Chapter 11.2, Problem 33PPS

a.

To determine

To graph: A kite if $x=2$ .

Explanation of Solution

Given information:A kite having diagonals of length 12 cm and 4 cm.

Graph: When $x=2$ 2 cm

  

b.

To determine

To graph: A kite for three x- values between 2 and 10 and for an x -value of 10.

Explanation of Solution

Given information:A kite having diagonals of length 12 cm and 4 cm.

Graph: When $x=3$

  

When $x=4$

  

When $x=5$

  

When $x=10$

  

c.

To determine

To calculate: Perimeter of each kite.

Explanation of Solution

Given information:A kite having diagonals of length 12 cm and 4 cm.

Formula used: Perimeter of kite is given by

  $\text{Perimeter of kite}=2(a+b)$

where, $a\&b$ are sides of kite.

Calculation:

In kite, longer diagonal cuts the shorter one in half.

When $x=2$

  $\begin{array}{l}a=\sqrt{{(2)}^2+{(2)}^2}\\ =\sqrt{4+4}\\ \text{ =}\sqrt{8}\text{ cm}\end{array}$

  $\begin{array}{l}b=\sqrt{{(10)}^2+{(2)}^2}\\ =\sqrt{100+4}\\ \text{ =}\sqrt{104}\text{ cm}\end{array}$

  $\begin{array}{l}\text{Perimeter of kite}=2(a+b)\\ =2(\sqrt{8}+\sqrt{104})\\ =26.05\text{ cm}\end{array}$

When $x=3$

  $\begin{array}{l}a=\sqrt{{(3)}^2+{(2)}^2}\\ =\sqrt{9+4}\\ \text{ =}\sqrt{13}\text{ cm}\end{array}$

  $\begin{array}{l}b=\sqrt{{(9)}^2+{(2)}^2}\\ =\sqrt{81+4}\\ \text{ =}\sqrt{85}\text{ cm}\end{array}$

  $\begin{array}{l}\text{Perimeter of kite}=2(a+b)\\ =2(\sqrt{13}+\sqrt{85})\\ =25.65\text{ cm}\end{array}$

When $x=4$

  $\begin{array}{l}a=\sqrt{{(4)}^2+{(2)}^2}\\ =\sqrt{16+4}\\ \text{ =}\sqrt{20}\text{ cm}\end{array}$

  $\begin{array}{l}b=\sqrt{{(8)}^2+{(2)}^2}\\ =\sqrt{64+4}\\ \text{ =}\sqrt{70}\text{ cm}\end{array}$

  $\begin{array}{l}\text{Perimeter of kite}=2(a+b)\\ =2(\sqrt{20}+\sqrt{70})\\ =25.68\text{ cm}\end{array}$

When $x=5$

  $\begin{array}{l}a=\sqrt{{(5)}^2+{(2)}^2}\\ =\sqrt{25+4}\\ \text{ =}\sqrt{29}\text{ cm}\end{array}$

  $\begin{array}{l}b=\sqrt{{(7)}^2+{(2)}^2}\\ =\sqrt{49+4}\\ \text{ =}\sqrt{53}\end{array}$

  $\begin{array}{l}\text{Perimeter of kite}=2(a+b)\\ =2(\sqrt{29}+\sqrt{53})\\ =25.33\text{ cm}\end{array}$

When $x=10$

  $\begin{array}{l}a=\sqrt{{(10)}^2+{(2)}^2}\\ =\sqrt{100+4}\\ \text{ =}\sqrt{104}\text{ cm}\end{array}$

  $\begin{array}{l}b=\sqrt{{(2)}^2+{(2)}^2}\\ =\sqrt{4+4}\\ \text{ =}\sqrt{8}\text{ cm}\end{array}$

  $\begin{array}{l}\text{Perimeter of kite}=2(a+b)\\ =2(\sqrt{104}+\sqrt{8})\\ =26.05\text{ cm}\end{array}$

In tabular form,

x- value	2	3	4	5	10
Perimeter of kite	26.05cm	25.65cm	25.68 cm	25.53 cm	26.05 cm

d.

To determine

To graph: The perimeter versus x -values.

Explanation of Solution

Given information:A kite having diagonals of length 12 cm and 4 cm.

Graph:

  

e.

To determine

To analyze: The value of x that will minimize the perimeter of kite.

Explanation of Solution

Given information:A kite having diagonals of length 12 cm and 4 cm.

From the graph, we can analyze that perimeter is minimum when $x=6$ .

Thus, we can draw a kite which will minimize the surface the kite covers when $x=6$ .

Hence, significance is we will get a kite of minimum perimeter having diagonals 12 cm and 4 cm.