Re: DE or 3 sqrt5
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04 Aug 2020, 07:10
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