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Right Triangle

by manik11 » Tue Nov 03, 2015 4:27 am
Experts...could you guys please help me with this one.

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

OA : E
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by GMATGuruNY » Tue Nov 03, 2015 5:42 am
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.
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by akash singhal » Tue Nov 03, 2015 11:05 pm
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
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by conquistador » Fri Mar 25, 2016 7:24 am
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?
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by DavidG@VeritasPrep » Fri Mar 25, 2016 8:33 am
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.
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