Chapter 10.8, Problem 19PSC

a)

To determine

To find : The value of $\text{x}$ .

Answer to Problem 19PSC

  $\text{x}=4$ units

Explanation of Solution

Given Information :

As figure given in question :

  

Formula Used:

Pythagoras Theorem : (hypotenuse)2 = (perpendicular)2 + (base)2

Calculation :

  $\begin{array}{l}{r}^2+8^2={(r+y)}^2\mathrm{.....}(i)\text{In}\Delta \text{OLP},\text{where}\text{LL}=90\\ \therefore {(y+r)}^2={(x+6)}^2+h^2\mathrm{.....}(\mathrm{ii})\text{In}\Delta \text{OLR}\\ \therefore r^2=h^2+6^2\mathrm{.....}(iii)\end{array}$

Now

Replacing the ${(y+r)}^2$ of equation (ii) and (iii)

  $r^2+8^2={(x+6)}^2+h^2\mathrm{...}(iv)$

Now putting the value of ${\text{r}}^2$ with the value of equation (iii)

  $\begin{array}{l}\because h^2+6^2+8^2=x^2+36+12x+h^2\\ \therefore x^2+12x-64=0\\ \Rightarrow (x+16)(x-4)=0\\ \therefore x=4\text{or}x=-16\text{ (not valid)}\end{array}$

b)

To determine

To find : Restrictions on $y$ .

Answer to Problem 19PSC

Explanation of Solution

Given Information :

In $\Delta$ ORQ where $\angle$ R = $90°$ (Tangent with radius)

Calculation :

From equation on (ii)

  $\begin{array}{l}{(y+r)}^2={(4+6)}^2+h^2{(y+r)}^2=100+h^{2}y+r=\sqrt{h^2+10}\\ y=\sqrt{h^2+100}-r\end{array}$