Chapter 9.7, Problem 20WE

(a)

To determine

To write: equations for area and perimeter.

Answer to Problem 20WE

Equations for area and perimeter are $xy=1$ and $2\sqrt{x^2+y^2}+2x=4$ respectively

Explanation of Solution

Given:

No isosceles triangle has perimeter 4 and area 1.

Calculation:

Show that no isosceles triangle has perimeter 4 and area 1.

The figure of an isosceles triangle is as follows:

  

Write the equations for area and perimeter.

Given that

  $Base=2x$ and $height=y$

The area of a triangle is $\frac{1}{2}\times base\times height$ .

Since Base $BC=2x$ and height $AD=y$ ,

  $\begin{array}{l}Area=\frac{1}{2}\times base\times height\\ 1=\frac{1}{2}\times 2x\times y\left[Areaisgivenas1\right]\end{array}$

Reversing the left and right sides,

  $xy=1$

The perimeter of a triangle is the summation of the three sides of that triangle.

For the given triangle the perimeter is $\left(AB+BC+CA\right)$ .

Again the given triangle is isosceles. Hence,

  $AB=AC$

Therefore the equation for the perimeter of the given triangle is,

Perimeter = $AB+BC+CA$

  $4=2AB+BC\left[SinceAB=AC\right]$

Reversing the left and right sides,

  $2AB+BC=4\mathrm{......}\left(1\right)$

Again from the triangle, it is found that $\Delta ABD$ is a right angled triangle and its sides are $AB$ .

  $BD$ and $DA$ , Using Pythagorean Theorem

  $\begin{array}{l}AB=\sqrt{B{D}^2+D{A}^2}\\ =\sqrt{x^2+y^2}\left[SinceBD=xandDA=y\right]\end{array}$

From equation (1),

  $\begin{array}{l}2AB+BC=4\\ \boxed{2\sqrt{x^2+y^2}+2x=4}\left[BC=2x\right]\mathrm{.........}\left(2\right)\end{array}$

Conclusion:

Equations for area and perimeter are $xy=1$ and $2\sqrt{x^2+y^2}+2x=4$ respectively

(b)

To determine

To simplify: the perimeter equation

Answer to Problem 20WE

The required result is $y^2+4x-4=0$

Explanation of Solution

Calculation:

Simplify the perimeter equation.

From equation (2) the perimeter equation is,

  $\begin{array}{l}2\sqrt{x^2+y^2}+2x=4\\ 2\sqrt{x^2+y^2}=4-2x\left[Taking2xtotherightside\right]\\ \sqrt{x^2+y^2}=2-x\end{array}$

By squaring both sides of the equation,

  $\begin{array}{l}{x}^2+y^2=4-4x+x^2\\ y^2=4-4x\left[Cancelling{x}^{2}frombothsides\right]\\ y^2+4x-4=0\left[Taking4-4xtotheleftside\right]\end{array}$

Conclusion:

Therefore, the required result is $y^2+4x-4=0$

(c)

To determine

To graph: the system and explain the conclusion.

Answer to Problem 20WE

   There are no solutions for this value of $xandy$ .

Explanation of Solution

Calculation:

The graph for the two solutions $xy=1$ and $y^2+4x-4=0$ are as follows:

  

From the graph, it is found that the graph of the two equations is intersecting at one point in the third quadrant. In third quadrant the value of both $xandy$ are negative. But for a triangle all the three sides are positive. Therefore, there are no solutions for this value of $xandy$ .

Thus, it is concluded that for an isosceles triangle with perimeter 4 and area 1 there is no solution.

Conclusion:

There are no solutions for this value of $xandy$ .