Lesson: Right Triangles

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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. However, when real world questions, we have to remember that the length of a side of a triangle cannot be negative. So, we must ignore the negative solution.

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 are not permitted in 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

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