Question

trapezoid abcd was dilated to create trapezoid abcd.
which statements are true about the trapezoids? select three options.
the length of side ad is 8 units.
the length of side ad is 4 units.
the image is larger than the pre - image.
sides cd and cd both have the same slope, 2.
the scale factor is \\(\frac{1}{2}\\).

Explanation:

Step1: Check AD length

Point A is at -4, D at 4. Length: \( 4 - (-4) = 8 \). So "AD is 8 units" is true.

Step2: Check A'D' length

A' at -2, D' at 2. Length: \( 2 - (-2) = 4 \). So "A'D' is 4 units" is true.

Step3: Check image size

Pre - image ABCD, image A'B'C'D'. A'D' (4) is half of AD (8), so image is smaller. "Image larger" is false.

Step4: Check slope of CD and C'D'

For CD: Let C be (2, y1), D be (4, 0). Wait, from graph, C is (2, 3)? Wait, B is (-2, 3), C is (2, 3), A is (-4, 0), D is (4, 0). For C'D': C' is (1, 1), D' is (2, 0)? Wait no, A' is (-2, 0)? Wait no, looking at the grid: A is at (-4, 0), D at (4, 0), so AD is from -4 to 4 on x - axis, y = 0. A' is at (-2, 0), D' at (2, 0), so A'D' is from -2 to 2 on x - axis, y = 0. For CD: C is (2, 3), D is (4, 0). Slope of CD: \(m=\frac{0 - 3}{4 - 2}=\frac{-3}{2}\). For C'D': C' is (1, 1), D' is (2, 0). Slope: \(m=\frac{0 - 1}{2 - 1}=-1\)? Wait, maybe I misread the y - coordinates. Wait, B is at (-2, 3), C at (2, 3), so B and C are on y = 3. A is (-4, 0), D is (4, 0). B' is at (-1, 1), C' is at (1, 1), A' is (-2, 0), D' is (2, 0). So for CD: from C(2, 3) to D(4, 0). Slope: \(\frac{0 - 3}{4 - 2}=\frac{-3}{2}\). For C'D': from C'(1, 1) to D'(2, 0). Slope: \(\frac{0 - 1}{2 - 1}=-1\). Wait, maybe my initial assumption is wrong. Wait, maybe B is at (-2, 2), C at (2, 2), A at (-4, 0), D at (4, 0). Then C is (2, 2), D is (4, 0). Slope of CD: \(\frac{0 - 2}{4 - 2}=\frac{-2}{2}=-1\). C' is (1, 1), D' is (2, 0). Slope of C'D': \(\frac{0 - 1}{2 - 1}=-1\). Oh, I see, I misread the y - coordinate of B and C. B and C are at y = 2. So C is (2, 2), D is (4, 0). Slope of CD: \(\frac{0 - 2}{4 - 2}=-1\). C' is (1, 1), D' is (2, 0). Slope of C'D': \(\frac{0 - 1}{2 - 1}=-1\). Wait, the option says slope 2, which is wrong. Wait, maybe the points are different. Wait, maybe B is at (-2, 4), C at (2, 4), A at (-4, 0), D at (4, 0). Then C is (2, 4), D is (4, 0). Slope of CD: \(\frac{0 - 4}{4 - 2}=-2\). C' is (1, 2), D' is (2, 0). Slope of C'D': \(\frac{0 - 2}{2 - 1}=-2\). Ah, that makes sense. So if B is at (-2, 4), C at (2, 4), A at (-4, 0), D at (4, 0), then B' is at (-1, 2), C' is at (1, 2), A' is (-2, 0), D' is (2, 0). Then slope of CD: from (2, 4) to (4, 0): \(m=\frac{0 - 4}{4 - 2}=\frac{-4}{2}=-2\)? Wait, no, slope formula is \(m = \frac{y_2 - y_1}{x_2 - x_1}\). If D is (4, 0) and C is (2, 4), then \(y_2 - y_1=0 - 4=-4\), \(x_2 - x_1 = 4 - 2 = 2\), so \(m=-2\). For C'D': C' is (1, 2), D' is (2, 0). \(y_2 - y_1=0 - 2=-2\), \(x_2 - x_1=2 - 1 = 1\), so \(m=-2\). Wait, the option says slope 2, but it's - 2, but maybe the question has a typo or I misread the direction. But the key is that the slope of CD and C'D' is the same (since dilation preserves slope). So if the slope is 2 (maybe absolute value or direction), but from calculation, if we take the rise over run as positive, maybe the points are different. Anyway, the main thing is that dilation preserves slope, so CD and C'D' have the same slope. Wait, but in our correct calculation, if C is (2, 3), D is (4, 0), slope is - 3/2, C' is (1, 1.5), D' is (2, 0), slope is - 1.5, which is the same as - 3/2 (since - 1.5=-3/2). Oh! I see, I misread C' and D' coordinates. If B is (-2, 3), C is (2, 3), so B and C are at y = 3. A is (-4, 0), D is (4, 0). Then A' is (-2, 0), D' is (2, 0), B' is (-1, 1.5), C' is (1, 1.5). Then for CD: from (2, 3) to (4, 0), slope is \(\frac{0 - 3}{4 - 2}=-\frac{3}{2}\). For C'D': from (1, 1.5) to (2, 0), slope is \(\frac{0 - 1.5}{2 - 1}=-\frac{3}{2}\). So the slope is \(-\frac{3}{2}\), but maybe the question considers the absolute value or a miscalculation, but the key is that CD and C'D' have the same slope. Wait, the option says "Sides CD and C'D' both have the same slope, 2". That's incorrect. Wait, maybe I made a mistake. Let's recast: AD is from (-4, 0) to (4, 0), length 8. A'D' is from (-2, 0) to (2, 0), length 4. So scale factor is \( \frac{4}{8}=\frac{1}{2}\). So "The scale factor is \(\frac{1}{2}\)" is true.

Step5: Check scale factor

AD length is 8, A'D' length is 4. Scale factor \(k=\frac{\text{length of image}}{\text{length of pre - image}}=\frac{4}{8}=\frac{1}{2}\). So "scale factor is \(\frac{1}{2}\)" is true.

Answer:

The length of side AD is 8 units, The length of side A'D' is 4 units, The scale factor is \(\frac{1}{2}\)