Chapter 7.1, Problem 27WE

(a)

To determine

To find: The ratio of the perimeter two rectangle.

Answer to Problem 27WE

The required ratio is $14:19$ .

Explanation of Solution

Given:

The dimension of the first rectangle is $8\text{cm}\times 6\text{cm}$ .

The dimension of the second rectangle is $10\text{cm}\times 9\text{cm}$ .

Calculation:

Consider the perimeter of the rectangle is,

  $P=2\left(L+B\right)$

Then, the perimeter of the first rectangle is,

  $\begin{array}{c}P=2\left(8\text{cm}+6\text{cm}\right)\\ =28\text{cm}\end{array}$

Then, the perimeter of the first rectangle is,

  $\begin{array}{c}P=2\left(10\text{cm}+9\text{cm}\right)\\ =38\text{cm}\end{array}$

Thus, the ratio of the perimeter is,

  $\frac{28\text{cm}}{38\text{cm}}=14:19$

(b)

To determine

To find: The ratio of the area two rectangle.

Answer to Problem 27WE

The required ratio is $8:15$ .

Explanation of Solution

Given:

The dimension of the first rectangle is $8\text{cm}\times 6\text{cm}$ .

The dimension of the second rectangle is $10\text{cm}\times 9\text{cm}$ .

Calculation:

Consider the area of the rectangle is,

  $A=LB$

Then, the area of the first rectangle is,

  $\begin{array}{c}A=\left(8\text{cm}\right)\left(6\text{cm}\right)\\ =48{\text{cm}}^2\end{array}$

Then, the area of the first rectangle is,

  $\begin{array}{c}A=\left(10\text{cm}\right)\left(9\text{cm}\right)\\ =90{\text{cm}}^2\end{array}$

Thus, the ratio of the perimeter is,

  $\frac{48{\text{cm}}^2}{90{\text{cm}}^2}=8:15$