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In the diagram, what is the value of x?
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Updated on: 09 Jul 2013, 15:16
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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. \(2sqrt(3)\) 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 306090 triangle within the triangle, but I think I'm getting stuck with the algebra now?
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Originally posted by essarr on 31 Mar 2012, 13:06.
Last edited by Bunuel on 09 Jul 2013, 15:16, edited 1 time in total.
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Re: find side length of isosceles triangle
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31 Mar 2012, 14:16
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 306090 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 16705 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 \(AD=ACDC=x\frac{x\sqrt{3}}{2}\). Next, from Pythagoras theorem \(AB^2=BD^2+AD^2\) > \(2=(\frac{x}{2})^2+(x\frac{x\sqrt{3}}{2})^2\) > \(2=x^2(\frac{1}{4}+12*\frac{\sqrt{3}}{2}+\frac{3}{4})\) > \(2=x^2(2\sqrt{3})\) > \(x^2=\frac{2}{2\sqrt{3}}\) > rationalize: \(x^2=\frac{2*(2+\sqrt{3})}{(2\sqrt{3})(2+\sqrt{3})}=4+2\sqrt{3}\) > \(x^2=4+2\sqrt{3}=(\sqrt{3})^2+2\sqrt{3}+1=(\sqrt{3}+1)^2\) > \(x=\sqrt{3}+1\). Answer: B. Hope it's clear.
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Re: find side length of isosceles triangle
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31 Mar 2012, 15:36
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 firstyear calculus when we had to rationalize like this.. thanks so much!



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Find length x of triangle
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14 Jan 2015, 15:57
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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Re: In the diagram, what is the value of x?
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14 Jan 2015, 17:03
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. Also, please read carefully and follow: rulesforpostingpleasereadthisbeforeposting133935.html Pay attention to rules 1, 2, 3, and 6. Thank you.
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Re: In the diagram, what is the value of x?
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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)



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In the diagram, what is the value of x?
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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 306090 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 \(AD=ACDC=x\frac{x\sqrt{3}}{2}\). Next, from Pythagoras theorem \(AB^2=BD^2+AD^2\) > \(2=(\frac{x}{2})^2+(x\frac{x\sqrt{3}}{2})^2\) > \(2=x^2(\frac{1}{4}+12*\frac{\sqrt{3}}{2}+\frac{3}{4})\) > \(2=x^2(2\sqrt{3})\) > \(x^2=\frac{2}{2\sqrt{3}}\) > rationalize: \(x^2=\frac{2*(2+\sqrt{3})}{(2\sqrt{3})(2+\sqrt{3})}=4+2\sqrt{3}\) > \(x^2=4+2\sqrt{3}=(\sqrt{3})^2+2\sqrt{3}+1=(\sqrt{3}+1)^2\) > \(x=\sqrt{3}+1\). Answer: B. 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?
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04 Feb 2016, 22:18
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 306090 triangle within the triangle, but I think I'm getting stuck with the algebra now? Here is how I approached this question: Create a 306090 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?
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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 306090 triangle within the triangle, but I think I'm getting stuck with the algebra now? Here is how I approached this question: Create a 306090 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?
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07 Feb 2016, 10:38
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 306090 triangle within the triangle, but I think I'm getting stuck with the algebra now? Here is how I approached this question: Create a 306090 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?
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20 Feb 2017, 06:51
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?
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15 Mar 2018, 09:21
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 \((\sqrt{2})^2\) = \(x^2 +x^2  2X^2* Cos 30\) \(2 = 2x^22x^2*(\sqrt{3}/2)\) 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
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