Chapter 8.1, Problem 68PFA

To determine

To calculate the third side of a triangle.

Answer to Problem 68PFA

  $x^2+2x-3$

Explanation of Solution

Given:

Perimeter of triangle: $3x^2+6x-9$

Sum of two sides: $2x^2+4x-6$

Formula used:

Perimeter of any triangle is the sum of the sides.

  $P=x+y+z$

Where x , y and z are the sides of the triangle.

Calculation:

Let X be the length of the third side.

Using the perimeter of triangle,

  $P=x+y+z$

The perimeter of triangle is $3x^2+6x-9$

  $\Rightarrow P=3x^2+6x-9$

The sum of two sides is $2x^2+4x-6$

  $\Rightarrow y+z=2x^2+4x-6$

Substituting the value of the perimeter and the sum of the sides of the triangle,

  $3x^2+6x-9=X+(2x^2+4x-6)$

Subtracting $2x^2+4x-6$ on both sides,

  $\begin{array}{l}(3x^2+6x-9)-(2x^2+4x-6)=X+(2x^2+4x-6)-(2x^2+4x-6)\\ (3x^2+6x-9)-(2x^2+4x-6)=X\end{array}$

Combining the like terms,

  $\begin{array}{l}3x^2+6x-9-2x^2-4x+6=X\\ (3x^2-2x^2)+(6x-4x)+(6-9)=X\\ x^2+2x-3=X\end{array}$

Conclusion:

The length of the third side is $x^2+2x-3.$