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06 Feb 2016, 17:17
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
88% (02:21) correct
13%
(02:06)
wrong
based on 32
sessions
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What is the area of the shaded portion in circle O, as pictured above, with m< O=60° and radius r = 10?
A) \(\frac{50}{3}\) \(\pi\) − \(50\sqrt{3}\)
B) \(\frac{50}{3}\) \(\pi\) − 50
C) \(\pi \sqrt{3}\)
D) \(\frac{50}{3}\) \(\pi\) − \(25\sqrt{3}\)
E) \(\frac{50}{3}\) \(\pi\)
A) \(\frac{50}{3}\) \(\pi\) − \(50\sqrt{3}\)
B) \(\frac{50}{3}\) \(\pi\) − 50
C) \(\pi \sqrt{3}\)
D) \(\frac{50}{3}\) \(\pi\) − \(25\sqrt{3}\)
E) \(\frac{50}{3}\) \(\pi\)
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06 Feb 2016, 18:31
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It is not a hard problem, if you know the Area of a circle segment and the value of sin for a 60 degree angle.
A=\(\frac{r^2}{2}\)*(\(\frac{\pi*\alpha}{180}\)-sin\(\alpha\))
Sin \(\alpha\) 60 = \(\frac{\sqrt{3}}{2}\)
Another way is to find the Area of the triangle and the Area of the sector of circle to find the Area of the Segment of the Circle.
Area Sector - Area Triangle = Area Segment.
Hope it helps.
Ans. D)
A=\(\frac{r^2}{2}\)*(\(\frac{\pi*\alpha}{180}\)-sin\(\alpha\))
Sin \(\alpha\) 60 = \(\frac{\sqrt{3}}{2}\)
Another way is to find the Area of the triangle and the Area of the sector of circle to find the Area of the Segment of the Circle.
Area Sector - Area Triangle = Area Segment.
Hope it helps.
Ans. D)
07 Feb 2016, 02:32
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As mentioned by FelixM, you simply need to take the area of the sector/circle slice and subtract the area of the interior triangle.
Area of the circle slice:
1. \(\frac{60°}{360°} * π 10 ^2\)
2. (1) reduces into \(\frac{{π 50}}{3}\)
Area of the triangle:
1. We see that the interior angle of our triangle is 60°. We are dealing with an equilateral triangle.
2. We bisect our triangle down the middle to find our height. When we bisect the triangle, that creates two sub-triangles, mirrored, each conforming to the 30-60-90 special triangle. Plugging in the magic numbers for this triangle gives us a triangle height of \(5\sqrt{3}\).
3. \(5 \sqrt{3} * 10 * \frac{1}{2} = 25 \sqrt{3}\)
To find the area of the shaded portion:
1. As mentioned above, subtract the area of the triangle from the area of the sector: \(\frac{{π 50}}{3} - 25\sqrt{3}\), which is option D.
Area of the circle slice:
1. \(\frac{60°}{360°} * π 10 ^2\)
2. (1) reduces into \(\frac{{π 50}}{3}\)
Area of the triangle:
1. We see that the interior angle of our triangle is 60°. We are dealing with an equilateral triangle.
2. We bisect our triangle down the middle to find our height. When we bisect the triangle, that creates two sub-triangles, mirrored, each conforming to the 30-60-90 special triangle. Plugging in the magic numbers for this triangle gives us a triangle height of \(5\sqrt{3}\).
3. \(5 \sqrt{3} * 10 * \frac{1}{2} = 25 \sqrt{3}\)
To find the area of the shaded portion:
1. As mentioned above, subtract the area of the triangle from the area of the sector: \(\frac{{π 50}}{3} - 25\sqrt{3}\), which is option D.
