Chapter 13.4, Problem 9PSC

(a)

To determine

To find: The coordinates of A,B, and C.

Answer to Problem 9PSC

The coordinates are $\text{A}\left(1,1\right);\text{B}\left(2,4\right);\text{C}\left(3,9\right)$ .

Explanation of Solution

Given:

The given graph is:

  

The table for the graph $y=x^2$ is given as:

$x$	$1$	$2$	$3$
$y$	$1$	$4$	$9$

The coordinates of A,B, and C are given as:

  $\text{A}\left(1,1\right);\text{B}\left(2,4\right);\text{C}\left(3,9\right)$ .

Therefore, the coordinates are $\text{A}\left(1,1\right);\text{B}\left(2,4\right);\text{C}\left(3,9\right)$ .

(b)

To determine

To find: The sum of $\Delta \text{AOD,}\text{trapezoid}\text{ABED,}\text{and}\text{trapezoid}\text{BCFE}$ .

Answer to Problem 9PSC

The sum of the area is $9\frac{1}{2}$ unit square.

Explanation of Solution

Given:

The given graph is:

  

Area of triangle and trapezoid are given as:

  $\begin{array}{l}\text{Area}=\frac{1}{2}\cdot \text{base}\cdot \text{altitude}\\ \text{Area}=\frac{1}{2}\left(\text{sum of the length of parallel side}\right)\cdot \text{h}\end{array}$

The area of $\Delta \text{AOD,}\text{trapezoid}\text{ABED,}\text{and}\text{trapezoid}\text{BCFE}$ are calculated as:

  $\begin{array}{c}\text{Area of }\Delta \text{AOD}=\frac{1}{2}\cdot 1\cdot 1\\ =\frac{1}{2}\\ \text{Area of trapezoid ABED}=\frac{1}{2}\left(1+4\right)\cdot 1\\ =2\frac{1}{2}\end{array}$

Further simplified as:

  $\begin{array}{c}\text{Area of trapezoid BCFE}=\frac{1}{2}\left(4+9\right)\cdot 1\\ =6\frac{1}{2}\end{array}$

The sum of these area is:

  $\frac{1}{2}+2\frac{1}{2}+6\frac{1}{2}=9\frac{1}{2}$ .

Therefore, the sum of the area is $9\frac{1}{2}$ unit square.