Lesson: Right Triangles

Comment on Right Triangles

The 2nd last ques in the "Reinforcement Activities" by greenlight prep, question - "Each edge of the above cube has length 1, If an ant walks from point A to point B along the OUTSIDE of the cube, what is shortest distance the ant must travel? "

Can't we use the formula AB = √(L² + W² + H²) ,as this is also used to find the shortest distance from A to point B in a cube. Plz explain
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/tricky-each-edge-of-the-above-cube-has-...

That formula applies to the direct distance between 2 opposite vertices.
So, if the ant could FLY in a straight line (through the INSIDE of the box) from A to B, then we could apply that formula.

However, we're told that the ant walks along the OUTSIDE of the cube. In other words, the ant is not travelling in a straight line directly from A to B

Does that help?

Cheers,
Brent

yes got it.
Thanks

Please explain why the answer is not square root 29

https://gre.myprepclub.com/forum/qotd-10-in-right-triangle-abc-the-ratio-of-the-lengths-of-2627.html
greenlight-admin's picture

https://gre.myprepclub.com/forum/qotd-10-in-right-triangle-abc-the-ratio-of-the-lengths-of-2627.html
Why did we introduce the variable x
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/qotd-10-in-right-triangle-abc-the-ratio...

Since we don't know the lengths of any sides of the triangle, we need to assign some variables so that we can create equations using the given information (e.g., the ratio of the lengths of the two legs is 2 to 5, and the area of triangle ABC is 20)

Does that help?

Cheers,
Brent

greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/qotd-10-in-right-triangle-abc-the-ratio...

Since we don't know the lengths of any sides of the triangle, we need to assign some variables so that we can create equations using the given information (e.g., the ratio of the lengths of the two legs is 2 to 5, and the area of triangle ABC is 20)

Does that help?

Cheers,
Brent

https://gre.myprepclub.com/forum/which-is-greater-n-2-or-p-2-m-12479.html

Shouldnt the sum of 2 sides be bigger than the 3rd side?
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/which-is-greater-n-2-or-p-2-m-12479.html

Yes, that rule still applies here. However, keep in mind that this question deals with the SQUARES of each length, not the actual lengths.

Cheers,
Brent

I had the same question...I took the square root of both sides and assumed that the 3rd side (n) cannot be greater than the sum of the other two sides (p and m)... This is annoying too because this is an 'official' question by the test makers! GRRR... They really are out to trick us...
greenlight-admin's picture

"They really are out to trick us"
I wholeheartedly agree!!!
AND they're very good at it!

https://gre.myprepclub.com/forum/o-is-the-center-of-the-circle-1800.html
Please can you explain this exercise a little more I don’t understand how ton get the value = 4 and how the perpendicular lines Help to find the radius.
Thank you
greenlight-admin's picture

https://gre.myprepclub.com/forum/the-area-of-triangle-pqr-or-the-area-of-triangle-psr-1866.html

Do you mind explaining this question further?
greenlight-admin's picture

Thank you for the quick response and excellent explanation!

https://gmatclub.com/forum/the-perimeter-of-a-certain-isosceles-right-triangle-is-39346.html

For this problem I got 2x + x√2 = 16 + 16√2.
I wanted to know why does it not work out to compare 2x to 16 and say x = 8 for instance? For this problem I would either back solve or factor out an x, but more than likely back solve given the answer choices.
greenlight-admin's picture

Question link: https://gmatclub.com/forum/the-perimeter-of-a-certain-isosceles-right-tr...
A lot of students will assume that 2x = 16 (i.e., x = 8)
However, if x = 8, then the perimeter (x + x + x√2) becomes 8 + 8 + 8√2, which simplifies to be 16 + 8√2, but the question tells us the perimeter is 16 + 16√2.
In this case, it's better to start with 2x + x√2 = 16 + 16√2 and go from there.

https://gre.myprepclub.com/forum/in-square-pqrs-above-t-is-the-midpoint-of-side-rs-if-pt-18877.html

Hey Brent can we use isosceles special properties for this i dont think its a good move?
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/in-square-pqrs-above-t-is-the-midpoint-...

No, we can't use any properties of isosceles triangles since there aren't any isosceles triangles in the diagram.

Or are you thinking of special triangle properties like those for 30-60-90 right triangles?
If so, that won't work either. Here's why:
Although triangle PST has one side that's twice as long as another side, this isn't a 30-60-90 triangle.
In a 30-60-90 triangle, the HYPOTENUSE is twice as long as the side opposite the 30 degree angle.
In triangle PST, one of the LEGS is twice as long as the other LEG.
So, triangle PST can't be a 30-60-90 triangle.

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