Chapter 7.3, Problem 8QR

To determine

To find: The formula to the area of the plane region in terms of the single variable $x$ .

Answer to Problem 8QR

The answer: Area $=\frac{x^2\sqrt{15}}{4}$

Explanation of Solution

Given information:

An isosceles triangle with two sides of length $2x$ and one side of length is $x$ .

Calculation:

Let consider the triangle $ABC$ that satisfies the given condition. The triangle's height which corresponds to the side $AB$ divide the side $AB$ into two segments of the same length $A{C}^{\prime }$ and $C^{\prime }B$ .

  

Finally,

Area $=\frac{C{C}^{\prime }\cdot AB}{2}$

Know that,

  $AC=2x$

And,

  $A{C}^{\prime }=\frac{x}{2}$

To use the Pythagorean theorem:

  $\begin{array}{c}{\left(C{C}^{\prime }\right)}^2={(}^A-{\left(A{C}^{\prime }\right)}^2\\ ={(}^2-{\left(\frac{x}{2}\right)}^2\\ =4x^2-\frac{x^2}{4}\\ =\frac{15x^2}{4}\end{array}$

Therefore,

The length of $C{C}^{\prime }\text{ =}\frac{x\sqrt{15}}{2}$

know the length of $AB$ and $C{C}^{\prime }$ , to use the formula $(1)$ obtain that the area is represented as:

Area $=\frac{x^2\sqrt{15}}{4}$

Therefore, the required formula to the area of the plane region in terms of the single variable $x$ is $\frac{x^2\sqrt{15}}{4}$ .