Lesson: Right Triangles

Comment on Right Triangles

Your answer is incorrect on the first Pythagorean Theorem example. The square root of 100 is actually two solutions: +10 and -10, not just 10 as you stated.
gmat-admin's picture

That's true; there ARE two solutions to the equation x² = 100. However, when answering real world questions, we must remember that the length of a side of a triangle cannot be negative. As such, we must ignore the negative square root.

I don't get it. Why not just use a calculator to solve the square root? eg: if we have a triangle with a length of 8m and a height of 3m, to find out the hypotenuse we would just go [square root of 8^2+3^2]. And to solve an angle that's not the hypotenuse, you just go [square root of c^2-b^2]=a
gmat-admin's picture

Calculators aren't permitted on the quantitative section of the GMAT.

Hi Brent,

First of all, I really like your videos and explanation of each question.

I am little skeptical about this question.

Link : https://gmatclub.com/forum/right-triangle-abc-has-sides-with-length-x-y-and-z-if-triangle-abc-236889.html

Right triangle ABC has sides with length x, y and z. If triangle ABC has perimeter 17, and x² + y² + z² = 98, then what is the area of triangle ABC?

A) 12.75
B) 13.25
C) 14
D) 14.5
E) 15.25

I saw your response on gmatclub. My confusion is, how can we say for sure that z is the hypotenuse. Couldn't it be either x or y?
gmat-admin's picture

Link: https://gmatclub.com/forum/right-triangle-abc-has-sides-with-length-x-y-...

Since triangle ABC is a right triangle, we know that one of the lengths (x, y or z) is the length of the hypotenuse. Since the question is asking for area (and not asking for the specific value of one of the variables), I just let z = the length of the hypotenuse

I could have just as easily chosen x or y to be the length of the hypotenuse, and it wouldn't have changed the solution. Try it, and you'll see.

Here is where my confusion lies - you know X + Y = 10 since X + Y + Z = 17 and Z = 7. We also know that X^2 + Y^2 = Z^2 = 49. Why is that when I change X + Y = 10 to Y = 10 - X and then plug that into X^2 + Y^2 = 49, the variables cancel out and I end up with something along the lines of 100 = 49 which is impossible?

I understand I can square the first equation and make it into quadratic and then solve, but i do not understand why my method above fails. Any advice?
gmat-admin's picture

You're referring to this question: https://gmatclub.com/forum/right-triangle-abc-has-sides-with-length-x-y-...

I'd need to see your calculations to see where things went wrong.

Here are my calculations:

Take x² + y² = 49 and replace y with (x - 10)
We get: x² + (10 - x)² = 49
Expand: x² + 100 - 20x + x² = 49
Simplify: 2x² - 20x + 100 = 49
Rearrange: 2x² - 20x + 51 = 0
The solution to this is not pretty.

IMPORTANT: our goal here is to find the value of xy/2. So, rather than try to solve this awful equation, we should probably look for different approach.

Aside: We COULD solve the above equation for x and then use that x-value to determine the value of y. At that point, we could determine the value of xy/2. The problem is that the numbers are very awful to work with.

Can you please answer this question (with diagrams if possible): https://gmatclub.com/forum/a-certain-right-triangle-has-sides-of-length-x-y-and-z-107872.html
gmat-admin's picture

Hi Brent,

I was referring to Bunuel's solution to this Q : https://gmatclub.com/forum/in-the-figure-each-side-of-square-abcd-has-length-1-the-length-of-li-54152.html#p1034198

In problems such as these, I fail to come up with the first step, viz. extending EC to EO, making the diagram favorable for application of known properties and theorems. Its not that I don't know the Geometry rules and properties well. I know about special right triangles, properties of square such as "Diagonals bisect at 90 deg", "Diagonals bisect vertical angles". Yet its really frustrating for me to not be able to come up with the first step as I am easily able to solve the problem once I have the reconstructed drawing.

I am currently at Q49 and it is such lack of intuition that is preventing me from achieving Q50-51. Hence I am seeking your advice as you are an expert and must have helped students going through a similar concern. How do I build the intuition required to solve such problems? Or do I simply concede that such Qs are beyond my ability and move on?

Thanks & Regards,
gmat-admin's picture

Hi Abhirub,

How to build intuition. TOUGH question!

It's important to keep in mind that, there is a finite number of mathematical strategies (even outside-the-box strategies) that are applicable to solving GMAT math questions.

So, as you study more and more practice questions AND more solutions (reviewing solutions is key here), you will add more and more strategies to your mathematical toolbox.


Hi Brent,

Please see link below a question that I have not understood the process to solve it.

Thanks a lot for your help

Best Regards


gmat-admin's picture

Hi Fatima-Zahra,

Here's my full solution: https://gmatclub.com/forum/in-the-figure-above-what-is-the-area-of-trian...


Hi Brent,

Thanks for the video, this is helpful!
gmat-admin's picture

Thanks Lidiia!

Hi Brent,

Are questions like these possible to on GMAT official examination



gmat-admin's picture

- https://gmatclub.com/forum/rectangle-abcd-is-comprised-of-4-right-triang...
- https://gmatclub.com/forum/in-the-figure-bellow-the-rectangular-at-the-c...

I think both questions are within the scope of the GMAT.
The success rates (42% and 48%) and the average times both seem reasonable.
Having said that, both questions are very tricky, and only those test-takers performing REALLY WELL on the Quant section (e.g., Q49-51) would see them.

Hi brent can you please clarify why the answer to this question is not A.

gmat-admin's picture

Question link: https://gmatclub.com/forum/if-a-is-the-area-of-a-triangle-with-sides-of-...

Statement 1: (1) z = 13
A lot of students will conclude that, if the hypotenuse has length 13, then this must be a 5-12-13 right triangle.
However, that is just ONE possible right triangle with a hypotenuse of length 13.

It could also be the case that x = 10, y = √69 and z = 13, since those values satisfy the Pythagorean theorem: 10² + (√69)² = 13²
Or it could be the case that x = √80, y = √89 and z = 13, since those values satisfy the Pythagorean theorem: (√80)² + (√89)² = 13²
In fact there are infinitely many right triangles with a hypotenuse of length 13
For each of these right triangles, the area will be different.

For that reason, Statement 1 is not sufficient,

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