What is the length of the hypotenuse of ΔABC ?

Is a hypotenuse present only in right triangles?
the measures could be 10-12-14

Hypotenuse is defined only for the right angled triangle.

Hypotenuse is the longest side of the right angled triangle.

When we say three consecutive even integers
2n, 2n+2, 2n+4 (longest side is the hypotenuse)

How do we solve it as per Pythagoras theorem and get to the definitive answer ?

Request you to please help !

Many thanks

Have you tried to solve this ?
This can be easily solved algebraically as given:

For statement 2, isn't the triplet (6, 8, 10) the only possible solution in which hypotenuse is 4 units more than its legs???

If yes, then the answer could be D.

nishant: I have the same doubt. Can anyone explain this?

Aren't you assuming that the lengths of the sides are integers? What if the sides are 9/8, 5 and 41/8?

Is it correct to conclude that only 3-4-5 is a valid condition for a right triangle w “consecutive integers” sides?

Posted from my mobile device

Yes. But notice that (1) says that "The lengths of the three sides of ΔABC are consecutive even integers." 6-8-10 triangle is the only one with side lengths of consecutive even integers.

Thanks Bunuel!
Yes, you are absolutely right. Here the main qualifier includes even integers.

I was thinking in case statement A said “consecutive integers”, that also would have been sufficient to infer the length of the hyp. Correct?

Posted from my mobile device

_______________
Correct.

Hello GMATBusters,

I was wondering whether you could help me on this one,

IF statement 2 said that the 3 sides must have an integer value , would then statement 2 be sufficient too? If yes how could we prove that there is only one triplet that fulfils the statement?

hi
You are right.

if the 3 sides are given to be integers in St2
The hypotenuse of ΔABC is 4 units longer than the shorter leg. , it would be sufficient to satisfy.

only (6,8,10) is satisfying it .

But if St2 had been sides are integers and Hypotenuse is 4 units more than one of the legs. then it would NOT be sufficient.

both (6,8,10) and (12, 16, 20) ( even more such triplets are there )satisfy it.

I have the same doubt. I request anyone to please explain why Statement 2 is insufficient. I mean, algebraically yes, it is insufficient but logically isn't this a sufficient condition?

Bunuel MartyTargetTestPrep JeffTargetTestPrep ScottTargetTestPrep

Statement 2 provides a relationship between only two sides, the hypotenuse and the shorter leg. There are an infinite number of triangles such that the difference in length between the hypotenuse and the shorter leg is 4.

You can think of it this way: 6, 8, 10 is the only triangle that fits that difference of 4 constraint and has side lengths that are integer values. However, by playing with the side lengths, we can see that other values are possible if we are not constrained to integer values.

For example, we could add 0.1 to 6 and 10 to have a triangle the lengths of whose shorter leg and hypotenuse are 6.1 and 10.1 respectively. We don't have to do the math to see that such a triangle is possible and that the length of the third side will be a decimal value just a little greater than 8.

We could add 0.2 and do the same thing.

So, Statement 2 is insufficient.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.