Chapter 1.6, Problem 26PPS

To determine

To graph figure and fine perimeter and area

Answer to Problem 26PPS

Area is 15 unit².

Perimeter is $11+\sqrt{61}$ units.

Explanation of Solution

Given information:

  $J(-3,-3),K(3,2),L(3,-3)$

Formula used:

Distance formula= $\sqrt{{(x_2^{}-x_1^{})}^2+{(y_2^{}-y_1^{})}^2}$

Graph:

  

Calculation:

The points are: $J(-3,-3),K(3,2),L(3,-3)$

Using distance formula to find JK , KL, LJ

JK ,

  $\begin{array}{l}{x}_2^{}=3,x_1^{}=-3\\ y_2^{}=2,y_1^{}=-3\end{array}$

Substitute values in formula,

JK = $\sqrt{{(3-(-3))}^2+{(2-(-3))}^2}$

Solving,

JK = $\sqrt{{(6)}^2+{(5)}^2}$

Hence, JK is $\sqrt{61}$

KL ,

  $\begin{array}{l}{x}_2^{}=-3,x_1^{}=3\\ y_2^{}=-3,y_1^{}=2\end{array}$

Substitute values in formula,

KL = $\sqrt{{(3-3)}^2+{(2-(-3))}^2}$

Solving,

KL = $\sqrt{0+{(5)}^2}$

Hence, KL is 5.

LJ ,

  $\begin{array}{l}{x}_2^{}=-3,x_1^{}=3\\ y_2^{}=-3,y_1^{}=-3\end{array}$

Substitute values in formula,

LJ = $\sqrt{{(-3-3)}^2+{(-3-(-3))}^2}$

Solving,

LJ = $\sqrt{{(-6)}^2+{(0)}^2}$

Hence, LJ is $\sqrt{36}=6$ .

Now,

Perimeter = sum of all sides

Substituting values,

Perimeter= $\sqrt{61}+5+6$

So, perimeter is $11+\sqrt{61}$ units,

Now, to find area,

Area= $\frac{1}{2}(base)(height)$

Substituting values,

Area= $\frac{1}{2}(6)(5)$

Solving,

Area= $(3)(5)$

Hence, area is 15 unit².