Chapter 9.1, Problem 9PSA

a

To determine

To calculate: The Value of xif $y=21$ , and $r=29$ .

Answer to Problem 9PSA

The Value of x is $20$ .

Explanation of Solution

Given information: $x^2+y^2=r^2$ ; $y=21$ and $r=29$ .

Formula used: Pythagoras Theorem: ${(Perpendicular)}^2+{(Base)}^2={(Hypotenuse)}^2$

Calculation: ${(x)}^2+{(y)}^2={(r)}^2$

  $\begin{array}{l}\Rightarrow {(}^x+{(}^{21}={(}^{29}\\ \Rightarrow x^2+441=841\\ \Rightarrow x^2=881-441\\ \Rightarrow x^2=400\\ \Rightarrow x=\sqrt{400}\\ \Rightarrow x=20\end{array}$

b

To determine

To calculate: The Value of yif $x=2$ ,and $r=4$ .

Answer to Problem 9PSA

Value of y is $2\sqrt{3}$ .

Explanation of Solution

Given information:$x^2+y^2=r^2$ ; $x=2$ and $r=4$ .

Formula used: Pythagoras Theorem: ${(Perpendicular)}^2+{(Base)}^2={(Hypotenuse)}^2$

Calculation: ${(x)}^2+{(y)}^2={(r)}^2$

  $\begin{array}{l}\Rightarrow {(}^2+{(}^y={(}^4\\ \Rightarrow y^2+4=16\\ \Rightarrow y^2=16-4\\ \Rightarrow y^2=12\\ \Rightarrow y=\sqrt{12}\\ \Rightarrow y=\sqrt{2\times 2\times 3}\\ \Rightarrow y=2\sqrt{3}\end{array}$

c

To determine

To calculate: The Value of rif $x=4.1$ ,and $y=7.1$ .

Answer to Problem 9PSA

Value of r to the nearest tenth is $10$ .

Explanation of Solution

Given information:$x^2+y^2=r^2$ ; $x=4.1$ and $y=7.1$ .

Formula used: Pythagoras Theorem: ${(Perpendicular)}^2+{(Base)}^2={(Hypotenuse)}^2$

Calculation: ${(x)}^2+{(y)}^2={(r)}^2$

  $\begin{array}{l}\Rightarrow {(}^{4.1}+{(}^{7.1}=r^2\\ \Rightarrow 16.81+50.41=r^2\\ \Rightarrow r^2=67.22\\ \Rightarrow r=\sqrt{67.22}\\ \Rightarrow r=\sqrt{\frac{6722}{100}}\\ \Rightarrow r=\frac{81.98}{10}\\ \Rightarrow r=8.198\end{array}$

Nearest tenth = $10$