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10 Feb 2016, 11:20
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
53% (03:04) correct
47%
(02:53)
wrong
based on 250
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Not Attempted Yet
In triangle ABC, if angle ABC is 30 degrees, \(AC = 2*\sqrt 2\) and AB = BC = X, what is the value of X?
(A) \(\sqrt 3 – 1\)
(B) \(\sqrt 3 + 2\)
(C) \(\frac{(\sqrt 3 – 1)}{2}\)
(D) \(\frac{(\sqrt 3 + 1)}{2}\)
(E) \(2*(\sqrt 3 + 1)\)
(A) \(\sqrt 3 – 1\)
(B) \(\sqrt 3 + 2\)
(C) \(\frac{(\sqrt 3 – 1)}{2}\)
(D) \(\frac{(\sqrt 3 + 1)}{2}\)
(E) \(2*(\sqrt 3 + 1)\)
12 Feb 2016, 07:55
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Bunuel
Hi,
Just saw some discussion on this Q, I too will put in my bit to it..
Most of the Qs can be answered by elimination and standard methods...
1) POE
So A straight forward way to eliminate all wrong choices will be
Take B as center of a circle and draw arc AC..
now Arc AC > line AC..
arc AC>2\(\sqrt{2}\)..
arc AC= circumference of circle * \(\frac{30}{360}= 2*\pi*x*\frac{30}{360}= 2*\pi*\frac{x}{12}\)..
so \(2*pi*\frac{x}{12}>2\sqrt{2}\)...
\(x> 2*7*12*2\frac{\sqrt{2}}{(2*22)}\)..
\(x> 84\frac{\sqrt{2}}{22}\)...
this is nearly equal to 4\(\sqrt{2}\) ..
so x is nearly equal to 5.6
lets see the choices..
(A) \(\sqrt{3}-1 \approx 0.73\)
(B) \(\sqrt{3}+2 \approx 3.73\)
(C) \(\frac{\sqrt{3}-1}{2} \approx 0.36\)
(D) \(\frac{\sqrt{3}+1}{2} \approx 1.36\)
(E) \(2(\sqrt{3}+1) \approx 5.46\)
clearly E is the answer
2)algebric way
Although rarely tested, If I see a Q on this line, where an angle and opposite sides are given, apply cosine rule..
This triangle would mean two sides of x with third side 2\(\sqrt{2}\) and opposite angle 30..
\(AC^2=BC^2+AB^2-2*BC*AB*cos 30\)...
\({(2\sqrt{2})}^2=x^2+x^2-2*x*x*\sqrt{3}/2\)
\(8=2x^2-x^2\sqrt{3}\)..
\(x^2=8/(2-\sqrt{3})\)..
we can simplify
E is the answer
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IMG_5544.JPG [ 1.79 MiB | Viewed 65258 times ]
Updated on: 08 Jul 2016, 11:44
Originally posted by davedekoos on 11 Feb 2016, 12:44.
Last edited by davedekoos on 08 Jul 2016, 11:44, edited 1 time in total.
Last edited by davedekoos on 08 Jul 2016, 11:44, edited 1 time in total.
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There are two approaches to this problem - we can solve it directly, or we can examine the answer choices and come to a conclusion about which one must be correct.
Algebraic approach:
Draw a perpendicular from point A to side BC, label it point D. Draw another perpendicular from point B to side AC, label it point E. Now we have a 30-60-90 triangle and a two similar 15-75-90 triangles. Let the length of CD = a
Isosceles Triangle.png [ 4.68 KiB | Viewed 46462 times ]
Now we can use some ratios to figure out x
Looking at triangle ABD, we know that \(\frac{BD}{AB} = \frac{x-a}{x} = \frac{\sqrt{3}}{2}\) (1)
And looking at triangles ADC and BEC (similar triangles) we can see that \(\frac{AC}{DC} = \frac{BC}{EC} = \frac{2\sqrt{2}}{a}=\frac{x}{\sqrt{2}}\) (2)
From (2) \(ax = 4\), or \(a=\frac{4}{x}\)
Cross multiply (1) and plug in \(a=\frac{4}{x}\)
\(2x-2a = \sqrt{3}x\)
\(2x-2(\frac{4}{x})=\sqrt{3}x\)
\(2x-\frac{8}{x}-\sqrt{3}=0\) --> multiply by \(x\)
\(2x^2-8-\sqrt{3}x^2=0\) --> gather terms
\((2-\sqrt{3})x^2=8\)
\(x=\sqrt{\frac{8}{2-\sqrt{3}}}\) --> Now this doesn't look like any of our answer choices, so we will have to manipulate it a bit
\(x=\frac{2\sqrt{2}}{\sqrt{2-\sqrt{3}}}\) --> multiply top and bottom by the conjugate of the denominator to rationalize the denominator
\(x=\frac{2\sqrt{2}*\sqrt{2+\sqrt{3}}}{(\sqrt{2-\sqrt{3}})*(\sqrt{2+\sqrt{3}})}\)
\(x=\frac{2\sqrt{2}*\sqrt{2+\sqrt{3}}}{1}\) --> bring the \(\sqrt{2}\) into the square root
\(x=2\sqrt{4+2\sqrt{3}}\) --> Now rearrange the expression to complete the square within the square root
\(x=2\sqrt{1+2\sqrt{3}+3} = 2\sqrt{(1+\sqrt{3})^2}\)
\(x=2(1+\sqrt{3})\)
Answer: E
WHEW! Ok, now the simpler approach that involves almost no calculation, just a few quick estimates and a good idea of what's going on in the triangle:
Because \(\angle{B}\) is \(30^{\circ}\), and the side opposite \(\angle{B}\) is \(2\sqrt{2}\), we know that the other two sides, x, will be longer than \(2\sqrt{2}\).
\(2\sqrt{2}\) is \(\approx 2.8\)
Now if we look at the answer choices:
(A) \(\sqrt{3}-1 \approx 0.73\)
(B) \(\sqrt{3}+2 \approx 3.73\)
(C) \(\frac{\sqrt{3}-1}{2} \approx 0.36\)
(D) \(\frac{\sqrt{3}+1}{2} \approx 1.36\)
(E) \(2(\sqrt{3}+1) \approx 5.46\)
Immediately we can eliminate A, C and D. To choose between B and E, we need to ask whether x should be almost twice as much as AC, or less than 1.5*AC. Go back to the diagram and look at triangle ABD. Since it is a 30-60-90 triangle, we know that AB (i.e. \(x\)) = 2*AD. Now, though the diagram is not to scale, it is accurate enough to surmise that AD is only slightly less than AC. Therefore, \(x\) should be only slightly less than 2*AC, which matches with answer E.
I recommend the second approach, but it only works well because the answer choices are spread out enough. That being said, I don't think the GMAT will expect you to do the algebraic approach, and will design the answer choices specifically to be able to use approximation.
Cheers
Algebraic approach:
Draw a perpendicular from point A to side BC, label it point D. Draw another perpendicular from point B to side AC, label it point E. Now we have a 30-60-90 triangle and a two similar 15-75-90 triangles. Let the length of CD = a
Attachment:
Isosceles Triangle.png [ 4.68 KiB | Viewed 46462 times ]
Now we can use some ratios to figure out x
Looking at triangle ABD, we know that \(\frac{BD}{AB} = \frac{x-a}{x} = \frac{\sqrt{3}}{2}\) (1)
And looking at triangles ADC and BEC (similar triangles) we can see that \(\frac{AC}{DC} = \frac{BC}{EC} = \frac{2\sqrt{2}}{a}=\frac{x}{\sqrt{2}}\) (2)
From (2) \(ax = 4\), or \(a=\frac{4}{x}\)
Cross multiply (1) and plug in \(a=\frac{4}{x}\)
\(2x-2a = \sqrt{3}x\)
\(2x-2(\frac{4}{x})=\sqrt{3}x\)
\(2x-\frac{8}{x}-\sqrt{3}=0\) --> multiply by \(x\)
\(2x^2-8-\sqrt{3}x^2=0\) --> gather terms
\((2-\sqrt{3})x^2=8\)
\(x=\sqrt{\frac{8}{2-\sqrt{3}}}\) --> Now this doesn't look like any of our answer choices, so we will have to manipulate it a bit
\(x=\frac{2\sqrt{2}}{\sqrt{2-\sqrt{3}}}\) --> multiply top and bottom by the conjugate of the denominator to rationalize the denominator
\(x=\frac{2\sqrt{2}*\sqrt{2+\sqrt{3}}}{(\sqrt{2-\sqrt{3}})*(\sqrt{2+\sqrt{3}})}\)
\(x=\frac{2\sqrt{2}*\sqrt{2+\sqrt{3}}}{1}\) --> bring the \(\sqrt{2}\) into the square root
\(x=2\sqrt{4+2\sqrt{3}}\) --> Now rearrange the expression to complete the square within the square root
\(x=2\sqrt{1+2\sqrt{3}+3} = 2\sqrt{(1+\sqrt{3})^2}\)
\(x=2(1+\sqrt{3})\)
Answer: E
WHEW! Ok, now the simpler approach that involves almost no calculation, just a few quick estimates and a good idea of what's going on in the triangle:
Because \(\angle{B}\) is \(30^{\circ}\), and the side opposite \(\angle{B}\) is \(2\sqrt{2}\), we know that the other two sides, x, will be longer than \(2\sqrt{2}\).
\(2\sqrt{2}\) is \(\approx 2.8\)
Now if we look at the answer choices:
(A) \(\sqrt{3}-1 \approx 0.73\)
(B) \(\sqrt{3}+2 \approx 3.73\)
(C) \(\frac{\sqrt{3}-1}{2} \approx 0.36\)
(D) \(\frac{\sqrt{3}+1}{2} \approx 1.36\)
(E) \(2(\sqrt{3}+1) \approx 5.46\)
Immediately we can eliminate A, C and D. To choose between B and E, we need to ask whether x should be almost twice as much as AC, or less than 1.5*AC. Go back to the diagram and look at triangle ABD. Since it is a 30-60-90 triangle, we know that AB (i.e. \(x\)) = 2*AD. Now, though the diagram is not to scale, it is accurate enough to surmise that AD is only slightly less than AC. Therefore, \(x\) should be only slightly less than 2*AC, which matches with answer E.
I recommend the second approach, but it only works well because the answer choices are spread out enough. That being said, I don't think the GMAT will expect you to do the algebraic approach, and will design the answer choices specifically to be able to use approximation.
Cheers
