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
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Isosceles Triangle.png [ 4.68 KiB | Viewed 37555 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
_________________
Dave de Koos
GMAT aficionado