Chapter 7.5, Problem 6CYP

To determine

To Write:

An equation for L in terms of x and simplify it.

Answer to Problem 6CYP

  $L=\frac{900}{12\sqrt{3}+x}$

Explanation of Solution

Given Information:

The area of the rectangle is $900{\text{ ft}}^2$ .

Concept Used :

• Rationalizing the denominator :

To eliminate radicals from a denominator or fractions from a radicand , use this process.

To rationalize a denominator, multiply the numerator and denominator by a quantity so that the radicand has an exact root.

• Area of a rectangle having length l and width w is lw
• In a right angled triangle ,
$\begin{array}{l}\tan \theta =\frac{perpendicular}{base}=\frac{side\text{ opposite }\theta }{side\text{ adjacent to }\theta }\\ \tan \left({60}^{\circ }\right)=\sqrt{3}\end{array}$

Calculation:

• Given area of the rectangle is 900
• 

  $\begin{array}{l}\text{In the right angled triangle, we have}\\ \text{tan}\left({60}^{\circ }\right)=\frac{\text{side opposite to the angle}}{\text{side adjacent to the angle}}=\frac{w-x}{12}\\ \Rightarrow \sqrt{3}=\frac{w-x}{12}\\ \Rightarrow 12\sqrt{3}=w-x\mathrm{...........}[\text{multiply 12 each side]}\\ \Rightarrow w-x+x=12\sqrt{3}+x\mathrm{...}[\text{add x each side]}\\ \Rightarrow w=12\sqrt{3}+x\mathrm{...........}(1)\end{array}$

• $\begin{array}{l}{\text{The area of the rectangle is 900 ft}}^2.\\ So,\\ \text{Lw = 900}\\ \text{From equation (1) , substitute }w=12\sqrt{3}+x\text{ above and solve for L:}\\ \text{L}\left(12\sqrt{3}+x\right)=900\\ \Rightarrow L=\frac{900}{12\sqrt{3}+x}\mathrm{......................}[\text{divide each side by }12\sqrt{3}+x]\\ \Rightarrow L=\frac{900}{12\sqrt{3}+x}\cdot \frac{12\sqrt{3}-x}{12\sqrt{3}-x}\mathrm{......}[\text{rationalize the denominator]}\\ \Rightarrow L=\frac{900\left(12\sqrt{3}-x\right)}{\left(12\sqrt{3}+x\right)\left(12\sqrt{3}-x\right)}\\ \Rightarrow L=\frac{900\left(12\sqrt{3}-x\right)}{432-x^2}\end{array}$
• So, $L=\frac{900}{12\sqrt{3}+x}=\frac{900\left(12\sqrt{3}-x\right)}{432-x^2}$