Chapter 12.3, Problem 15IP

(a)

To determine

To find: Total area of given structure

Answer to Problem 15IP

  $\text{73}{\text{.14in}}^{\text{2}}$

Explanation of Solution

Given:

Dimension of triangle and semi-circle are given

  

Formula: $\text{area of triangle =}\frac{\text{1}}{\text{2}}\text{×base×height}$

Calculation: total area = ${(area)}_1+{(area)}_2$

  $\begin{array}{l}$ =\frac{1}{2}\times 8\times 12+\frac{1}{2}\pi {(4)}^2$=48+\frac{1}{2}\times \frac{22}{7}\times 4\times 4\\ =73.14i{n}^2\end{array}$

(b)

To determine

To find: Find total area of given structure

Answer to Problem 15IP

  $\text{104}{\text{.57in}}^{\text{2}}$

Given:

Dimension of triangle and semi-circle are given

Formula:

  $\text{area of triangle = }\frac{\text{1}}{\text{2}}\text{×base×height}$

Calculation: ${\text{area=(area)}}_{\text{1}}{\text{+(area)}}_{\text{2}}$

  $\begin{array}{l}\text{=}\frac{\text{1}}{\text{2}}\text{×8×12+}\frac{\text{1}}{\text{2}}{\text{π(6)}}^{\text{2}}\\ \text{=48+}\frac{\text{1}}{\text{2}}\text{×}\frac{\text{22}}{\text{7}}\text{×6×6}\\ \text{=104}{\text{.57in}}^{\text{2}}\end{array}$

Explanation of Solution

Given:

Dimension of triangle and semi-circle are given

Formula:

  $\text{area of triangle = }\frac{\text{1}}{\text{2}}\text{×base×height}$

Calculation: ${\text{area=(area)}}_{\text{1}}{\text{+(area)}}_{\text{2}}$

  $\begin{array}{l}\text{=}\frac{\text{1}}{\text{2}}\text{×8×12+}\frac{\text{1}}{\text{2}}{\text{π(6)}}^{\text{2}}\\ \text{=48+}\frac{\text{1}}{\text{2}}\text{×}\frac{\text{22}}{\text{7}}\text{×6×6}\\ \text{=104}{\text{.57in}}^{\text{2}}\end{array}$