Question

the trapezoid ( pqrs ) is a dilation of the trapezoid ( pqrs ). what is the scale factor of the dilation?
simplify your answer and write it as a proper fraction, an improper fraction, or a whole number.

Explanation:

Step1: Find corresponding side lengths

Take side \(PP'\) and \(SS'\). Coordinates of \(P(8,8)\) and \(P'(6,6)\), \(S(- 8,8)\) and \(S'(-6,6)\).
The length of \(PP'\) (using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), but since \(y\) - values are same, for \(P(8,8)\) and \(P'(6,6)\), the horizontal distance (since dilation is uniform) from \(P\) to \(P'\) along \(x\) - axis: \(|8 - 6| = 2\), from \(S\) to \(S'\) along \(x\) - axis: \(|-8-(-6)|=2\).
Original length (for example, consider the horizontal length of the side of the original trapezoid \(PQRS\) from \(P(8,8)\) to \(S(-8,8)\)): \(|8-(-8)| = 16\).
Length of the corresponding side of the dilated trapezoid \(P'Q'R'S'\) (from \(P'(6,6)\) to \(S'(-6,6)\)): \(|6-(-6)| = 12\).
Another way: Use a single - point dilation. If we consider a point \(P(8,8)\) and its image \(P'(6,6)\).

Step2: Calculate the scale factor \(k\)

The scale factor \(k\) of a dilation is given by the formula \(k=\frac{\text{length of image segment}}{\text{length of original segment}}\).
If we use the \(x\) - coordinates (or \(y\) - coordinates since the dilation is uniform), for a point \((x,y)\) and its image \((x',y')\) after dilation, \(k=\frac{x'}{x}=\frac{y'}{y}\).
Take \(x = 8,y = 8\) (coordinates of \(P\)) and \(x'=6,y' = 6\) (coordinates of \(P'\)). Then \(k=\frac{6}{8}=\frac{3}{4}\).

Answer:

\(\frac{3}{4}\)