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In the diagram, what is the value of x? [#permalink ]
11 Sep 2011, 09:44

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61% (01:58) wrong

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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) OPEN DISCUSSION OF THIS QUESTION IS HERE: in-the-diagram-what-is-the-value-of-x-129962.html

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Re: isoceles triangle -- find x [#permalink ]
11 Sep 2011, 09:52

i understood the OE provided but can someone help me find out what's wrong with my approach described in the attached?

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Re: isoceles triangle -- find x [#permalink ]
11 Sep 2011, 15:02
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How can you say A is the mid-point of DE? It is not the midpoint.

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Re: isoceles triangle -- find x [#permalink ]
11 Sep 2011, 16:19

jamifahad wrote:

How can you say A is the mid-point of DE? It is not the midpoint.

Posted from my mobile device yeah, you are right. Angle bisector AB will not bisect the opposite side as DBE is a right triangle.

Thanks!

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Re: isoceles triangle -- find x [#permalink ]
11 Sep 2011, 17:40
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let the triangle be ABC where AB = AC = x and BC = sqrt(2) angle A of triangle ABC is 30 degrees. = > angle B = angle C = 75 (as its an isosceles triangle). draw a perpendicular BD from B to AC . this forms two right angle triangles ABD and BCD. ABD forms 30-60-90 triangle , where AB = x, AD = x * sqrt(3)/2 and BD = x/2 -------------equation 1 now consider BCD triangle, where BD = x/2 ( from equation 1) BC = 2 CD = x-(x*sqrt(3)/2) using Pythagorean theorem we have BC^2 = CD^2+BD^2 Solving this will give x^2 = 2/(2-sqrt(3)) = 2(2+sqrt(3)) = 4 + 2 sqrt(3) = (sqrt(3)^2) +1+ 2 * sqrt(3) = (1+sqrt(3)^2) => x = 1+sqrt(3) Answer is B.

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Re: isoceles triangle -- find x [#permalink ]
12 Sep 2011, 00:53

Great question. I took nearly 6 minutes to answer this. Similar procedure explained by Spidy..

Is there a best way make an educated 'correct' guess for these questions and move on (when there is no sufficient time to solve)?

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Re: isoceles triangle -- find x [#permalink ]
12 Sep 2011, 04:37

Please advise where i am wrong with this Sin 30 = Opp side/Hyp side. 1/2 = sq(2)/x x = sq(2)*2

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Re: isoceles triangle -- find x [#permalink ]
13 Sep 2011, 04:06
praveenvino wrote:

Please advise where i am wrong with this Sin 30 = Opp side/Hyp side. 1/2 = sq(2)/x x = sq(2)*2

The trigonometric relations hold for right triangles.

When you say Sin 30 = 1/2, you are assuming that there is a RIGHT angle there. That's how you get the hypotenuse. This triangle has no right angle yet. How did you decide that hypotenuse is x? Hypotenuse is the side opposite the 90 degrees angle. There is no angle that is 90 degrees here.

BTW, good explanation by Spidy001.

I am a little curious - what's the OE? Do they use trigonometry or something similar?

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Re: isoceles triangle -- find x [#permalink ]
13 Sep 2011, 09:35

I guess this cannot be solved without the use of Trigonometry!!

Spidy001, Thanks for the explanation!!

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Re: isoceles triangle -- find x [#permalink ]
15 Sep 2011, 12:37

Excellent post , Spidy ..learnt a lot !!

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Re: isoceles triangle -- find x [#permalink ]
15 Sep 2011, 13:15

Used my Trigonometry skills

The two isoscales base angles are 75 degrees each.

Cos (45+30) = Cos (45) Cos (30) - Sin(45) Sin(30)

Cos(75) = (\sqrt{3} - 1)/2 = 1/\sqrt{2}x

Hence, x = \sqrt{3}+1

Answer B.

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Re: isoceles triangle -- find x [#permalink ]
16 Sep 2011, 10:56

B is the answer. it took me 4 minutes. Great Question and there is no need to use Cos and Sin.

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Re: isoceles triangle -- find x [#permalink ]
27 Sep 2011, 09:58

raghavakumar85 wrote:

Great question. I took nearly 6 minutes to answer this. Similar procedure explained by Spidy.. Is there a best way make an educated 'correct' guess for these questions and move on (when there is no sufficient time to solve)?

i see,the best way to solve this is using cosine law or sin law.

2=2x^2-2x^2cos30

1=x^2(1-cos30)

x=1+sqrt3

or

x=sqrt2*sin75/sin30 =1+sqrt3

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Re: isoceles triangle -- find x [#permalink ]
16 Oct 2011, 13:25

arzad wrote:

B is the answer. it took me 4 minutes. Great Question and there is no need to use Cos and Sin.

How did you solve this without Cos and Sin?

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Re: isoceles triangle -- find x [#permalink ]
21 Oct 2011, 01:12

\sqrt{2}/sin30 = \frac{x}{sin75} sin30 = 1/2 sin75 = (\sqrt{6}+\sqrt{2})/4 solve for x

x = 1+\sqrt{3} _________________

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Re: In the diagram, what is the value of x? [#permalink ]
14 Nov 2013, 07:17

Re: In the diagram, what is the value of x?
[#permalink ]
14 Nov 2013, 07:17