Chapter 5.1, Problem 15PSC

a.

To determine

To show: Diagonals $\overline{CS}$ and $\overline{OI}$ of isosceles trapezoid CISO do not bisect each other.

Explanation of Solution

Given information:

The coordinates of point C are $\left(-3,8\right)$ .

The coordinates of point I are $\left(-5,0\right)$ .

The coordinates of point S are $\left(5,0\right)$ .

The coordinates of point O are $\left(3,8\right)$ .

Proof:

  

The coordinates of point Care$\left(-3,8\right)$ .

The coordinates of point S are $\left(5,0\right)$ .

The coordinates of midpoints of $\overline{CS}$ are $\left(\left(\frac{-3+5}{2}\right),\left(\frac{8+0}{2}\right)\right)=\left(1,4\right)$ .

The coordinates of midpoints of $\overline{OI}$ are $\left(\left(\frac{3-5}{2}\right),\left(\frac{8+0}{2}\right)\right)=\left(-1,4\right)$ .

Diagonals $\overline{CS}$ and $\overline{OI}$ do not bisect each other, since midpoints are not the same points.

b.

To determine

To check:That Diagonals $\overline{CS}$ and $\overline{OI}$ of isosceles trapezoid CISO are perpendicular.

Answer to Problem 15PSC

Yes,diagonals $\overline{CS}$ and $\overline{OI}$ are perpendicular.

Explanation of Solution

Given information:

The coordinates of point C are $\left(-3,8\right)$ .

The coordinates of point I are $\left(-5,0\right)$ .

The coordinates of point S are $\left(5,0\right)$ .

The coordinates of point O are $\left(3,8\right)$ .

Formula used:

Slope of a line:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=slope\end{array}$

  $(x_1,y_1)$ are coordinates of first point in the line.

  $(x_2,y_2)$ are coordinates of second point in the line.

  

The coordinates of point C are $\left(-3,8\right)$ .

The coordinates of point S are $\left(5,0\right)$ .

We now check the slopes of isosceles trapezoid CISO.

Slope of $\overline{CS}=\frac{0-8}{5-\left(-3\right)}=\frac{-8}{8}=-1$

The coordinates of point O are $\left(3,8\right)$ .

The coordinates of point I are $\left(-5,0\right)$ .

Slope of $\overline{OI}=\frac{0-8}{-5-3}=\frac{-8}{-8}=1$

Since the product of slopes $\overline{CS}$ and $\overline{OI}$ is -1, diagonals $\overline{CS}$ and $\overline{OI}$ are perpendicular.

c.

To determine

To Check:That the Diagonals of every isosceles trapezoid are perpendicular.

Answer to Problem 15PSC

No, diagonals are not necessarily perpendicular.

Explanation of Solution

Given information:

The coordinates of point C are $\left(-3,8\right)$ .

The coordinates of point I are $\left(-5,0\right)$ .

The coordinates of point S are $\left(5,0\right)$ .

The coordinates of point O are $\left(3,8\right)$ .

Formula used:

Slope of a line:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=slope\end{array}$

  $(x_1,y_1)$ are coordinates of first point in the line.

  $(x_2,y_2)$ are coordinates of second point in the line.

  

The coordinates of point C are $\left(-3,8\right)$ .

The coordinates of point S are $\left(5,0\right)$ .

We now check the slopes of isosceles trapezoid CISO.

Slope of $\overline{CS}=\frac{0-8}{5-\left(-3\right)}=\frac{-8}{8}=-1$

The coordinates of point O are $\left(3,8\right)$ .

The coordinates of point I are $\left(-5,0\right)$ .

Slope of $\overline{OI}=\frac{0-8}{-5-3}=\frac{-8}{-8}=1$

Since the product of slopes $\overline{CS}$ and $\overline{OI}$ is -1, diagonals $\overline{CS}$ and $\overline{OI}$ are perpendicular.

If I and S are moved to $\left(-6,0\right)$ and $\left(6,0\right)$ respectively.

The coordinates of point C are $\left(-3,8\right)$ .

The coordinates of point S are $\left(6,0\right)$ .

We now check the slopes of isosceles trapezoid CISO.

Slope of $\overline{CS}=\frac{0-8}{6-\left(-3\right)}=\frac{-8}{9}$

The coordinates of point O are $\left(3,8\right)$ .

The coordinates of point I are $\left(-6,0\right)$ .

Slope of $\overline{OI}=\frac{0-8}{-6-3}=\frac{-8}{-9}=\frac{8}{9}$

Since the product of slopes $\overline{CS}$ and $\overline{OI}$ is not -1, diagonals $\overline{CS}$ and $\overline{OI}$ are not perpendicular.

d.

To determine

To find:The figure to draw to obtain a value of $\overline{OS}$ .

Answer to Problem 15PSC

The value of $\overline{OS}$ is $8.25$ .

Explanation of Solution

Given information:

The coordinates of point C are $\left(-3,8\right)$ .

The coordinates of point I are $\left(-5,0\right)$ .

The coordinates of point S are $\left(5,0\right)$ .

The coordinates of point O are $\left(3,8\right)$ .

Formula used:

The distance between two points by using the distance formula, which is an application of the Pythagoras theorem.

D$=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Calculation:

  

Draw b perpendicular to $\overline{IS}$ from point O to make a right triangle and then use Pythagoras Theorem.

The point A lies on x axis. Its y coordinate is 0.The x coordinate will be same as of x coordinate of point O.

The coordinates of point A are $\left(3,0\right)$ .

The distance OA can be calculated by applying Pythagoras Theorem.

  $\begin{array}{l}OA=\sqrt{{\left(3-3\right)}^2+{\left(0-8\right)}^2}\\ OA=\sqrt{{\left(0\right)}^2+{\left(8\right)}^2}\\ OA=\sqrt{64}\\ OA=8\end{array}$

The distance AS can be calculated by applying Pythagoras Theorem.

  $\begin{array}{l}AS=\sqrt{{\left(5-3\right)}^2+{\left(0-0\right)}^2}\\ AS=\sqrt{{\left(2\right)}^2+{\left(0\right)}^2}\\ AS=\sqrt{4+0}\\ AS=\sqrt{4}\\ AS=2\end{array}$

In right angle triangle OAS, we get

  $\begin{array}{l}{\left(OS\right)}^2={\left(OA\right)}^2+{\left(AS\right)}^2\\ {\left(OS\right)}^2={\left(8\right)}^2+{\left(2\right)}^2\\ {\left(OS\right)}^2=64+4\\ {\left(OS\right)}^2=68\\ OS=\sqrt{68}\\ OS\approx 8.25\end{array}$