Chapter 6.5, Problem 36WE

a.

To determine

The value of $\left(x\right)$ for the right angle triangle having perimeter $18$ and sides $x,\sqrt{3}x\text{ and }2x$ respectively.

Answer to Problem 36WE

The value of $\left(x\right)$ is equal to $9-3\sqrt{3}$ .

Explanation of Solution

Given information: It is given that the three sides of an right angle triangle are $x,\sqrt{3}x\text{ and }2x$ respectively, such that

  $2x>\sqrt{3}x>x\mathrm{....}\left(1\right)$

Formula used: The perimeter of an right angle triangle is equal to sum of its three sides.

Calculation: According to question and apply formula for perimeter of an right angle triangle.

  $\begin{array}{c}x+\sqrt{3}x\text{+}2x=18\text{ gives,}\\ \left(3+\sqrt{3}\right)x=18\\ x=\frac{18}{3+\sqrt{3}}\\ x=\left(\frac{18}{3+\sqrt{3}}\right)\left(\frac{3-\sqrt{3}}{3-\sqrt{3}}\right)\mathrm{......}\left(\text{Rationalized}\right)\end{array}$

Further solving,

  $\begin{array}{c}x=\frac{18\left(3-\sqrt{3}\right)}{9-3}\\ x=\frac{18\left(3-\sqrt{3}\right)}{6}\\ x=3\left(3-\sqrt{3}\right)\\ x=9-3\sqrt{3}\end{array}$

b.

To determine

Show that $x=4\sqrt[4]{3}$ for the right angle triangle having area equal to $24$ and sides $x,\sqrt{3}x\text{ and }2x$ respectively.

Explanation of Solution

Given information: It is given that the three sides of an right angle triangle are $x,\sqrt{3}x\text{ and }2x$ respectively and area of this triangle is $24$ .

Mathematically,

  $2x>\sqrt{3}x>x\mathrm{....}\left(2\right)$

  $\text{Area}\left(A\right)\text{ of triangle}=24$

Formula used: The area $\left(A\right)$ of a right angle triangle is given as,

  $\text{Area}\left(A\right)\text{ of triangle}=\frac{1}{2}\left(\text{base}\right)\left(\text{height}\right)$

Proof: According to question and apply formula for area of a right angle triangle.

  $\begin{array}{c}24=\frac{1}{2}\left(\sqrt{3}x\right)\left(x\right)\\ \text{or, }\sqrt{3}{x}^2=48\\ \text{or, }{x}^2=\frac{\left(16\right)\left(\sqrt{3}\right)\left(\sqrt{3}\right)}{\sqrt{3}}\mathrm{......}\left(\text{Factorized 48}\right)\\ x^2=16\sqrt{3}\\ x=4\sqrt[4]{3}\mathrm{.....}\left(\text{Proved}\right)\end{array}$