Chapter 10, Problem 45SGR

To determine

To find: the perimeter of $\Delta DEF$ .

Answer to Problem 45SGR

The perimeter of $\Delta DEF$ is 15.41

Explanation of Solution

Given:

  

Formula used:

The distance between the points are calculated as

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

Calculation:

The coordinates are

  $D\left(2,3\right),F\left(0,-3\right)\&E\left(3,-3\right)$

Apply distance formula by keeping the values of the two coordinates

Hence

For Side DF

  $d=\sqrt{{\left(0-2\right)}^2+{\left(\left(-3\right)-3\right)}^2}$

Or, $d=\sqrt{{\left(-2\right)}^2+{\left(-6\right)}^2}$

Or, $d=\sqrt{4+36}$

Or, $d=\pm \sqrt{40}$

Or, $d=\sqrt{40}$ (distance can't be negative so omit the negative root found)

Similarly for side DE

  $d=\sqrt{{\left(3-2\right)}^2+{\left(\left(-3\right)-3\right)}^2}$

Or, $d=\sqrt{{\left(1\right)}^2+{\left(-6\right)}^2}$

Or, $d=\sqrt{1+36}$

Or, $d=\pm \sqrt{37}$

Or, $d=\sqrt{37}$ (distance can't be negative so omit the negative root found)

Lastly for side EF

  $d=\sqrt{{\left(3-0\right)}^2+{\left(\left(-3\right)-\left(-3\right)\right)}^2}$

Or, $d=\sqrt{{\left(3\right)}^2+{\left(0\right)}^2}$

Or, $d=\sqrt{9+0}$

Or, $d=\pm \sqrt{9}$

Or, $d=3$ (distance can't be negative so omit the negative root found)

Hence the measure of all the three sides is

  $\begin{array}{l}DF=\sqrt{40}\\ DE=\sqrt{37}\&\\ EF=3\end{array}$

The perimeter of $\Delta DEF$ is

  $\begin{array}{l}\text{perimeter =}\sqrt{40}+\sqrt{37}+3\\ =15.407\end{array}$

Conclusion:

Therefore, the perimeter of $\Delta DEF$ is 15.41