Chapter 7.3, Problem 106E

To determine

To write: a system of linear equations to find an equation of the form $y=a{x}^2+bx+c$ that represents the relationship between x and y for the three polygons.

Answer to Problem 106E

The equation of the polygons is, $y=\frac{1}{2}{x}^2-\frac{1}{2}x.$

Explanation of Solution

Given information: The number of sides x and the combined number of sides and

diagonals y for each of three regular polygons are (x=3, y=3) for triangle, (x=4, y=6) for square, (x=5, y=10) for pentagon.

Calculation:

Consider the given equation that represents the relationship between number of sides and combined number of sides and diagonal of the figures.

  $y=a{x}^2+bx+c$ ...... (1)

For triangle, substitute 3 for x and 3 for y in equation (1),

  $\begin{array}{l}$ a{(3)}^2+3b+c=3$9a+3b+c=\mathrm{3................}(2)\end{array}$

For square, substitute 4 for x and 6 for y in equation (1),

  $\begin{array}{l}$ a{(4)}^2+4b+c=6$16a+4b+c=\mathrm{6................}(3)\end{array}$

For pentagon, substitute 5 for x and 10 for y in equation (1),

  $\begin{array}{l}$ a{(5)}^2+5b+c=10$25a+5b+c=\mathrm{10................}(4)\end{array}$

Solving equation (2),(3) and (4), get

  $\begin{array}{c}a=\frac{1}{2}.\\ b=\frac{-1}{2}.\\ c=0.\end{array}$

Substitute the value of a, b, c in equation (1), and the equation of the polygons is,

  $\begin{array}{l}y=\frac{1}{2}{x}^2-\frac{1}{2}x+0=\frac{1}{2}{x}^2-\frac{1}{2}x\\ y=\frac{1}{2}{x}^2-\frac{1}{2}x.\end{array}$