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In the diagram, what is the value of x? [#permalink]

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31 Mar 2012, 13:06

16

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A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

38% (01:35) correct
63% (03:08) wrong based on 168 sessions

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Attachment:

isofindx.jpg [ 7 KiB | Viewed 13219 times ]

In the diagram, what is the value of x?

A. \(1+sqrt(2)\) B. \(1+sqrt(3)\) C. \(2sqrt(2)\) D. \(sqrt(2)+sqrt(3)\) E. \(2sqrt(3)\)

I know we aren't supposed to solve this through any use of trigonometry rules, but I'm not sure how to go about solving this within 2 minutes. I've tried dropping perpendicular from one of the vertices at the base to form a 30-60-90 triangle within the triangle, but I think I'm getting stuck with the algebra now?

A. \(1+sqrt(2)\) B. \(1+sqrt(3)\) C. \(2sqrt(2)\) D. \(sqrt(2)+sqrt(3)\) E. \(2sqrt(3)\)

I know we aren't supposed to solve this through any use of trigonometry rules, but I'm not sure how to go about solving this within 2 minutes. I've tried dropping perpendicular from one of the vertices at the base to form a 30-60-90 triangle within the triangle, but I think I'm getting stuck with the algebra now?

Rotate the diagram as shown below:

Attachment:

Ttriangle.png [ 23.26 KiB | Viewed 13169 times ]

Triangle BDC is a 30°-60°-90° right triangle. Now, in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio \(1 : \sqrt{3}: 2\), the leg opposite 30° (BD) corresponds with \(1\), thethe leg opposite 60° (DC) corresponds with \(\sqrt{3}\), and the hypotenuse \(BC=x\) corresponds with 2. So, since \(BC=x\) then \(BD=\frac{x}{2}\) and \(DC=\frac{x\sqrt{3}}{2}\).

Also notice that \(AD=AC-DC=x-\frac{x\sqrt{3}}{2}\).

I got up to the \(x^2 = 4 + 2sqrt(3)\)... lol, i don't think i'd ever be able to recognize the way you rationalized it after that in a test situation, although it does make perfect sense.. bringing back memories of first-year calculus when we had to rationalize like this..

Re: In the diagram, what is the value of x? [#permalink]

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15 Nov 2014, 09:28

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I was looking at the advanced quant book for mgmat and I came across this problem. In my head I figured that it is an isosceles triangle, with the remaining angles as 75 degrees. Since the 30 degree angle matches up with a √2 (~1.4), and 75 degrees is 2.5x bigger than the 30 degrees, then the triangle side X must be 2.5 (~1.4). That is around 3.5, which is answer E. The thing is, that isn't the answer and I am confused why that wouldn't work.

I appreciate the assistance!

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Screen Shot 2015-01-14 at 2.52.52 PM.png [ 31.34 KiB | Viewed 9446 times ]

I was looking at the advanced quant book for mgmat and I came across this problem. In my head I figured that it is an isosceles triangle, with the remaining angles as 75 degrees. Since the 30 degree angle matches up with a √2 (~1.4), and 75 degrees is 2.5x bigger than the 30 degrees, then the triangle side X must be 2.5 (~1.4). That is around 3.5, which is answer E. The thing is, that isn't the answer and I am confused why that wouldn't work.

I appreciate the assistance!

Merging topics. please refer to the discussion above.

Re: In the diagram, what is the value of x? [#permalink]

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15 Jan 2015, 02:28

TheLostOne wrote:

I was looking at the advanced quant book for mgmat and I came across this problem. In my head I figured that it is an isosceles triangle, with the remaining angles as 75 degrees. Since the 30 degree angle matches up with a √2 (~1.4), and 75 degrees is 2.5x bigger than the 30 degrees, then the triangle side X must be 2.5 (~1.4). That is around 3.5, which is answer E. The thing is, that isn't the answer and I am confused why that wouldn't work.

I appreciate the assistance!

I tackled this problem in the same way...

this was my approach....

as it's an isosceles triangle =

30 + a + a = 180 ---------> a = 75

Now area of triangle is given by

1/2(x*x) sin30 and 1/2(x* sqrt2) sin75

as area will be same

(x*x) sin30 = (x*sqrt2) sin75

sin 30 = 0.5 sin 75 < 1 (close to 0.9)

on solving

x < 2 * sqrt2

option A and Option B left

Safe pick is Option B (as 2*sqrt2 = 2.8 and 1 + sqrt3 = 2.73)

Re: In the diagram, what is the value of x? [#permalink]

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04 Feb 2016, 12:19

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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In the diagram, what is the value of x? [#permalink]

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04 Feb 2016, 18:14

Bunuel wrote:

essarr wrote:

In the diagram, what is the value of x?

A. \(1+sqrt(2)\) B. \(1+sqrt(3)\) C. \(2sqrt(2)\) D. \(sqrt(2)+sqrt(3)\) E. \(2sqrt(3)\)

I know we aren't supposed to solve this through any use of trigonometry rules, but I'm not sure how to go about solving this within 2 minutes. I've tried dropping perpendicular from one of the vertices at the base to form a 30-60-90 triangle within the triangle, but I think I'm getting stuck with the algebra now?

Rotate the diagram as shown below:

Attachment:

Ttriangle.png

Triangle BDC is a 30°-60°-90° right triangle. Now, in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio \(1 : \sqrt{3}: 2\), the leg opposite 30° (BD) corresponds with \(1\), thethe leg opposite 60° (DC) corresponds with \(\sqrt{3}\), and the hypotenuse \(BC=x\) corresponds with 2. So, since \(BC=x\) then \(BD=\frac{x}{2}\) and \(DC=\frac{x\sqrt{3}}{2}\).

Also notice that \(AD=AC-DC=x-\frac{x\sqrt{3}}{2}\).

A. \(1+sqrt(2)\) B. \(1+sqrt(3)\) C. \(2sqrt(2)\) D. \(sqrt(2)+sqrt(3)\) E. \(2sqrt(3)\)

I know we aren't supposed to solve this through any use of trigonometry rules, but I'm not sure how to go about solving this within 2 minutes. I've tried dropping perpendicular from one of the vertices at the base to form a 30-60-90 triangle within the triangle, but I think I'm getting stuck with the algebra now?

Here is how I approached this question: Create a 30-60-90 triangle by dropping an altitude from one of the base angles. The sides are in the ratio \(1:\sqrt{3}:2\). Since side x corresponds to 2 in the ratio, the length of the altitude will be x/2. Say the altitude from 30 degree angle to side \(\sqrt{2}\) is of length y.

Area of triangle = \((\frac{1}{2})*y*\sqrt{2} = (\frac{1}{2})*(\frac{x}{2})*x\) \(y = \frac{x^2}{2\sqrt{2}}\)

Re: In the diagram, what is the value of x? [#permalink]

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05 Feb 2016, 08:58

VeritasPrepKarishma wrote:

essarr wrote:

Attachment:

isofindx.jpg

In the diagram, what is the value of x?

A. \(1+sqrt(2)\) B. \(1+sqrt(3)\) C. \(2sqrt(2)\) D. \(sqrt(2)+sqrt(3)\) E. \(2sqrt(3)\)

I know we aren't supposed to solve this through any use of trigonometry rules, but I'm not sure how to go about solving this within 2 minutes. I've tried dropping perpendicular from one of the vertices at the base to form a 30-60-90 triangle within the triangle, but I think I'm getting stuck with the algebra now?

Here is how I approached this question: Create a 30-60-90 triangle by dropping an altitude from one of the base angles. The sides are in the ratio \(1:\sqrt{3}:2\). Since side x corresponds to 2 in the ratio, the length of the altitude will be x/2. Say the altitude from 30 degree angle to side \(\sqrt{2}\) is of length y.

Area of triangle = \((\frac{1}{2})*y*\sqrt{2} = (\frac{1}{2})*(\frac{x}{2})*x\) \(y = \frac{x^2}{2\sqrt{2}}\)

Re: In the diagram, what is the value of x? [#permalink]

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07 Feb 2016, 10:38

1

This post received KUDOS

vinny87 wrote:

VeritasPrepKarishma wrote:

essarr wrote:

Attachment:

isofindx.jpg

In the diagram, what is the value of x?

A. \(1+sqrt(2)\) B. \(1+sqrt(3)\) C. \(2sqrt(2)\) D. \(sqrt(2)+sqrt(3)\) E. \(2sqrt(3)\)

I know we aren't supposed to solve this through any use of trigonometry rules, but I'm not sure how to go about solving this within 2 minutes. I've tried dropping perpendicular from one of the vertices at the base to form a 30-60-90 triangle within the triangle, but I think I'm getting stuck with the algebra now?

Here is how I approached this question: Create a 30-60-90 triangle by dropping an altitude from one of the base angles. The sides are in the ratio \(1:\sqrt{3}:2\). Since side x corresponds to 2 in the ratio, the length of the altitude will be x/2. Say the altitude from 30 degree angle to side \(\sqrt{2}\) is of length y.

Area of triangle = \((\frac{1}{2})*y*\sqrt{2} = (\frac{1}{2})*(\frac{x}{2})*x\) \(y = \frac{x^2}{2\sqrt{2}}\)

Re: In the diagram, what is the value of x? [#permalink]

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19 Feb 2017, 03:33

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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