Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 21 May 2013, 11:19

# isoceles triangle -- find x

Author Message
TAGS:
Manager
Joined: 24 Nov 2010
Posts: 216
Location: United States (CA)
Concentration: Technology, Entrepreneurship
Schools: Ross '15, Duke '15
Followers: 2

Kudos [?]: 28 [0], given: 7

isoceles triangle -- find x [#permalink]  11 Sep 2011, 10:44
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)
Attachments

geometry.png [ 7.69 KiB | Viewed 1912 times ]

Manager
Joined: 24 Nov 2010
Posts: 216
Location: United States (CA)
Concentration: Technology, Entrepreneurship
Schools: Ross '15, Duke '15
Followers: 2

Kudos [?]: 28 [0], given: 7

Re: isoceles triangle -- find x [#permalink]  11 Sep 2011, 10:52
i understood the OE provided but can someone help me find out what's wrong with my approach described in the attached?
Attachments

geometry1.gif [ 10.75 KiB | Viewed 1903 times ]

Senior Manager
Joined: 03 Mar 2010
Posts: 461
Followers: 3

Kudos [?]: 74 [1] , given: 21

Re: isoceles triangle -- find x [#permalink]  11 Sep 2011, 16:02
1
KUDOS
How can you say A is the mid-point of DE? It is not the midpoint.

Posted from my mobile device
_________________

My dad once said to me: Son, nothing succeeds like success.

Manager
Joined: 24 Nov 2010
Posts: 216
Location: United States (CA)
Concentration: Technology, Entrepreneurship
Schools: Ross '15, Duke '15
Followers: 2

Kudos [?]: 28 [0], given: 7

Re: isoceles triangle -- find x [#permalink]  11 Sep 2011, 17:19
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!
Director
Joined: 01 Feb 2011
Posts: 792
Followers: 11

Kudos [?]: 62 [7] , given: 42

Re: isoceles triangle -- find x [#permalink]  11 Sep 2011, 18:40
7
KUDOS
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)

Manager
Status: Prepping for the last time....
Joined: 28 May 2010
Posts: 202
Location: Australia
Concentration: Technology, Strategy
GMAT 1: 630 Q47 V29
GPA: 3.2
Followers: 0

Kudos [?]: 8 [0], given: 21

Re: isoceles triangle -- find x [#permalink]  12 Sep 2011, 01:53
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)?
_________________

Two great challenges: 1. Guts to Fail and 2. Fear to Succeed

Intern
Joined: 06 Nov 2010
Posts: 23
Followers: 0

Kudos [?]: 1 [0], given: 16

Re: isoceles triangle -- find x [#permalink]  12 Sep 2011, 05:37

Sin 30 = Opp side/Hyp side.

1/2 = sq(2)/x

x = sq(2)*2
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 3109
Location: Pune, India
Followers: 568

Kudos [?]: 2002 [0], given: 92

Re: isoceles triangle -- find x [#permalink]  13 Sep 2011, 05:06
praveenvino wrote:

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?
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save 10% on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Manager
Joined: 25 Aug 2008
Posts: 238
Location: India
WE 1: 3.75 IT
WE 2: 1.0 IT
Followers: 2

Kudos [?]: 19 [0], given: 5

Re: isoceles triangle -- find x [#permalink]  13 Sep 2011, 10:35
I guess this cannot be solved without the use of Trigonometry!!

Spidy001, Thanks for the explanation!!
_________________

Cheers,
Varun

If you like my post, give me KUDOS!!

Manager
Joined: 06 Feb 2011
Posts: 87
GMAT 1: Q V
GMAT 2: Q V
GMAT 3: Q V
WE: Information Technology (Computer Software)
Followers: 0

Kudos [?]: 6 [0], given: 11

Re: isoceles triangle -- find x [#permalink]  15 Sep 2011, 13:37
Excellent post , Spidy ..learnt a lot !!
Senior Manager
Joined: 11 May 2011
Posts: 383
Location: US
Followers: 1

Kudos [?]: 46 [0], given: 46

Re: isoceles triangle -- find x [#permalink]  15 Sep 2011, 14:15
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

Cheers!
_________________

-----------------------------------------------------------------------------------------
What you do TODAY is important because you're exchanging a day of your life for it!
-----------------------------------------------------------------------------------------

Senior Manager
Status: MBAing!!!!
Joined: 24 Jun 2011
Posts: 314
Location: United States (FL)
Concentration: Finance, Real Estate
GPA: 3.65
WE: Project Management (Real Estate)
Followers: 5

Kudos [?]: 41 [0], given: 56

Re: isoceles triangle -- find x [#permalink]  16 Sep 2011, 11:56
B is the answer. it took me 4 minutes. Great Question and there is no need to use Cos and Sin.
Intern
Joined: 11 Sep 2011
Posts: 9
Followers: 0

Kudos [?]: 1 [0], given: 0

Re: isoceles triangle -- find x [#permalink]  27 Sep 2011, 10:58
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
Intern
Joined: 14 Oct 2011
Posts: 7
GMAT Date: 12-01-2011
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: isoceles triangle -- find x [#permalink]  16 Oct 2011, 14:25
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?
Senior Manager
Status: mba here i come!
Joined: 07 Aug 2011
Posts: 271
Location: Pakistan
Concentration: Strategy, Marketing
GMAT 1: 680 Q46 V37
GMAT 2: Q V
Followers: 13

Kudos [?]: 458 [0], given: 48

Re: isoceles triangle -- find x [#permalink]  21 Oct 2011, 02:12
\sqrt{2}/sin30 = \frac{x}{sin75}

sin30 = 1/2
sin75 = (\sqrt{6}+\sqrt{2})/4

solve for x
x = 1+\sqrt{3}
_________________

press +1 Kudos to appreciate posts

Re: isoceles triangle -- find x   [#permalink] 21 Oct 2011, 02:12
Similar topics Replies Last post
Similar
Topics:
If an isoceles right triangle has an area of 2x^2 + 2x + 11 01 Mar 2005, 03:53
Diagram of Triangle ABC Is triangle ABC an isoceles 7 19 May 2005, 21:05
The perimeter of a certain isoceles right triangle is 16 + 1 11 Sep 2006, 18:38
The perimeter of a certain isoceles right triangle is 16 and 14 26 Feb 2007, 12:14
Isoceles right triangle 3 08 Oct 2009, 21:37
Display posts from previous: Sort by