Chapter 9.5, Problem 15PSB

To determine

To find:ThePerimeter of quadrilateral to the nearest tenth and the most descriptive name of the figure.

Answer to Problem 15PSB

Perimeter of quadrilateral is $24.9$ units and the figure would be kite.

Explanation of Solution

Given information:Here, we have given that vertices of parallelogram are A $(2,1)$ , B $(7,3)$ , C $(12,1)$ and D $(7,-4)$ .

Formula used: Distance formula: d = $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$ and Perimeter of quadrilateral = Sum of its all sides.

Calculation: AB = $\sqrt{{(7-2)}^2+{(3-1)}^2}$

  $\begin{array}{l}\Rightarrow \sqrt{{(5)}^2+{(2)}^2}\\ \Rightarrow \sqrt{25+4}\\ \Rightarrow \sqrt{29}\end{array}$

Now, BC = $\sqrt{{(12-7)}^2+{(1-3)}^2}$

  $\begin{array}{l}\Rightarrow \sqrt{{(5)}^2+{(-2)}^2}\\ \Rightarrow \sqrt{25+4}\\ \Rightarrow \sqrt{29}\end{array}$

Similarly, CD = $\sqrt{{(7-12)}^2+{(-4-1)}^2}$

  $\begin{array}{l}\Rightarrow \sqrt{{(-5)}^2+{(-5)}^2}\\ \Rightarrow \sqrt{25+25}\\ \Rightarrow \sqrt{50}\end{array}$

And, AD = $\sqrt{{(7-2)}^2+{(-4-1)}^2}$

  $\begin{array}{l}\Rightarrow \sqrt{{(5)}^2+{(-5)}^2}\\ \Rightarrow \sqrt{25+25}\\ \Rightarrow \sqrt{50}\end{array}$

Now, Perimeter of quadrilateral = AB + BC + AC + AD

  $\begin{array}{l}\Rightarrow \sqrt{29}+\sqrt{29}+\sqrt{50}+\sqrt{50}\\ \Rightarrow 2(\sqrt{29})+2(\sqrt{50})\\ \Rightarrow 2(5.38)+2(7.07)\\ \Rightarrow 10.76+14.14\\ \Rightarrow 24.9\end{array}$

As, Adjacent sides of quadrilateral are equal,

  $\Rightarrow$ Quadrilateral is a kite.