Chapter 4.3, Problem 37WE

To determine

To calculate: The perimeter and area of the shaded region of the solid.

Answer to Problem 37WE

The perimeter is $18a$ and area is $10a^2$ units.

Explanation of Solution

Given information:

The solid is provided below,

Formula used:

The perimeter of any solid is the sum of all its sides.

The area of rectangle with length l , width bis $A=lb$ .

The area of square with side sis $A=s^2$ .

Calculation:

Consider thesolid provided below,

Recall that the perimeter of any solid is the sum of all its sides.

Sum the lengths of all sides,

  $\begin{array}{c}P=2\left(4a\right)+2a+4\left(2a\right)\\ =8a+2a+8a\\ =18a\end{array}$

Next to compute the area of the solid, solid has 4 squares at corner and one rectangle in middle.

For 4 squares, side is a units.

Recall that the area of square with side sis $A=s^2$ .

Area of 4 squares is,

  $\begin{array}{c}A=4\left(a^2\right)\\ =4a^2\end{array}$

Next, there is a rectangle in middle.

Here, $l=2a\text{ and }b=3a$ .

Apply it,

  $\begin{array}{c}A=\left(2a\right)\left(3a\right)\\ =6a^2\end{array}$

Therefore, total area of the solid is,

  $\begin{array}{c}A=4a^2+6a^2\\ =10a^2\end{array}$

Thus, the perimeter is $18a$ and area is $10a^2$ units.