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Dilate quadrilateral ABCD using P as the center and the following scale factors:

IN RED: Scale factor of 1/2. Label the image A'B'C'D'.

IN BLUE: Scale factor of 2. Label the image A"B"C"D"

Dilate quadrilateral ABCD using P as the center and the following scale factors IN RED Scale factor of 12 Label the image ABCD IN BLUE Scale factor of 2 Label t class=


Answer :

semsee45

Answer:

A' (5.5, 4.5)

B' (8.5, 4.5)

C' (8.5, 1.5)

D' (5.5, 1.5)

A'' (1, 9)

B'' (13, 9)

C'' (13, -3)

D'' (1, -3)

Step-by-step explanation:

The vertices of the quadrilateral ABCD in the given diagram are:

  • A = (4, 6)
  • B = (10, 6)
  • C = (10, 0)
  • D = (4, 0)

The sides of quadrilateral ABCD are congruent, therefore quadrilateral ABCD is a square.

The x-coordinate of the center P is halfway between the x-coordinates of A and B, and the y-coordinate of the center P is halfway between the y-coordinates of A and D.  Therefore, the center of ABCD is:

  • P = (7, 3)

To dilate an image when the center of dilation is not the origin, observe the horizontal and vertical distances of each vertex from the center of dilation.

From inspection of the diagram:

  • A is 3 units to the left and 3 units above the center P.
  • B is 3 right to the left and 3 units above the center P.
  • C is 3 units to the right and 3 units below the center P.
  • D is 3 units to the left and 3 units below the center P.

Under a scale factor of 1/2, points A', B', C' and D' need to be half as far away from the center as A, B, C and D:

⇒ 3 × 1/2 = 1.5

Therefore:

  • A' is 1.5 units to the left and 1.5 units above the center P.
  • B' is 1.5 units to the right and 1.5 units above the center P.
  • C' is 1.5 units to the right and 1.5 units below the center P.
  • D' is 1.5 units to the left and 1.5 units below the center P.

So the vertices of A'B'C'D' are:

  • A' = (7-1.5, 3+1.5) = (5.5, 4.5)
  • B' = (7+1.5, 3+1.5) = (8.5, 4.5)
  • C' = (7+1.5, 3-1.5) = (8.5, 1.5)
  • D' = (7-1.5, 3-1.5) = (5.5, 1.5)

Similarly, under a scale factor of 2, points A'', B'', C'' and D'' need to be twice as far away from the center as A, B, C and D:

⇒ 3 × 2 = 6

Therefore:

  • A'' is 6 units to the left and 6 units above the center P.
  • B'' is 6 units to the right and 6 units above the center P.
  • C'' is 6 units to the right and 6 units below the center P.
  • D'' is 6 units to the left and 6 units below the center P.

So the vertices of A''B''C''D'' are:

  • A'' = (7-6, 3+6) = (1, 9)
  • B'' = (7+6, 3+6) = (13, 9)
  • C'' = (7+6, 3-6) = (13, -3)
  • D'' = (7-6, 3-6) = (1, -3)
View image semsee45

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