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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
57% (02:50) correct
43%
(03:25)
wrong
based on 42
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A semicircle is inscribed in equilateral triangle ABC, as shown above. If the area of the triangle is 16, what is the area of the semicircle?
A. \(\frac{3\pi}{8*\sqrt{3}}\)
B. \(\sqrt{3}\pi\)
C. \(2\pi\)
D. \(\sqrt{6}\pi\)
E. \(2\sqrt{3}\pi\)
Attachment:
1.png [ 8.88 KiB | Viewed 7812 times ]
23 May 2021, 07:00
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Bunuel
The area of equilateral triangle = \(side^2*\frac{\sqrt{3}}{4}=16\) --> \(side=\frac{8}{\sqrt[4]{3}}\).
Now, the half of the side (\(\frac{4}{\sqrt[4]{3}}\)) is the hypotenuse of 30-60-90 triangle, where the radius is the leg opposite 60° angle. In such triangles the sides are always in the ratio\(1 : \sqrt{3}: 2.\). So, if hypotenuse is \(\frac{4}{\sqrt[4]{3}}\), then the \(\frac{r}{(\frac{4}{\sqrt[4]{3}})}=\frac{\sqrt{3}}{2}\) --> \(r=2\sqrt[4]{3}\).
Therefore the area of the semicircle is \(\frac{\pi{r^2}}{2}=2\sqrt{3}\pi\).
Answer: E.
Attachment:
2.png [ 7.49 KiB | Viewed 7416 times ]
General Discussion
Updated on: 20 May 2021, 01:12
Originally posted by Fdambro294 on 19 May 2021, 23:43.
Last edited by Fdambro294 on 20 May 2021, 01:12, edited 1 time in total.
Last edited by Fdambro294 on 20 May 2021, 01:12, edited 1 time in total.
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E - 2 * sqrt(3) * (pi)
-Draw a full circle - complete the other half of the semi-circle.
Since the diameter of the circle lies on the side of the equilateral triangle, the center of the circle will be the same point that bisects the bottom side of the triangle. Call this Point Center O
-connect a radius from the center of the circle to the point of tangency. Call this Point X.
-call the vertex of the equilateral triangle A.
Rule: any radius of a circle drawn to a tangent line will be perpendicular to that tangent line at the point of tangency.
Thus, we will have a 90 degree angle at Point X
Further, since it is an equilateral triangle, the angle at Vertex A is 60 degrees
Triangle AXO is a 30-60-90 Right Triangle and Side XO = Hypotenuse = Radius of the Circle
-next, find the side of the equilateral triangle:
Set 16 = (s)^2 * sqrt(3) * (1/4)
(S) = side of equilateral triangle =
(8) / (3)^1/4
-Since center O is also the midpoint of the side of the equilateral triangle, AO = (1/2) (side) = (4) / (3)^1/4
-using the ratios of 30-60-90 right triangles, we can find the radius = side across from 60 degree Angle at Vertex A =
[(2) * sqrt(3) ]
___________
(3)^1/4
-now you can find the full area of the circle by SQUARING the radius (expression above) and multiplying the result by (pi)
= (4) (3) / (3)^2/4
= 12 / sqrt(3)
Conjugate the fraction by multiplying the NUM and DEN by sqrt(3) and then multiply the result by (pi)
Area of Full Circle = (4) * sqrt(3) * (pi)
Area of the Semi-Circle will be (1/2) or this area or
(E)
2 * sqrt(3) * (pi)
Posted from my mobile device
-Draw a full circle - complete the other half of the semi-circle.
Since the diameter of the circle lies on the side of the equilateral triangle, the center of the circle will be the same point that bisects the bottom side of the triangle. Call this Point Center O
-connect a radius from the center of the circle to the point of tangency. Call this Point X.
-call the vertex of the equilateral triangle A.
Rule: any radius of a circle drawn to a tangent line will be perpendicular to that tangent line at the point of tangency.
Thus, we will have a 90 degree angle at Point X
Further, since it is an equilateral triangle, the angle at Vertex A is 60 degrees
Triangle AXO is a 30-60-90 Right Triangle and Side XO = Hypotenuse = Radius of the Circle
-next, find the side of the equilateral triangle:
Set 16 = (s)^2 * sqrt(3) * (1/4)
(S) = side of equilateral triangle =
(8) / (3)^1/4
-Since center O is also the midpoint of the side of the equilateral triangle, AO = (1/2) (side) = (4) / (3)^1/4
-using the ratios of 30-60-90 right triangles, we can find the radius = side across from 60 degree Angle at Vertex A =
[(2) * sqrt(3) ]
___________
(3)^1/4
-now you can find the full area of the circle by SQUARING the radius (expression above) and multiplying the result by (pi)
= (4) (3) / (3)^2/4
= 12 / sqrt(3)
Conjugate the fraction by multiplying the NUM and DEN by sqrt(3) and then multiply the result by (pi)
Area of Full Circle = (4) * sqrt(3) * (pi)
Area of the Semi-Circle will be (1/2) or this area or
(E)
2 * sqrt(3) * (pi)
Posted from my mobile device
