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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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Difficulty:   95% (hard)

Question Stats: 33% (02:26) correct 67% (02:38) wrong based on 148 sessions

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Attachment: isofindx.jpg [ 7 KiB | Viewed 19025 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.
Math Expert V
Joined: 02 Sep 2009
Posts: 61508
Re: find side length of isosceles triangle  [#permalink]

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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 19004 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: 16
Re: find side length of isosceles triangle  [#permalink]

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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!
Senior Manager  Joined: 23 Jul 2013
Posts: 302
Find length x of triangle  [#permalink]

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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!
Attachments Screen Shot 2015-01-14 at 2.52.52 PM.png [ 31.34 KiB | Viewed 15279 times ]

Math Expert V
Joined: 02 Sep 2009
Posts: 61508
Re: In the diagram, what is the value of x?  [#permalink]

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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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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
Posts: 6
In the diagram, what is the value of x?  [#permalink]

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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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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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Intern  Joined: 17 Aug 2015
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Re: In the diagram, what is the value of x?  [#permalink]

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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!
Intern  Joined: 17 Aug 2015
Posts: 6
Re: In the diagram, what is the value of x?  [#permalink]

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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 Manager  S
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GMAT 1: 620 Q46 V29 Re: In the diagram, what is the value of x?  [#permalink]

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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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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 = $$1+\sqrt{3}$$
Attachments sine-and-cosine-rule-19-638.jpg [ 35.4 KiB | Viewed 4925 times ]

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

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The MGMAT book also talks about estimating in a problem like this. Assume the diagram is drawn with the altitude added as above to make a 30:60:90 triangle:

√2 ≈ 1.4 (a bit more than) Now, obviously the perpendicular line is a bit less than this. However, if we assume that its length is also 1.4 then we can use the 30:60:90 triangle ratios right away and eliminate some answers.

If 1.4 = s (angle with 30) then x = 2s, or 2.8 So we know our maximum possible value is a bit less than this. Right away we can eliminate C, D and E.

C) 2*1.4 = 2.8
D) 1.4 + 1.7 = 3.1
E) 2*1.7 = 3.4

We're left with A and B.
A) 1+1.4 = 2.4
B) 1+1.7 = 2.7

Here I would just guess and move on, we assumed the line is just a bit less than 1.4 so B) is closest. You can also note that the answers vary by .3 except for B and C which can potentially be a hint (or not...) Re: In the diagram, what is the value of x?   [#permalink] 17 Apr 2019, 08:02
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