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Re: DE or 3 sqrt5 [#permalink]
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Unless the book missed it....the question is complete

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Re: DE or 3 sqrt5 [#permalink]
Carcass wrote:
Unless the book missed it....the question is complete
Regards


ok so can u explain it I didn't get how DE is greater
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Re: DE or 3 sqrt5 [#permalink]
Can someone please post the solution of the question?
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Re: DE or 3 sqrt5 [#permalink]
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Official Explanation

If you’re not sure how to solve a complex geometry problem, a good first step is to fill in as many deductions as you can and then reassess the situation. First, note that any triangle formed by the diameter of a circle and a point on the circle will always be a right triangle. Because triangles AEB and CED each have a right angle and a 30 degree angle, it follows that they are both 30-60-90 right triangles. Next, since you’re given the area of the semicircle, you can find the diameter of semicircle O— which is also the hypotenuse of ΔABE. If the area of half a circle is 16π, then the area of the whole circle is 32π:

\(A=\pi r^2\)

\(32 \pi=\pi r^2\)

\(32=r^2\)

\(r=\sqrt{32}=4\sqrt{2}\)


This is the radius. The diameter is double the radius, or \(8 \sqrt{2} \) which is also the hypotenuse of Δ ABE. Recall that the sides of a 30-60-90 triangle are in the proportion \(x:x\sqrt{3}:2x\) Since the hypotenuse is \(8\sqrt{2}\) it follows that the short leg (AE) is and the long leg (AE) is \(4 \sqrt{2}\) The last piece of information you haven’t used yet is the fact that \(AC=\sqrt{2}\)

Subtracting this from AE gives CE:


\(CE=AE-AC=4\sqrt{2}-\sqrt{2}=3\sqrt{2}
\)

CE is the shorter leg of a 30-60-90 triangle, and DE (the value in Quantity A) is the longer leg. Thus,

\(DE=CE\sqrt{3}=3 \sqrt{2} \sqrt{3}=3 \sqrt{6}\)

A is the answer
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Re: DE or 3 sqrt5 [#permalink]
1
Because triangles AEB and CED each have a right angle and a 30 degree angle, it follows that they are both 30-60-90 right triangles. Next, since you’re given the area of the semicircle, you can find the diameter of semicircle O— which is also the hypotenuse of ΔABE. If the area of half a circle is 16π, then the area of the whole circle is 32π:

Missing info:
- triangles AEB and CED each have a right angle and a 30 degree angle
- the area of half a circle is 16π

Cheers.
Brent
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Re: DE or 3 sqrt5 [#permalink]
Took me like 5 mins to solve this question. It's relatively hard to solve in less than 2 minutes!
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Re: DE or 3 sqrt5 [#permalink]
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No worries. It is a NON official question.

It is there to training your skills.

More, see our quant book https://gre.myprepclub.com/forum/gre-prep- ... html#p5077

Regards
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Re: DE or 3 sqrt5 [#permalink]
Is it possible to know based on geometrical axioms that angle AEB is 90 degrees or is it missing info?
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Re: DE or 3 sqrt5 [#permalink]
AEB is 90 degree. If this is not spotted then this question cannot be answered. Its formula based
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Re: DE or 3 sqrt5 [#permalink]
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Re: DE or 3 sqrt5 [#permalink]
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