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Perimeter of a triangle

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by didieravoaka » Mon Mar 07, 2016 1:28 pm
I got the answer right, but got confused in my calculations. Could you explain? Do we need to consider the notion of triangle 3-4-5 to get to the answer ?

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by Brent@GMATPrepNow » Mon Mar 07, 2016 1:31 pm
didieravoaka wrote:I got the answer right, but got confused in my calculations. Could you explain? Do we need to consider the notion of triangle 3-4-5 to get to the answer ?

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What question are you referring to?
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by didieravoaka » Mon Mar 07, 2016 1:37 pm
Sorry Brent, I forgot to attach the image. There it is...

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Can I have similar problems?

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by Brent@GMATPrepNow » Mon Mar 07, 2016 2:22 pm
There's no reason to believe that this is a right triangle, so we don't necessarily have to consider various Pythagorean triplets (like 3-4-5 or 5-12-13 triangles).

Instead, we need to use the distance formula to determine the length of each side.
Here's a video for finding the distance between two points: https://www.gmatprepnow.com/module/gmat ... /video/992

And here's a discussion of the question: https://www.beatthegmat.com/triangle-in- ... 13706.html

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by Matt@VeritasPrep » Thu Mar 17, 2016 9:10 pm
It might be easier to think of side, since two of them are the hypotenuses of right triangles and one of them is just a straight line from (0,0) to (7,0).

First side: (0,0) to (7,0) = 7

Second side: hypotenuse of (3,3) to (3,0) and (3,0) to (7,0). Since this is a 3-4-5 ∆, our side = 5.

Third side: hypotenuse of (0,0) to (3,0) and (3,0) to (3,3). This is a 45-45-90 ∆, so its sides are 3, 3, and 3√2, and our hypotenuse is 3√2.
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