Chapter 3.2, Problem 86PS

Solution

To show that the hypotenuse of an isosceles right triangle with legs of length x is $x\sqrt{2}$ .

Given:

The isosceles right triangle with legs of length x .

Concept Used:

• Power of a power property: ${\left(x^a\right)}^b=x^{a\cdot b}$
• Product of power property: $a^m\cdot a^n=a^{m+n}$
• Quotient of power property: $\frac{a^m}{a^n}=a^{m-n}$
• Product property of radicals: $\sqrt[n]{a}\cdot \sqrt[n]{b}=\sqrt[n]{a\cdot b}$
• Quotient property of radicals: $\frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac{a}{b}}$
• The exponent form of the nth root of a number a is given by $\sqrt[n]{a}=a^{\frac{1}{n}}$

Calculation:

By the definition of Pythagorean theorem, the sum of the square of the legs of a right-triangle is equal to the square of the hypotenuse of the right-triangle.

So, using Pythagorean theorem in the given isosceles triangle, it follows

  $\begin{array}{c}{\left(\text{Hypotenuse}\right)}^2=x^2+x^2\\ {\left(\text{Hypotenuse}\right)}^2=\left(1+1\right)x^2\\ {\left(\text{Hypotenuse}\right)}^2=2x^2\\ \text{Hypotenuse}=\pm \sqrt{2x^2}\\ \text{Hypotenuse}=\pm x\sqrt{2}\end{array}$

Since the hypotenuse of the right-triangle cannot be negative, the length of the hypotenuse of the isosceles right-triangle is $x\sqrt{2}$ .