Question

Triangle CAB is the image of triangle CMN.
What point is the center of dilation?
What point is a pre-image point?
What point is an image point?
http://i.imgur.com/GgOX8aL.png


What is the length of segment VT?
ZP = 16
PR = 4
VX = 20
http://i.imgur.com/1zaKyZi.png

Find x, which is the same as length BA
http://i.imgur.com/hIebtn4.png

Answer 1

@ganeshie8 Could you help me?

Answer 2

for #1 :
center of dilation is the only invariant point under dilation - everything else expands/contracts around that point.

Answer 3

Look at the both triangles and see if u can figure out a point that was not moved even after dilation

Answer 4

I don't understand :(

Answer 5

invariant means fixed,
you just need to find out the point that stays fixed after dilation - thats the center of dilation

Answer 6

Notice that both triangles have a point C in common, and doesnt that look fixed ?

Answer 7

Yes, so Point C is the center of dilation?

Answer 8

Yep ! also you should observe that everything else is expand about that point C...

Answer 9

*expanding

Answer 10

for the remaining questions, use below theorem :
http://ceemrr.com/Geometry1/ParallelSimilar/ParallelSimilar4.html

Answer 11

So, would the image point would also be C

Answer 12

Yes ! C is fixed right there as it is the center of dilation
so image = preimage = C

Answer 13

N goes to B

Answer 14

and M goes to A

Answer 15

after dilation^

Answer 16

Okay could you show me how to solve the other 2 with the theorem?

Answer 17

setup a proportion :

\(\large \dfrac{ZP}{PR} = \dfrac{VX}{XT}\)

Answer 18

plugin the given values and solve \(XT\) first

Answer 19

16/4 = 20/?

Answer 20

how do i solve for XT
@jim_thompson5910 could you help me?

Answer 21

\(\large \dfrac{16}{4} = \dfrac{20}{XT}\)

\(\large \dfrac{4}{1} = \dfrac{20}{XT}\)

\(\large \dfrac{1}{1} = \dfrac{5}{XT}\)


\(\large XT = 5\)

Answer 22

\(\large VT = VX+XT = 20+5 = 25\)

Answer 23

you are an angle @ganeshie8