reto wrote:
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ABC is an equilateral triangle of area 3, and arc DE is centered at C. If E is the midpoint of AC, what is the area of the shaded region?
A. \(3 - \frac{\sqrt(3)*\pi}{2}\)
B. \(3 - \frac{\pi}{\sqrt(3)}\)
C. \(3 - \frac{\pi}{2}\)
D. \(3 - \frac{\pi}{2*\sqrt(3)}\)
E. \(3 - \frac{\pi}{6}\)
2 ways to solve this:
Method 1: Solving by calculating areas of the triangle and the circle.
Let r be the radius of the arc.
From area of the triangle = \(\frac{\sqrt{3}*a^2}{4}\) , a = side of the triangle.
Thus, \(\frac{\sqrt{3}*a^2}{4}\) = 3 ---> \(a^2 = 4*\sqrt{3}\)
Now, given that r=a/2 --> area of the sector ( = 60 degree arc) = \((60/360)* \pi*r^2\) = \(\frac{\pi*\sqrt{3}}{6}\)
Thus the area of the shaded region = area of the triangle - area of the sector = \(3 - \frac{\pi*\sqrt{3}}{6}\) = option D.
Method 2: Approximations
*Per the attached picture, the shaded area is 3/4 ths of the triangle area \(\approx{2.25}\).
Start plugging in \(\pi=3 and \sqrt{3} = 1.7\) and see which one gives you the closest value.
All values below are approximate
A: 0.75
B: 1.2
C: 1.5
D: 2.2
E: 2.5
Thus option D is the correct one.
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