∆ABC has vertices A(0, 0), B(2, 4), and C(4, 2). ∆RST has vertices R(0, 3), S(-1, 5), and T(-2, 4). Prove that ∆ABC ∼ ∆RST. (Hint: Graph ∆ABC and ∆RST in the coordinate plane.)

- anonymous

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- chestercat

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- dm91

@steve816
he can help you i think.

- steve816

Well, do you know how to plot points on the coordinate plane? All you do is draw the points and connect the dots.

- dm91

|dw:1443590087569:dw|

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## More answers

- anonymous

omg thank you

- steve816

No problem :)

- anonymous

so how do i do this problem?

- anonymous

@steve816

- dm91

ok so as steve put it do you have a graph?

- anonymous

ive already graphed the points but i dont know how to prove it similar

- steve816

If the triangles are the same size once you graphed, then they are similar.

- anonymous

i thought you had to do the distance formula?

- anonymous

they werent the same size

- steve816

Yes, you can use the distance formula

- steve816

For each of the sides

- anonymous

okay how can i do it with the distance formula?

- steve816

Plug in your (x, y) coordinates into this formula:
http://2.bp.blogspot.com/-GMrbM3fg4NU/VHljne7DleI/AAAAAAAABec/3emecyWG4ew/s1600/Slide1.jpg

- anonymous

ok give me a second

- anonymous

wait which numbers do i put into it?

- anonymous

i did AB and got 20 is that correct?

- anonymous

?

- anonymous

@steve816 is that correct?

- anonymous

hello?

- anonymous

i need help fast can someone help?

- dm91

ok now what?

- anonymous

is that correct?

- anonymous

i got the squareroot of 20 for AB

- dm91

yes

- anonymous

okay thanks

- dm91

yep

- anonymous

for rs i got the sqaureroot of 5 is that correct?

- dm91

yes

- anonymous

and for CT i got the sqareroot of 40 is that correct?

- dm91

yes

- anonymous

ok what do i do now?

- dm91

no that one would have to be up to Steve because he gave you the square root idea but i might be able to get you someone else that can help because Steve is helping other pp at the time.

- anonymous

okay thanks!

- dm91

@dan815

- dm91

yep

- dm91

@Koikkara

- Koikkara

Hai, how can i help you ?

- anonymous

i was doing the distance formula for the problem and dont know what to do next

- Koikkara

|dw:1443591908295:dw| According to your question..

- anonymous

yes

- anonymous

i did Ab and got the squareroot of 20 then i did RS and got the sqareroot of 5 then i did CT and got the squareroot of 40

- Koikkara

hmm... better, this user @dan815 can help you more clearly.

- zpupster

this is cheating a bit becasue this program gives the lentgh of the sides
but you need to use the SSS postulate and this is easier because 2 side length are the same
\[\frac{ BC}{ EF} = \frac{ AB }{DF} \]

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