A right triangle has sides of a, b, and 11, respectively,

Question

A right triangle has sides of a, b, and 11, respectively, where a and b are both integers. What is the value of (a + b)?

A. 15 
B. 57 
C. 93 
D. 109 
E. 121

MKS · Accepted Answer

Here is the explanation -

Having sides a, b, and 11 is a RIGHT angle triangle , Hence using Pythagoras theorem -
CASE 1 -> a*a + b*b = 11*11 (Assuming 11 is the longest side)
or CASE 2 -> 11*11 + b*b = a*a (Assuming a is the longest side)

Do not forget that we are also given all sides are integer.

Consider CASE 1 now - Can you think of any set of two integers (both lest than 11 ) such as
a*a + b*b = 11*11 = 121

we have perfect squares less than 121 as - 1, 4, 9, 16, 25, 36, 49, 64, 81, 100
If you add any of the above two you will never get sum as 121 .
Hence CASE 1 is VOID

Consider CASE 2 now -

11*11 + b*b = a*a
then 11*11 = a*a + b*b => 121 = (a+b)(a-b) - Equation 1

As we are given All sides are integer -> a, b are integers -> 'a+b' and 'a-b' are also integers.
Equation 1 states that 121 is [color=#0072bc]product of two integers and 121 is a very interesting number in the way that it has only two combination of
product of two integers
Combination 1 : a+b = 11 also a-b = 11 -> This case is invalid as it will result that one side of triangle b =0
Combination 2 : a+b = 121 and a-b=1 -> This case is Valid and gives us the answer that 'a+b' = 121
[/color]

Hence option E
Tricky question indeed ! 8-)

KarishmaB · Answer

The logic of "11 is not the hypotenuse" is quite simple. Note that all three sides of the right triangle are integers so the sides will form a pythagorean triplet. 
We know the first two pythagorean triplets are 3-4-5 and 5-12-13. If there were a pythagorean triplet such as a-b-11, we would know it as the second pythagorean triplet and 5-12-13 would be the third. So 11 will not be the hypotenuse but would be one of the legs. 
So a^2 + 11^2 = c^2
121 = (c+a)*(c-a)
There are only 2 ways to factorize 121 into pairs of 2 factors each: (1 and 121) , (11, 11)
Sum and difference of two integers cannot both be 11 so c+a = 121 and c-a = 1

Answer (E)

AV086 · Answer

It is debatable if a three sided figure with one zero length is a triangle. Is this question GMAT worthy?

Bunuel · Answer

The length must be greater than 0. The sides are {11, 60, 61}: each sides is greater than 0. Don't see any problem with this question.

bhatiavai · Answer

Veritas Prep Official Solution:

Since the first two right triangles with integer sides are the 3-4-5, the 6-8-10, and the 5-12-13, 11 cannot be the hypotenuse. Hence 11 is one of the legs, and our equation is  11^2 + a^2 = b^2.  Subtracting a^2 from both sides gives us  11^2 = b^2 - a^2 , or  121 = (b + a)(b - a) . Since b and a are both positive integers, (b + a) must = 121 and (b - a) must = 1. (The only other option, (b+a)=(b-a)=11 is impossible with positive a.) Hence  (b + a) = 121 .

GMATinsight · Answer

The trick for this question is

If the side of a right angle triangle is a prime number then other two sides will be

Second Side =  /2

Third Side =  /2

i.e. for One side = 11

Second Side = (11^2 -1)/2 = (121-1)/2 = 60
Third Side = (11^2 +1)/2 = (121+1)/2 = 61

I hope this helps!

I personally find it an UNSUITABLE question for GMAT... GMAT Doesn't expect such tricks from students.

IanStewart · Answer

First, if all you know is that the length of one side is prime, there are infinitely many possible lengths for the other sides. It's only if you know the other sides are integers that there might be a formula.

But you also mean to say "if the  shortest  side of a right triangle is prime, and the other sides are of integer length, the other sides will be..." If all you know is that one side is of length 5, say, using your formulas you'd think the other sides need to be 12 and 13. They could also be 3 and 4.

No one will ever need the formulas you posted in a GMAT question, and it would be pointless for a test taker to memorize them. But you certainly don't need those formulas here. You only need to know the difference of squares factorization, and that there is only one way to write 11^2 as a product of two distinct positive integers. The question is well within the scope of the GMAT.

GMATinsight · Answer

You seem of have less understanding of the word "OPINION"

I hope you prove me wrong by bringing ONE question from authentic source of GMAT questions if you have enough exposure of GMAT.(Refer OG, GMAT PREP etc.)

Another point, whether you take (3, 4, 5) or (5, 12, 13) it fit in both cases. and the forward application doesn't approve of backward application as well. (1 is a natural no. doesn't mean that a natural no. has to be 1)

And yes, It fit's to the cases when the smallest side is prime Number and the all Sides are of Integer length so I must admit you are good at finding faults and not as good in appreciating a new concept and an opinion.

IanStewart · Answer

I'm afraid I don't understand your reply. In your first post, you said that when "the side of a right angle triangle is a prime number then other two sides will be" x and y. That is, you said we can determine the other two sides from one side, when the length of that one side is prime. That's not true, but if a test taker believed it was, he or she might answer questions incorrectly on the GMAT - for example, a DS question like the following: If the lengths of the three sides of a right triangle are integers, what is the perimeter of the triangle? 1. The length of one side is 5 2. Two of the lengths are prime numbers A test taker who thinks that a single prime length is sufficient to determine the other two will think Statement 1 is sufficient, and will pick A, which is not the right answer - the answer is E, since the triangle might be a 3-4-5 or 5-12-13 triangle. I will always post a clarification if an expert here posts something which might be misinterpreted, and which might lead a test taker to answer test questions incorrectly, and I'd always be grateful to anyone who posts a clarification of one of my posts if I were ever to make an error. I also don't understand why you think I confused your statement with its converse (what you're suggesting in your comment about natural numbers). I did not. Nor do I understand why you consider it relevant that the 3-4-5 triangle fits your formula, since we don't find the lengths 3 and 4 by plugging in '5', but regardless there are other triangles with a prime length that do not satisfy those formulas - the 8-15-17 triangle, for example. I don't think it often is a matter of opinion whether a question falls within the scope of the GMAT. It's not clear to me precisely why you think this question is out of the scope of the test. I agree with you when you say, referring to your formulas, that "GMAT Doesn't expect such tricks from students." That seemed to me the only reason you described this question as "unsuitable" for the test. But those formulas aren't at all necessary here, so I disagreed with your initial premise. The main mathematical issue in this question, when solving it using the difference of squares factorization, is recognizing that when b > c > 0, and b and c are positive integers, the equation (b+c)(b-c) = 121 has only one possible solution. If you want to see an official question that relies on very similar logic, see this one from GMATFocus: if-x-and-y-are-integers-such-that-x-y-0-what-is-x-156464.html

bazu · Answer

Hi, could you then please elaborate a bit more on the solution proposed above? How can we reach the conclusion that 11 is not the hipotenuse?

subrataroy0210 · Answer

A right triangle has sides of a, b, and 11, respectively, where a and b are both integers. What is the value of (a + b)?

Case 1 :  a^2 + b^2 = 11^2 = 121  (unit digit 1)

Perfect squares end with any one of the following: 0; 1; 4; 9; 6; 5.
To have unit digit 1, we have to use either of the combos: 0 and 1; 6 and 5. But none satisfies.   X

Case 2 :  a^2 + 121 = b^2

Subcase (a) 
=>  a^2 = b^2 - 121 
=>  a^2 = (b+11) (b-11) 
Since,  (b+11)  is not equal to  (b-11) and also  (b+11) > (b-11) therefore 
 b-11 = 1  and  b+11 = a^2 
 b=12  and  a^2=23  (a is not a integer)   X

Subcase (b)

=>  121 = b^2 - a^2 
=>  (b-a) (b+a) = 121 
Therefore,  b-a=1  and  b+a=121

Purnank · Answer

Tricky question,

And there is one shortcut i can think of

Pythagorian triplets are always in the form of n, (n^2 - 1)/2, (n^2 + 1)/2.

you will notice that you cant place 11 anywhere else other than n (because only integers value exist)

So, a + b = n^2 i. e. 11^2 = 121.

Hence E.

A right triangle has sides of a, b, and 11, respectively,

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