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In the figure above, an equilateral triangle is inscribed in
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12 Aug 2018, 10:12
12 Aug 2018, 10:12
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69% (01:54) correct
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#GREpracticequestion In the figure above, an equilateral triangle is.jpg [ 20.08 KiB | Viewed 13456 times ]
In the figure above, an equilateral triangle is inscribed in a circle. If the arc bounded by adjacent corners of the triangle is between \(4\pi\) and \(6\pi\) long, which of the following could be the diameter of the circle?
(A) 6.5
(B) 9
(C) 11.9
(D) 15
(E) 23.5
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In the figure above, an equilateral triangle is inscribed in
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10 Dec 2021, 10:26
10 Dec 2021, 10:26
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temp-3.JPG [ 22.92 KiB | Viewed 6812 times ]
Since the Arc made by two adjacent corners of the triangle is making 120˚ angle at the center so length of the arc is given by \(\frac{120˚}{360˚} *2 \pi r\)
[ Watch this video to know how ]
=> Arc length = \(\frac{1}{3} *2 \pi r\) and it is between \(4\pi\) and \(6\pi\)
=> \(4\pi\) <\(\frac{1}{3} *2 \pi r\) < \(6\pi\)
=> 4 < \(\frac{d}{3}\) < 6 (as Diameter (d) = 2r)
=> 4*3 < d < 3*6
=> 12 < d < 18
Only Option D (15) is in this range
So, Answer will be D
Hope it helps!
Watch the following video to Learn Properties of Equilateral Triangle inscribed in a Circle
General Discussion
Re: In the figure above, an equilateral triangle is inscribed in
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18 Aug 2018, 20:09
18 Aug 2018, 20:09
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Carcass wrote:
Attachment:
circle.jpg
In the figure above, an equilateral triangle is inscribed in a circle. If the arc bounded by adjacent corners of the triangle is between \(4\pi\) and \(6\pi\) long, which of the following could be the diameter of the circle?
(A) 6.5
(B) 9
(C) 11.9
(D) 15
(E) 23.5
Here we need to consider the both the arc length i.e. \(4\pi\) and \(6\pi\)
Since all the angles of equilateral triangle = 60°
Therefore
\(\frac{{central angle}}{360}= \frac{{Arc length}}{circumference}\)
\(\frac{120}{360}= \frac{{4\pi}}{{2\pi*radius}}\)
or \(radius = 6\)
or \(diameter = 12\)
Now if we consider arc length as \(6\pi\) then,
\(\frac{{central angle}}{360} = \frac{{6\pi}}{{2\pi*radius}}\)
\(\frac{120}{360}= \frac{{6\pi}}{{2\pi*radius}}\)
or \(radius = 9\)
or \(diameter = 18\)
So only Option D satisfy
