Chapter 6.5, Problem 36PPS

To determine

To Find :

The surface area of the sphere.

Answer to Problem 36PPS

  $(84+32\sqrt{5})\pi {\text{ in}}^2$

Explanation of Solution

Given Information:

  $\begin{array}{l}Theradiusr ofthesphereis:\\ r=4+\sqrt{5}\text{ inches}\end{array}$

Concept Used :

• Definition of nth root:
• Words : For any real numbers a and b , and any positive integer n , if $a^n=b$ then a is the nth root of b . That is $a=\sqrt[n]{b}$

    $\begin{array}{l}\sqrt[n]{b}\\ Here\\ b=radicand\\ n=index\\ \sqrt{}=radicalsign\end{array}$

  Product Property of Radicals: $\begin{array}{l}\text{For any real number a and b and any integer n > 1 , }\\ \sqrt[n]{ab}=\sqrt[n]{a}\cdot \sqrt[n]{b}\text{ , if n is even and a \& b are both nonnegative }\\ \text{or if n is odd}\text{.}\end{array}$

• Like radical expressions : If the radicals have both the index and radicand identical , then they are like radicals.
• In order to add or subtract , the radicals must be like radicals.
• $\begin{array}{l}\text{The surface area of a shpere of radius r is }\\ \text{S=4}\pi {\text{r}}^2\end{array}$

Calculation:

  $\begin{array}{l}Given\text{ radius of sphere is , r = 4+}\sqrt{5}\text{ inches}\\ \text{Surface area of sphere , S = 4}\pi r^2\\ $ \Rightarrow S=4\pi {(4+\sqrt{5})}^2\mathrm{.................}[r=\text{4+}\sqrt{5}\text{ inches}]$$ \Rightarrow S=4\pi (16+5+8\sqrt{5})\mathrm{..........}[{(a+b)}^2=a^2+b^2+2ab]$\Rightarrow S=4\pi (21+8\sqrt{5})\\ \Rightarrow S=(84+32\sqrt{5})\pi {\text{ in}}^2\end{array}$

So, the surface area of the sphere = $(84+32\sqrt{5})\pi {\text{ in}}^2$