Chapter 7.5, Problem 35PPS

To determine

To Find :

The area of the rectangle.

Answer to Problem 35PPS

  $Area=8\sqrt{6}+3\sqrt{2}{\text{ ft}}^2$

Explanation of Solution

Given Information:

Given rectangle is :

  

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.
• Area of a rectangle with length l and width w is lw .

Calculation:

• Given

$\begin{array}{l}\text{Length of the rectangle, }l=8+\sqrt{3}\text{ ft}\text{.}\\ \text{Width of the rectangle, w}=\sqrt{6}\text{ ft}\text{.}\\ \text{Area of the rectangle, A = }lw\\ \Rightarrow A=(8+\sqrt{3})\cdot \sqrt{6}\\ \Rightarrow A=8\sqrt{6}\text{+}\sqrt{3}\cdot \sqrt{6}\mathrm{..................}\text{[Distributive Property of Multiplication over addition]}\\ \Rightarrow A=8\sqrt{6}+\sqrt{18}\mathrm{.......................}\text{[Product Property of radicals]}\\ \Rightarrow A=8\sqrt{6}+\sqrt{3^2\cdot 2}\mathrm{...................}[\text{Factors using squares]}\\ \Rightarrow Area=8\sqrt{6}+3\sqrt{2}{\text{ ft}}^2\end{array}$

• So, $Area=8\sqrt{6}+3\sqrt{2}{\text{ ft}}^2$