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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

So if a and b both are integers, then a+b should also be integer. Now we are asked to find the value of (a+b)^2. I started POE. None of the values is a perfect square (which one square root result into a integer, which should be the value of base a+b) EXCEPT 121.

I am pretty sure it cant be that simple, what am i missing here

How do you figure that we are looking for (a + b)^2? We are looking for the value of (a + b) only.
_________________

Re: A right triangle has sides of a, b, and 11, respectively, [#permalink]

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24 Jan 2017, 07:07

VeritasPrepKarishma wrote:

neeraj609 wrote:

anceer wrote:

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

So if a and b both are integers, then a+b should also be integer. Now we are asked to find the value of (a+b)^2. I started POE. None of the values is a perfect square (which one square root result into a integer, which should be the value of base a+b) EXCEPT 121.

I am pretty sure it cant be that simple, what am i missing here

How do you figure that we are looking for (a + b)^2? We are looking for the value of (a + b) only.

Actually i reading (a+b)? wrongly as (a+b)^2. Thanks alot!

Re: A right triangle has sides of a, b, and 11, respectively, [#permalink]

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24 Jan 2017, 08:38

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