Chapter 11.4, Problem 9PSB

a

To determine

The Areas of region I region II and region III

Answer to Problem 9PSB

The Areas of region I region II and region III are 8, 28 and 18

Explanation of Solution

Given information :

Coordinates of points O (0, 0), A (12, 0), B(12, 3), C (12, 7), D (8, 7) and E (0, 7)

Formula used :

Area of triangle = $\frac{1}{2}\times \text{base}\times \text{height}$

Calculation :

Area of region I = $\frac{1}{2}\times \text{BC}\times \text{CD}$

Here

  $\begin{array}{l}\text{BC}=\sqrt{{\left(12-12\right)}^2+{\left(7-3\right)}^2}=4\text{units and }\\ \text{CD }=\sqrt{{\left(12-8\right)}^2+{\left(7-7\right)}^2}=4\text{units }\end{array}$

Required area = $\frac{1}{2}\times 4\times 4=8{\text{ unit}}^2$

Area of region II = $\frac{1}{2}\times \text{DE}\times \text{OE}$

Here

  $\begin{array}{l}\text{DE}=\sqrt{{\left(8-0\right)}^2+{\left(7-7\right)}^2}=8\text{units and }\\ \text{OE }=\sqrt{{\left(0-0\right)}^2+{\left(7-0\right)}^2}=7\text{units }\end{array}$

Required area = $\frac{1}{2}\times 8\times 7=28{\text{ unit}}^2$

Area of region III = $\frac{1}{2}\times \text{AB}\times \text{OA}$

Here

  $\begin{array}{l}\text{AB}=\sqrt{{\left(12-12\right)}^2+{\left(3-0\right)}^2}=3\text{units and }\\ \text{OA }=\sqrt{{\left(12-0\right)}^2+{\left(0-0\right)}^2}=12\text{units }\end{array}$

Required area = $\frac{1}{2}\times 3\times 12=18{\text{ unit}}^2$

b

To determine

The Areas of triangle OBD

Answer to Problem 9PSB

The Areas of OBD is 30

Explanation of Solution

Given information :

Coordinates of points O (0, 0), A (12, 0), B(12, 3), C (12, 7), D (8, 7) and E (0, 7)

Formula used :

Area of triangle OBD= Area of rectangle OACE − Sum of areas of region I, II & III

Calculation :

Area of triangle OBD= Area of rectangle − Sum of areas of region I, II & III

Area of rectangle OACE = $\text{OC}\times \text{OE}$

Here

  $\begin{array}{l}\text{OC}=\sqrt{{\left(12-0\right)}^2+{\left(0-0\right)}^2}=12\text{units and }\\ \text{OE }=\sqrt{{\left(0-0\right)}^2+{\left(0-7\right)}^2}=7\text{units }\end{array}$

Therefore,

Area of rectangle OACE = $\text{OC}\times \text{OE=12}\times {\text{7=84 unit}}^2$

Sum of areas of region I, II and III = 8 + 18 + 28 = 54 ${\text{unit}}^2$

Required area = 84 − 54 = 30 ${\text{unit}}^2$