Chapter 11, Problem 31RP

a.

To determine

To find: Thecoordinate of a point R.

Answer to Problem 31RP

The coordinates of the point $R$ are $\left(12,13\right).$

Explanation of Solution

Given Information:

  $TRAP$ is an isosceles trapezoid

  $AR\parallel PT$

  $RT=AP$

Formula used:

Distance between two points $\left(x_1,y_1\right)\text{and}\left(x_2,y_2\right)$ is $\left(D\right)=\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$

Calculation:

As $TRAP$ is a trapezoid

  $\therefore RT=AP$ … (i)

Also, $AR\parallel PT$

Then, x-coordinate of $R$ will be 12.

Let the y-coordinate of point $R$ be $b$

Now, distance between two points $\left(x_1,y_1\right)\text{and}\left(x_2,y_2\right)$ is $\left(D\right)=\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$

As $RT=AP$

  $\begin{array}{l}\therefore \sqrt{{\left(12-0\right)}^2+{\left(b-18\right)}^2}=\sqrt{{\left(12-0\right)}^2+{\left(5-0\right)}^2}\\ \therefore \sqrt{{12}^2+{\left(b-18\right)}^2}=\sqrt{{12}^2+5^2}\end{array}$

Squaring both the sides, we get

  $\begin{array}{l}{12}^2+{\left(b-18\right)}^2={12}^2+5^2\\ b^2-36b+324=25\\ b^2-36b+299=0\\ b^2-23b-13b+299=0\\ \left(b-23\right)\left(b-13\right)=0\\ b=23\text{or}b=13\end{array}$

According to the graph of $TRAP,b=13$

Therefore, y-coordinate of $R$ will be 13.

Hence, the coordinates of $R$ are $\left(12,13\right).$

b.

To determine

To find: The area of an isosceles trapezoid TRAP.

Answer to Problem 31RP

Area of the trapezoid $TRAP$ is 156.

Explanation of Solution

Given Information:

Vertices of the trapezoid $TRAP$ are

  $T\left(0,18\right),R\left(12,13\right),A\left(12,5\right)\text{and}P\left(0,0\right)$

Formula used:

Distance between two points $\left(x_1,y_1\right)\text{and}\left(x_2,y_2\right)$ is $\left(D\right)=\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$

Area of a trapezoid $=\frac{1}{2}\times \left(b_1+b_2\right)h$

  $b_1\text{and}{b}_2$ are the bases of the trapezoid

  $h$ is the height of the trapezoid

Calculation:

Vertices of the trapezoid $TRAP$ are

  $T\left(0,18\right),R\left(12,13\right),A\left(12,5\right)\text{and}P\left(0,0\right)$

We know that, Area of a trapezoid $=\frac{1}{2}\times \left(b_1+b_2\right)h$

Where, $b_1\text{and}{b}_2$ are the bases of the trapezoid

  $h$ is the height of the trapezoid

Now, $b_1=ARand{b}_2=PT$

Using distance formula, distance between any two points $\left(x_1,y_1\right)\text{and}\left(x_2,y_2\right)$ is

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

  $\begin{array}{l}\therefore b_1=\sqrt{{\left(12-12\right)}^2+{\left(5-13\right)}^2}\\ \therefore b_1=\sqrt{8^2}\\ \therefore b_1=8\end{array}$

Also,

  $\begin{array}{l}\therefore b_2=\sqrt{{\left(0-0\right)}^2+{\left(0-18\right)}^2}\\ \therefore b_2=\sqrt{{18}^2}\\ \therefore b_2=18\end{array}$

Now, drawing a line parallel to x-axis from A meeting y-axis at B

  

We get the coordinates of point B as $\left(0,5\right)$

  $\begin{array}{l}\therefore h=AB=\sqrt{{\left(12-0\right)}^2+{\left(5-5\right)}^2}\\ \therefore h=\sqrt{{12}^2}\\ \therefore h=12\end{array}$

  $\therefore$ Area of $TRAP=\frac{1}{2}\times \left(8+18\right)\times 12$

  $\begin{array}{l}\therefore \text{Area}\text{of}TRAP=26\times 6\\ \therefore \text{Area}\text{of}TRAP=156\end{array}$

Hence, area of the trapezoid $TRAP$ is 156.