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

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

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18 Feb 2015, 04:30
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Difficulty:

45% (medium)

Question Stats:

70% (01:24) correct 30% (01:51) wrong based on 120 sessions

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gpp-vitac_img2.png [ 5.63 KiB | Viewed 2037 times ]
In the rectangle above, the ratio of (length):(height) = A, which is a number greater than one. Which of the following expresses the ratio of (diagonal):(length) in terms of A?

A. A

B. $$\sqrt{A^2-1}$$

C. $$\sqrt{A^2+1}$$

D. $$\sqrt{1 - \frac{1}{A^2}}$$

E. $$\sqrt{1 + \frac{1}{A^2}}$$

Kudos for a correct solution.

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Re: In the rectangle above, the ratio of (length):(height) = A, which is a  [#permalink]

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23 Oct 2017, 09:15
5
Applying Pythagorean Theorem:

(Diagonal)^2 = (Length)^2 + (Height)^2
Dividing both sides by (Length)^2
(Diagonal)^2 / (Length)^2 = 1 + [(Height)^2 / (Length)^2]

Since Length / Height = A
So Height / Length = 1/A and (Height)^2 / (Length)^2 = 1/A^2

So

(Diagonal)^2 / (Length)^2 = 1 + (1/A^2)

Taking the square root of both sides gives:

Diagonal / Length = √[1 + (1/A^2)]

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Re: In the rectangle above, the ratio of (length):(height) = A, which is a  [#permalink]

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18 Feb 2015, 05:52
1
It is given that, l/h = A (>1) - l =length of rectangle; and h = height of rectangle
Then diagonal d = sqrt(l^2 + h^2)

therefore, d/l = sqrt(l^2 + h^2)/l
= sqrt((l^2 + h^2)/l^2) ; taking denominator into square root
Diving by h^2 and considering l^2/h^2 = A^2 ,
we get = sqrt(A^2+1/A^2) = sqrt(1 + 1/A^2)

This is equivalent to (E).
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In the rectangle above, the ratio of (length):(height) = A, which is a  [#permalink]

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18 Feb 2015, 23:37
Attachment:

gpp-vitac_img2.png [ 5.84 KiB | Viewed 1707 times ]

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 $$= \frac{2\sqrt{a^2 + 1}}{2a} = \sqrt{1 + \frac{1}{a^2}}$$
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Re: In the rectangle above, the ratio of (length):(height) = A, which is a  [#permalink]

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18 Feb 2015, 23:51
Bunuel wrote:
Attachment:
gpp-vitac_img2.png
In the rectangle above, the ratio of (length):(height) = A, which is a number greater than one. Which of the following expresses the ratio of (diagonal):(length) in terms of A?

A. A

B. $$\sqrt{A^2-1}$$

C. $$\sqrt{A^2+1}$$

D. $$\sqrt{1 - \frac{1}{A^2}}$$

E. $$\sqrt{1 + \frac{1}{A^2}}$$

Kudos for a correct solution.

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)
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Re: In the rectangle above, the ratio of (length):(height) = A, which is a  [#permalink]

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22 Feb 2015, 12:25
Bunuel wrote:

In the rectangle above, the ratio of (length):(height) = A, which is a number greater than one. Which of the following expresses the ratio of (diagonal):(length) in terms of A?

A. A

B. $$\sqrt{A^2-1}$$

C. $$\sqrt{A^2+1}$$

D. $$\sqrt{1 - \frac{1}{A^2}}$$

E. $$\sqrt{1 + \frac{1}{A^2}}$$

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION

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

gpp-vitac_img9.png [ 540 Bytes | Viewed 1715 times ]

Divide both sides by L squared.
Attachment:

gpp-vitac_img10.png [ 850 Bytes | Viewed 1717 times ]

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

gpp-vitac_img11.png [ 859 Bytes | Viewed 1717 times ]

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

gpp-vitac_img12.png [ 1.15 KiB | Viewed 1718 times ]

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.
Attachment:

gpp-vitac_img13.png [ 14.63 KiB | Viewed 1718 times ]

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

- See more at: http://magoosh.com/gmat/2014/gmat-pract ... QEkpA.dpuf
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Re: In the rectangle above, the ratio of (length):(height) = A, which is a  [#permalink]

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22 Feb 2015, 22:53
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...

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.

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Rich
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Re: In the rectangle above, the ratio of (length):(height) = A, which is a &nbs [#permalink] 22 Feb 2015, 22:53
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