In the rectangle above, the ratio of (length):(height) = A, which is a

Answer = E. \(\sqrt{1 + \frac{1}{A^2}}\)

Let the length = 2a, then height = 2

Diagonal \(= \sqrt{4a^2 + 4} = 2\sqrt{a^2 + 1}\)

Ratio of diagonal to length \(= \frac{2\sqrt{a^2 + 1}}{2a} = \sqrt{1 + \frac{1}{a^2}}\)

Again, you can plug in values for A. Say A = 1 (The questions says that A is greater than 1 but assume it is infinitesimally greater than 1)
If length:height = 1:1, the diagonal will be \(\sqrt{2}\). So diagonal:length = \(\sqrt{2}\)
When you put A = 1, only (C) and (E) give \(\sqrt{2}\).

Put A = 2 which will give diagonal \(= \sqrt{2^2 + 1} = \sqrt{5}\). So diagonal:length = \(\sqrt{5}/2\)
If you put A = 2 in (C), it gives \(\sqrt{5}\) which is wrong. So answer must be (E)

MAGOOSH OFFICIAL SOLUTION

Algebraic Solution: Let length = L, height = H, and diagonal = D. Clearly, these three are related by the Pythagorean Theorem

Divide both sides by L squared.

The right side is the square of the answer we seek. The left side is the square of the reciprocal of A.

Take a square root of both sides; remember that we cannot distribute the square root across addition:

Answer = (E)

Numerical Solution: Let’s choose an easy 3-4-5 triangle for the height-length-diagonal. Then A = 4/3, and (diagonal):(length) = 5/4. We want to plug in A = 4/3 and get an answer of 5/4.

This choice of numbers eliminated four answers and leaves only one, so (E) must be the answer.

- See more at: https://magoosh.com/gmat/2014/gmat-pract ... QEkpA.dpuf

Hi All,

This question is perfect for TESTing VALUES. It also has a few fantastic Number Property shortcuts in the answer choices, which can save you a LOT of "calculation time" if you recognize them.

Given the right triangle with no values (the one restriction is that the length is GREATER than the height, we can use ANY right triangle that we choose. The obvious choice would be to use a 3/4/5 right triangle (length = 4, height = 3).

We're told that A = length/height, so A = 4/3. We're asked for the value of diagonal/length, so we're looking for 5/4 when we plug A = 4/3 into the answer choices.

Now, here's where the Number Property shortcuts come into play. Four of the answer choices use the square root sign - for the square root of a number to = 5/4, the number MUST be GREATER than 1. We can use that to our advantage to quickly eliminate some of the answers.

Answer A: 4/3 This is NOT a match

Answer B: Root[(4/3)^2 - 1] = Root[16/9 - 1] = Root[7/9]. Stop working. This is LESS than 1. This is NOT a match.

Answer C: Root[(4/3)^2 + 1] = Root[16/9 + 1] = Root[25/9] = 5/3. This is NOT a match.

Answer D: Root[1 - a fraction]. Stop working. This is going to be LESS than 1. This is NOT a match.

There's only 1 answer left, and we've eliminated the others...

Final Answer:

Here's the proof though:

Answer E: Root[1 + 1/(4/3)^2] = Root[1+ 9/16] = Root[25/16] = 5/4. This IS a match.

GMAT assassins aren't born, they're made,
Rich

Let the length be A so the height will be 1.

So the diagonal will be √(A^2 + 1)

So the ratio of the diagonal and the length will be √(A^2 + 1)/ A = √(A^2 + 1)/ √A^2 = √(1 + (1/ A^2))

Option: E