Rephrase = Find a perfect square

When the measurement is ODD

[(Number)^2]/2

(11)^2 = 121/2

Now, consider the INTEGER and INTEGER+1 Quotient

==> 60 & 61

SO, y = 60

Answer [spoiler]{A}[/spoiler]

A much easier way, IMHO:

11² + y² = (y+1)²

so

11² = (y+1)² - y²

so

11² = (y+1 + y) * (y+1 - y)

so

11² = (2y + 1) * 1

so

121 = 2y + 1, and y = 60.

Using Pythagoras theorem:
(y+1)^2 - y^2 = 11^2

Expanding the brackets...
y^2 + 2x + 1 - y^2 = 121

So 2x + 1 = 121

2x = 120

x = 60

Answer = (A) 60

gmattesttaker2 wrote:Hello,

Can you please assist with this:

A certain right triangle has integer sides; if the lengths of its sides are 11, y, and (y + 1), which of the following could be the value of y?

(A) 60
(B) 50
(C) 40
(D) 30
(E) 20

OA: A


Thanks,
Sri

The length of the longest side -- the hypotenuse -- must be the greatest of 11, y and y+1.
Since the answer choices -- which represent the value of y -- are all greater than 11, the length of the hypotenuse must be equal to y+1.
Applying the Pythagorean theorem, we get:
11² + y² = (y+1)²
(y+1)² - y² = 121.

We can plug in the answers for the value of y.
Remember the following identity:
a² - b² = (a+b)(a-b).

Answer choice C: y=40
41² - 40² = (41+40)(41-40) = 81.
The difference is TOO SMALL, implying that the value of y must be GREATER than 40.
Eliminate C, D and E.

Answer choice B: y=50
51² - 50² = (51+50)(51-50) = 101.
The difference is still too small.
Eliminate B.

The correct answer is A.

Answer choice A: y=60
61² - 60² = (61+60)(61-60) = 121.
Success!