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# In the diagram, what is the value of x?

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Intern
Joined: 22 Jan 2012
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In the diagram, what is the value of x?  [#permalink]

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Updated on: 09 Jul 2013, 14:16
17
00:00

Difficulty:

95% (hard)

Question Stats:

33% (02:32) correct 68% (02:41) wrong based on 185 sessions

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isofindx.jpg [ 7 KiB | Viewed 17254 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?

Originally posted by essarr on 31 Mar 2012, 12:06.
Last edited by Bunuel on 09 Jul 2013, 14:16, edited 1 time in total.
Edited the question.
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Joined: 02 Sep 2009
Posts: 52337
Re: find side length of isosceles triangle  [#permalink]

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31 Mar 2012, 13:16
7
5
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 [ 23.26 KiB | Viewed 17199 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 $$AD=AC-DC=x-\frac{x\sqrt{3}}{2}$$.

Next, from Pythagoras theorem $$AB^2=BD^2+AD^2$$ --> $$2=(\frac{x}{2})^2+(x-\frac{x\sqrt{3}}{2})^2$$ --> $$2=x^2(\frac{1}{4}+1-2*\frac{\sqrt{3}}{2}+\frac{3}{4})$$ --> $$2=x^2(2-\sqrt{3})$$ --> $$x^2=\frac{2}{2-\sqrt{3}}$$ --> rationalize: $$x^2=\frac{2*(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}=4+2\sqrt{3}$$ --> $$x^2=4+2\sqrt{3}=(\sqrt{3})^2+2\sqrt{3}+1=(\sqrt{3}+1)^2$$ --> $$x=\sqrt{3}+1$$.

Hope it's clear.
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Intern
Joined: 22 Jan 2012
Posts: 20
Re: find side length of isosceles triangle  [#permalink]

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31 Mar 2012, 14:36
1
Bunuel wrote:
In the diagram, what is the value of x?

...
--> $$x^2=4+2\sqrt{3}=(\sqrt{3})^2+2\sqrt{3}+1=(\sqrt{3}+1)^2$$ --> $$x=\sqrt{3}+1$$.

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..

thanks so much!
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Joined: 23 Jul 2013
Posts: 304
Find length x of triangle  [#permalink]

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14 Jan 2015, 14:57
1
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 13467 times ]

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Posts: 52337
Re: In the diagram, what is the value of x?  [#permalink]

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14 Jan 2015, 16:03
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!

Merging topics. please refer to the discussion above.

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

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15 Jan 2015, 01:28
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)
Intern
Joined: 17 Aug 2015
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In the diagram, what is the value of x?  [#permalink]

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04 Feb 2016, 17:14
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 $$AD=AC-DC=x-\frac{x\sqrt{3}}{2}$$.

Next, from Pythagoras theorem $$AB^2=BD^2+AD^2$$ --> $$2=(\frac{x}{2})^2+(x-\frac{x\sqrt{3}}{2})^2$$ --> $$2=x^2(\frac{1}{4}+1-2*\frac{\sqrt{3}}{2}+\frac{3}{4})$$ --> $$2=x^2(2-\sqrt{3})$$ --> $$x^2=\frac{2}{2-\sqrt{3}}$$ --> rationalize: $$x^2=\frac{2*(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}=4+2\sqrt{3}$$ --> $$x^2=4+2\sqrt{3}=(\sqrt{3})^2+2\sqrt{3}+1=(\sqrt{3}+1)^2$$ --> $$x=\sqrt{3}+1$$.

Hope it's clear.
Being naive here... How do you define Triangle BDC as 30 60 90 ? isn't the entire angle B 60?
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Re: In the diagram, what is the value of x?  [#permalink]

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04 Feb 2016, 21:18
1
1
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 = $$(\frac{1}{2})*y*\sqrt{2} = (\frac{1}{2})*(\frac{x}{2})*x$$
$$y = \frac{x^2}{2\sqrt{2}}$$

Now use the pythagorean theorem:

$$y^2 + (\frac{\sqrt{2}}{2})^2 = x^2$$

$$\frac{x^4}{8} + \frac{1}{2} = x^2$$

$$x^4 - 8x^2 + 4 = 0$$

Put $$x^2 = a$$

$$a^2 - 8a + 4 = 0$$

$$a = 4 + 2\sqrt{3}$$

$$x^2 = 3 + 1 + 2\sqrt{3} = \sqrt{3}^2 + 1 + 2\sqrt{3}$$

$$x = \sqrt{3} + 1$$
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Re: In the diagram, what is the value of x?  [#permalink]

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05 Feb 2016, 07:58
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 = $$(\frac{1}{2})*y*\sqrt{2} = (\frac{1}{2})*(\frac{x}{2})*x$$
$$y = \frac{x^2}{2\sqrt{2}}$$

Now use the pythagorean theorem:

$$y^2 + (\frac{\sqrt{2}}{2})^2 = x^2$$

$$\frac{x^4}{8} + \frac{1}{2} = x^2$$

$$x^4 - 8x^2 + 4 = 0$$

Put $$x^2 = a$$

$$a^2 - 8a + 4 = 0$$

$$a = 4 + 2\sqrt{3}$$

$$x^2 = 3 + 1 + 2\sqrt{3} = \sqrt{3}^2 + 1 + 2\sqrt{3}$$

$$x = \sqrt{3} + 1$$
Again sorry how do you substitute x= sq rt 2 when you initially define x corresponds to 2...coming back after a break..sorry I'm being rusty here!
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Re: In the diagram, what is the value of x?  [#permalink]

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07 Feb 2016, 09:38
1
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 = $$(\frac{1}{2})*y*\sqrt{2} = (\frac{1}{2})*(\frac{x}{2})*x$$
$$y = \frac{x^2}{2\sqrt{2}}$$

Now use the pythagorean theorem:

$$y^2 + (\frac{\sqrt{2}}{2})^2 = x^2$$

$$\frac{x^4}{8} + \frac{1}{2} = x^2$$

$$x^4 - 8x^2 + 4 = 0$$

Put $$x^2 = a$$

$$a^2 - 8a + 4 = 0$$

$$a = 4 + 2\sqrt{3}$$

$$x^2 = 3 + 1 + 2\sqrt{3} = \sqrt{3}^2 + 1 + 2\sqrt{3}$$

$$x = \sqrt{3} + 1$$
Again sorry how do you substitute x= sq rt 2 when you initially define x corresponds to 2...coming back after a break..sorry I'm being rusty here!

Never mind...got it
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Re: In the diagram, what is the value of x?  [#permalink]

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20 Feb 2017, 05:51
The hardest thing here is to rationalize like did Bunuel
I picked B just doing approx calculation...after I got 4+2(3)^1/2...
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In the diagram, what is the value of x?  [#permalink]

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15 Mar 2018, 08:21
Hi
A small formula might save u a lot of precious time in the real GMAT.
For a triangle with sides a,b and c and angles A, B and C the cosine rule can be written as: a^2 = b^2 + c^2 - 2bc cos A.
According to this
We get $$(\sqrt{2})^2$$ = $$x^2 +x^2 - 2X^2* Cos 30$$
$$2 = 2x^2-2x^2*(\sqrt{3}/2)$$
hence $$x^2 = 2/(2-\sqrt{3})$$
rationalizing we get $$x^2 = 4+2\sqrt{3}$$
or x = $$1+\sqrt{3}$$
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In the diagram, what is the value of x? &nbs [#permalink] 15 Mar 2018, 08:21
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