Right Triangle

manik11 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)?

15

57

93

109

121

The average test-taker will assume that 11 is the length of the hypotenuse.
DON'T be the average test-taker.
Test whether it's possible for 11 to be the length of a LEG.

Let a and 11 = the legs and let b = the hypotenuse.
Then:
a² + 11² = b²
b² - a² = 11²
(b+a)(b-a) = 121.

Implication:
Since a and b are INTEGER VALUES, (b+a) and (b-a) must constitute a FACTOR PAIR of 121.
Factor pairs of 121:
1*121
11*11.

The first factor pair would imply that b+a = 121 and that b-a = 1.
Adding together these two equations, we get:
(b+a) + (b-a) = 121 + 1
2b = 122
b = 61.
Since b=61 and b-a = 1, a=60.
Success!
a and b are both integers.
Thus, a+b = 60+61 = 121.

The correct answer is E.

GMATGuruNY wrote:

manik11 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)?

15

57

93

109

121

The average test-taker will assume that 11 is the length of the hypotenuse.
DON'T be the average test-taker.
Test whether it's possible for 11 to be the length of a LEG.

Let a and 11 = the legs and let b = the hypotenuse.
Then:
a² + 11² = b²
b² - a² = 11²
(b+a)(b-a) = 121.

Implication:
Since a and b are INTEGER VALUES, (b+a) and (b-a) must constitute a FACTOR PAIR of 121.
Factor pairs of 121:
1*121
11*11.

The first factor pair would imply that b+a = 121 and that b-a = 1.
Adding together these two equations, we get:
(b+a) + (b-a) = 121 + 1
2b = 122
b = 61.
Since b=61 and b-a = 1, a=60.
Success!
a and b are both integers.
Thus, a+b = 60+61 = 121.

The correct answer is E.

As it is clear that sum of squares of no two integers equal to the square of 11
Therefore either a^2 + 11^2 = b^2 or b^2 + 11^2 = a^2

Now next solving the equation we get
(a-b)(a+b)=121
or (a-b)= 121/(a+b)

Now, Since a and b are two integers their difference will also be an integer.
Thus from the options only E yields (a-b) as an integer
Hence E

GMATGuruNY wrote:

manik11 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)?

15

57

93

109

121

The average test-taker will assume that 11 is the length of the hypotenuse.
DON'T be the average test-taker.
Test whether it's possible for 11 to be the length of a LEG.

Let a and 11 = the legs and let b = the hypotenuse.
Then:
a² + 11² = b²
b² - a² = 11²
(b+a)(b-a) = 121.

Implication:
Since a and b are INTEGER VALUES, (b+a) and (b-a) must constitute a FACTOR PAIR of 121.
Factor pairs of 121:
1*121
11*11.

The first factor pair would imply that b+a = 121 and that b-a = 1.
Adding together these two equations, we get:
(b+a) + (b-a) = 121 + 1
2b = 122
b = 61.
Since b=61 and b-a = 1, a=60.
Success!
a and b are both integers.
Thus, a+b = 60+61 = 121.

The correct answer is E.

can you just explain why we cannot imagine 11 as hypotenuse?
Is there any solid reason?

can you just explain why we cannot imagine 11 as hypotenuse?
Is there any solid reason?

If 11 were the hypotenuse, then a^2 + b^2 = 11^2 or a^2 + b^2 = 121. We know that a and b are integers. So a^2 and b^2 are both perfect squares. List out the perfect squares less than 121: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100. So that's our sample space for potential values of a^2 and b^2. No two of those will sum to 121. If a and b are integer values, 11 cannot be the length of the hypotenuse, because a^2 + b^2 can never be 121.