Chapter 3.2, Problem 15Q

Solution

The radical expression for the perimeter of the given triangle.

  $6\sqrt{5}+10$

Given:

The figure,

  

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:

In order to find the perimeter of the given triangle, first find the side lengths using the Pythagorean theorem and then find the sum of all sides.

Pythagorean theorem states that in a right-triangle the sum of the square of the legs is equal to the square of the hypotenuse. So, using this concept with the right triangle with legs 8 and 4, the lower side of the triangle is

  $\begin{array}{c}\sqrt{8^2+4^2}=\sqrt{64+16}\\ =\sqrt{80}\end{array}$

Similarly, the length of the side corresponding to the triangle with legs 2 and 4 is

  $\begin{array}{c}\sqrt{2^2+4^2}=\sqrt{4+16}\\ =\sqrt{20}\end{array}$

Note that the length of the legs of square corresponding to the third side of the triangle are 8 and 6, so the length of third side is

  $\begin{array}{c}\sqrt{8^2+6^2}=\sqrt{64+36}\\ =\sqrt{100}\\ =\sqrt{{10}^2}\\ =10\end{array}$

Thus, the perimeter of the triangle is

  $\begin{array}{c}P=\sqrt{80}+\sqrt{20}+10\\ =\sqrt{16\cdot 5}+\sqrt{4\cdot 5}+10\\ =\sqrt{16}\cdot \sqrt{5}+\sqrt{4}\cdot \sqrt{5}+10\\ =4\cdot \sqrt{5}+2\cdot \sqrt{5}+10\\ =6\sqrt{5}+10\end{array}$

Thus, the perimeter of the triangle is $6\sqrt{5}+10$ units.