Question

>>> http://prntscr.com/1za3dl

(See image)

Erhm. I've drawn a blank! (BTW! This is a PRACTICE TEST!) I just want to assess everything before I take the real test.

Answer 1

@Luigi0210 @GodIsMySavior @aaronq

Answer 2

@MattGibson @BulletWithButterflyWings @eashmore

Answer 3

Someone told me this " Think of AD as the new FI. Everything is getting 3 times bigger. The old base was 4 so the new base will be 12, so go 12 unit directly to the right from D and you come to the end corner at the right of the new base, and that point is (6,-5)."

BUT HOW! do we know it got 3 times bigger? Curious.

Answer 4

This is a similar shapes problem. Recall that a triangle with sides 3,4, and 5 will be similar to a triangle with sides 6, 8, and 10. The larger triangle is TWICE as big as the smaller triangle because (3*2=6).

The same principle applies here. Notice that F is three units above I.
If we look at the new points (D and A), we see that A is 9 units above D. We can establish the ratio of scaling as \[R = 9/3\]

Nine divided by three is equal to three. Therefore, the enlarged parallelogram is three times larger than the original.

Answer 5

So (-6, 5)

Answer 6

H is four units to the right of I. Because we have a scaling of 3, we need to move 12 units to the right of either A or D.

If we move 12 units to the right of D, we reach the point (6,-5).

If we move 12 units to the right of A, we reach a points of (12,4).

Answer 7

I believe it's the first one. If we move 12 units to the right of D, that is.

Answer 8

That is correct. Moving to the right of A isn't an answer choice.

Answer 9

@Compassionate