Hi Amit_BTG,
What is the source of this question? I think that the reason why you don't have 5 answer choices to choose from is because this is NOT a GMAT question. It appears to be from the Geometry section of a math textbook. While you will likely see 1 multi-shape Geometry question on the GMAT, it won't involve the "math" that this question requires.
GMAT assassins aren't born, they're made,
Rich
Circle and Equilateral Triangle Help
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I disagree with Rich's unnecessary quotations - this is real math, not "math" - but he's right that this question is straight from a geometry text (or a math test that expects you to know a lot more geometry than the GMAT requires). This question seems to be testing the Power of a Point Theorem, which has never been tested on the GMAT (to my knowledge).
Pretty cool question, though! Here's how I'd solve it:
By Power of a Point, AE * AD = AF * AG, so 4 * (4 + 26) = AF * (AF + 14). This gives us AF = 6.
Since AE + ED + DC = a side of the equilateral ∆, we know the sides are each 32.
AF + FG + GB also = a side of the equilateral ∆, so 6 + 14 + GB = 32, or GB = 12.
Now we can use Power of a Point twice more, from the other vertices of the triangle.
BG * BF = BH * BI, or 12 * 26 = BH * (BH + HI)
and
CD * CE = CI * CH, or 2 * 28 = CI * (CI + IH)
Letting BH = y, CI = z, and HI = x, we thus have
12 * 26 = y * (y + x)
2 * 28 = z * (z + x)
We also know that BH + CI + IH = 32, since together they are a side of the equilateral ∆.
Now all we have to do is solve the system:
312 = y² + xy
56 = z² + xz
32 = x + y + z
The solution is messy, but x = 4√22. If that's the diameter of the circle, its radius is 2√22, and its area is (2√22)² * π, or 88π.
I might have screwed up a couple of the numbers above -- I don't have any paper, so I can't draw it here in front of me -- but the process should work.
Pretty cool question, though! Here's how I'd solve it:
By Power of a Point, AE * AD = AF * AG, so 4 * (4 + 26) = AF * (AF + 14). This gives us AF = 6.
Since AE + ED + DC = a side of the equilateral ∆, we know the sides are each 32.
AF + FG + GB also = a side of the equilateral ∆, so 6 + 14 + GB = 32, or GB = 12.
Now we can use Power of a Point twice more, from the other vertices of the triangle.
BG * BF = BH * BI, or 12 * 26 = BH * (BH + HI)
and
CD * CE = CI * CH, or 2 * 28 = CI * (CI + IH)
Letting BH = y, CI = z, and HI = x, we thus have
12 * 26 = y * (y + x)
2 * 28 = z * (z + x)
We also know that BH + CI + IH = 32, since together they are a side of the equilateral ∆.
Now all we have to do is solve the system:
312 = y² + xy
56 = z² + xz
32 = x + y + z
The solution is messy, but x = 4√22. If that's the diameter of the circle, its radius is 2√22, and its area is (2√22)² * π, or 88π.
I might have screwed up a couple of the numbers above -- I don't have any paper, so I can't draw it here in front of me -- but the process should work.
