimeter of the second rectangle is $30\text{cm}$ .

Calculation:

The ratio of the perimeter of the first and the second rectangle is,

  $\frac{30\text{cm}}{30\text{cm}}=1:1$

(b)

To determine

To find: The ratio of the area two rectangle.

Answer to Problem 28WE

The required ratio is $18:25$ .

Explanation of Solution

Given:

The length of the first rectangle is $12\text{cm}$ .

The length of the second rectangle is $10\text{cm}$ .

Calculation:

Consider the breadth of the rectangle is,

  $B=\frac{P}{2}-L$

The breadth of the first rectangle is,

  $\begin{array}{c}B=\frac{30\text{cm}}{2}-12\text{cm}\\ =3\text{cm}\end{array}$

The breadth of the second rectangle is,

  $\begin{array}{c}B=\frac{30\text{cm}}{2}-10\text{cm}\\ =5\text{cm}\end{array}$

Consider the area of the rectangle is,

  $A=LB$

Then, the area of the first rectangle is,

  $\begin{array}{c}A=\left(12\text{cm}\right)\left(3\text{cm}\right)\\ =36{\text{cm}}^2\end{array}$

Then, the area of the first rectangle is,

  $\begin{array}{c}A=\left(10\text{cm}\right)\left(5\text{cm}\right)\\ =50{\text{cm}}^2\end{array}$

Thus, the ratio of the perimeter is,

  $\frac{36{\text{cm}}^2}{50{\text{cm}}^2}=18:25$