Question

the parallelogram bcde is a dilation of the parallelogram bcde. 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 coordinates of corresponding points

First, identify coordinates of \( B \) and \( B' \). From the graph, \( B(-4, 4) \) and \( B'(-1, 1) \)? Wait, no, let's check again. Wait, the original parallelogram \( BCDE \): let's find length of \( BC \). \( B(-4, 4) \), \( C(4, 4) \), so length \( BC = 4 - (-4) = 8 \)? Wait, no, distance between \( x \)-coordinates: \( 4 - (-4) = 8 \)? Wait, no, \( B(-4,4) \), \( C(4,4) \), so horizontal distance is \( 4 - (-4) = 8 \)? Wait, no, the grid: each square is 1 unit. Wait, \( B \) is at \( (-4, 4) \), \( C \) is at \( (4, 4) \)? Wait, no, looking at the graph, \( B \) is at \( (-4, 4) \), \( C \) is at \( (4, 4) \)? Wait, no, the green parallelogram \( B'C'D'E' \): \( B'(-1, 1) \)? Wait, no, let's check the coordinates again. Wait, \( B' \) is at \( (-1, 1) \)? No, looking at the grid, \( B' \) is at \( (-1, 1) \)? Wait, no, the green points: \( B' \) is at \( (-1, 1) \)? Wait, no, the original pink parallelogram: \( B(-4, 4) \), \( C(4, 4) \), so length \( BC = 4 - (-4) = 8 \)? Wait, no, \( x \)-coordinate of \( B \) is -4, \( x \)-coordinate of \( C \) is 4, so distance is \( 4 - (-4) = 8 \). For the green parallelogram \( B'C'D'E' \): \( B'(-1, 1) \), \( C'(1, 1) \)? Wait, no, the graph shows \( B' \) at \( (-1, 1) \)? Wait, no, looking at the grid, \( B' \) is at \( (-1, 1) \), \( C' \) is at \( (1, 1) \)? Wait, no, the green parallelogram: \( B' \) is at \( (-1, 1) \), \( C' \) is at \( (1, 1) \), so length \( B'C' = 1 - (-1) = 2 \)? Wait, no, that can't be. Wait, maybe I made a mistake. Let's find the length of \( B'C' \) and \( BC \).

Wait, original \( BC \): \( B(-4, 4) \), \( C(4, 4) \), so length \( BC = 4 - (-4) = 8 \)? Wait, no, horizontal distance: \( 4 - (-4) = 8 \), so length \( BC = 8 \). For \( B'C' \): \( B'(-1, 1) \), \( C'(1, 1) \), so length \( B'C' = 1 - (-1) = 2 \)? Wait, no, that's not right. Wait, maybe the coordinates are different. Wait, looking at the graph, \( B \) is at \( (-4, 4) \), \( C \) is at \( (4, 4) \), so \( BC \) length is \( 8 \) (from \( x=-4 \) to \( x=4 \), 8 units). The green parallelogram \( B'C'D'E' \): \( B' \) is at \( (-1, 1) \), \( C' \) is at \( (1, 1) \), so \( B'C' \) length is \( 2 \) (from \( x=-1 \) to \( x=1 \), 2 units). Wait, but dilation scale factor is \( \frac{\text{length of image}}{\text{length of original}} \). Wait, but maybe I got the direction wrong. Wait, maybe the original is \( B'C'D'E' \) and the image is \( BCDE \)? No, the problem says \( B'C'D'E' \) is a dilation of \( BCDE \), so \( BCDE \) is the original, \( B'C'D'E' \) is the image. So scale factor \( k = \frac{\text{length of } B'C'}{\text{length of } BC} \).

Wait, let's recalculate. \( B(-4, 4) \), \( C(4, 4) \): distance between \( B \) and \( C \) is \( |4 - (-4)| = 8 \) (since y-coordinates are same, horizontal distance). \( B'(-1, 1) \), \( C'(1, 1) \): distance between \( B' \) and \( C' \) is \( |1 - (-1)| = 2 \). So scale factor \( k = \frac{2}{8} = \frac{1}{4} \)? Wait, no, that can't be. Wait, maybe I messed up the coordinates. Wait, let's check \( B \) and \( B' \) again. Wait, \( B \) is at \( (-4, 4) \), \( B' \) is at \( (-1, 1) \)? No, looking at the grid, \( B' \) is at \( (-1, 1) \)? Wait, no, the green parallelogram: \( B' \) is at \( (-1, 1) \), \( E' \) is at \( (-1, -1) \)? Wait, no, the green points: \( B' \) is at \( (-1, 1) \), \( C' \) is at \( (1, 1) \), \( D' \) is at \( (1, -1) \), \( E' \) is at \( (-1, -1) \). So length \( B'C' = 1 - (-1) = 2 \). Original \( BC \): \( B(-4, 4) \), \( C(4, 4) \), length \( 4 - (-4) = 8 \). So scale factor is \( \frac{2}{8} = \frac{1}{4} \)? Wait, no, that seems small. Wait, maybe the original is \( B'C'D'E' \) and the image is \( BCDE \), but the problem says \( B'C'D'E' \) is a dilation of \( BCDE \), so \( BCDE \) is pre-image, \( B'C'D'E' \) is image. So scale factor is \( \frac{\text{image length}}{\text{pre-image length}} \).

Wait, another approach: find the ratio of corresponding sides. Let's take \( B(-4, 4) \) and \( B'(-1, 1) \). The vector from origin? Wait, dilation center: is it the origin? Let's check if \( B' \) is \( \frac{1}{4} \) of \( B \): \( (-4 \times \frac{1}{4}, 4 \times \frac{1}{4}) = (-1, 1) \), which matches \( B' \). Similarly, \( C(4, 4) \): \( (4 \times \frac{1}{4}, 4 \times \frac{1}{4}) = (1, 1) \), which matches \( C' \). \( D(0, -8) \): \( (0 \times \frac{1}{4}, -8 \times \frac{1}{4}) = (0, -2) \), which matches \( D' \) (wait, \( D' \) is at \( (0, -2) \)? Wait, the green \( D' \) is at \( (0, -2) \)? Yes, looking at the graph, \( D' \) is at \( (0, -2) \), and \( D \) is at \( (0, -8) \). So \( D(0, -8) \), \( D'(0, -2) \): ratio is \( \frac{-2}{-8} = \frac{1}{4} \). So scale factor is \( \frac{1}{4} \)? Wait, no, that's not right. Wait, \( D \) is at \( (0, -8) \)? Wait, original \( D \) is at \( (0, -8) \)? Let's check the pink parallelogram: \( D \) is at \( (0, -8) \), \( E \) is at \( (-8, -8) \)? Wait, no, \( E \) is at \( (-8, -8) \)? Wait, \( B(-4, 4) \), \( E(-8, -8) \)? No, that can't be. Wait, maybe I misread the coordinates. Let's start over.

Original parallelogram \( BCDE \): let's find coordinates of \( B, C, D, E \). \( B(-4, 4) \), \( C(4, 4) \), \( D(0, -8) \), \( E(-8, -8) \)? No, that's not a parallelogram. Wait, no, in a parallelogram, opposite sides are equal and parallel. So \( BC \) is horizontal (y=4), so \( DE \) should also be horizontal. \( B(-4, 4) \), \( C(4, 4) \): length \( BC = 8 \) (from x=-4 to x=4). \( D(0, -8) \), \( E(-8, -8) \): length \( DE = 8 \) (from x=-8 to x=0). So that's a parallelogram. Now, the dilated parallelogram \( B'C'D'E' \): \( B'(-1, 1) \), \( C'(1, 1) \), \( D'(0, -2) \), \( E'(-2, -2) \). Now, length \( B'C' = 1 - (-1) = 2 \) (x from -1 to 1), length \( BC = 4 - (-4) = 8 \) (x from -4 to 4). So scale factor is \( \frac{B'C'}{BC} = \frac{2}{8} = \frac{1}{4} \). Alternatively, length \( D'E' = 0 - (-2) = 2 \) (x from -2 to 0), length \( DE = 0 - (-8) = 8 \) (x from -8 to 0). So scale factor \( \frac{2}{8} = \frac{1}{4} \). Also, vertical length: \( B \) to \( D \): from y=4 to y=-8, distance is \( 12 \)? No, \( B(-4,4) \) to \( D(0,-8) \): vertical change is \( -12 \), horizontal change is \( 4 \). \( B'(-1,1) \) to \( D'(0,-2) \): vertical change is \( -3 \), horizontal change is \( 1 \). So ratio of vertical change: \( \frac{-3}{-12} = \frac{1}{4} \), horizontal change: \( \frac{1}{4} \). So scale factor is \( \frac{1}{4} \). Wait, but that seems too small. Wait, maybe the original is \( B'C'D'E' \) and the image is \( BCDE \), but the problem says \( B'C'D'E' \) is a dilation of \( BCDE \), so \( BCDE \) is pre-image, \( B'C'D'E' \) is image. So scale factor is \( \frac{\text{image length}}{\text{pre-image length}} = \frac{2}{8} = \frac{1}{4} \).

Step2: Confirm with another side

Check \( D \) and \( D' \). \( D(0, -8) \), \( D'(0, -2) \). The length from \( D \) to \( C \): \( C(4,4) \), \( D(0,-8) \): distance is \( \sqrt{(4-0)^2 + (4 - (-8))^2} = \sqrt{16 + 144} = \sqrt{160} = 4\sqrt{10} \). \( D'(0,-2) \), \( C'(1,1) \): distance is \( \sqrt{(1-0)^2 + (1 - (-2))^2} = \sqrt{1 + 9} = \sqrt{10} \). So ratio is \( \frac{\sqrt{10}}{4\sqrt{10}} = \frac{1}{4} \). So scale factor is \( \frac{1}{4} \).

Wait, but that seems correct. So the scale factor is \( \frac{1}{4} \).

Answer:

\(\frac{1}{4}\)