Chapter 6.5, Problem 35WE

To determine

If the perimeter of a right isosceles triangle is 10 and three sides are $x,x\text{ and }\sqrt{2}x$ respectively then evaluate $\left(x\right)$ .

Answer to Problem 35WE

The value of $\left(x\right)$ is equal to $5\left(2-\sqrt{2}\right)$ .

Explanation of Solution

Given information: It is given that the three sides of an isosceles triangle are $x,x\text{ and }\sqrt{2}x$ respectively and perimeter of this triangle is $10$ .

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

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

  $\begin{array}{c}x+x+\sqrt{2}x=10\text{ gives,}\\ \left(2+\sqrt{2}\right)x=10\\ x=\frac{10}{2+\sqrt{2}}\\ x=\left(\frac{10}{2+\sqrt{2}}\right)\left(\frac{2-\sqrt{2}}{2-\sqrt{2}}\right)\mathrm{......}\left(\text{Rationalized}\right)\end{array}$

Further solving,

  $\begin{array}{c}x=\frac{10\left(2-\sqrt{2}\right)}{{\left(2\right)}^2-{\left(\sqrt{2}\right)}^2}\\ =\frac{10\left(2-\sqrt{2}\right)}{4-2}\\ =\frac{10\left(2-\sqrt{2}\right)}{2}\\ =5\left(2-\sqrt{2}\right)\end{array}$

To determine

If the area of a right isosceles triangle is 12 and three sides are $x,x\text{ and }\sqrt{2}x$ respectively then evaluate $\left(x\right)$ .

Answer to Problem 35WE

The value of $\left(x\right)$ is equal to $2\sqrt{6}$ .

Explanation of Solution

Given information: It is given that the three sides of an isosceles triangle are $x,x\text{ and }\sqrt{2}x$ respectively and area of this triangle is $12$ .

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

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

Calculation: According to question and apply formula for area of an isosceles triangle.

  $\begin{array}{c}12=\frac{1}{2}\left(x\right)\left(x\right)\mathrm{......}\left(\begin{array}{l}\because \text{}\text{ Two equal}\\ \text{sides are }x,x\end{array}\right)\\ \text{or, }{x}^2=24\\ x=\pm \sqrt{24}\\ x=\pm 2\sqrt{6}\\ x=2\sqrt{6}\mathrm{......}\left(\begin{array}{l}\text{Neglect negative value, since}\\ \text{ side can not be negative}\end{array}\right)\end{array}$