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An equilateral triangle is inscribed in a circle. If the perimeter of
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06 Sep 2021, 10:00
06 Sep 2021, 10:00
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An equilateral triangle is inscribed in a circle. If the perimeter of the triangle is z inches and the area of the circle is y square inches, which of the following equations must be true?
A) 9z2−πy=0
B) 3z2−πy=0
C) πz2−3y=0
D) πz2−9y=0
E) πz2−27y=0
A) 9z2−πy=0
B) 3z2−πy=0
C) πz2−3y=0
D) πz2−9y=0
E) πz2−27y=0
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An equilateral triangle is inscribed in a circle. If the perimeter of
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30 Nov 2021, 06:25
30 Nov 2021, 06:25
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Attachment:
temp-2.JPG [ 30.92 KiB | Viewed 2673 times ]
Side of an equilateral triangle (a) inscribed in a circle with radius r is given by a = r√𝟑
[ Watch this video to know how ]
=> Perimeter of triangle = 3a = 3 * √𝟑 r = z (given) => r = z3√𝟑
Area of circle with radius r = πr2 = y (given)
=> π * (z3√𝟑)2 = y (substituting value of r from (1) )
=> π * z29∗3 = y
=> π * z227 = y
=> π * z2 = 27 * y
=> πz2−27y=0
So, Answer will be E
Hope it helps!
Watch the following video to Learn Properties of Equilateral Triangle inscribed in a Circle
General Discussion
Re: An equilateral triangle is inscribed in a circle. If the perimeter of
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06 Sep 2021, 17:42
06 Sep 2021, 17:42
3
Answer is E) pi *z^2−27y=0
Let s be the side of the triangle and r be the radius of the circle, the relation between them is s = r * sqrt(3).
So z = 3 * s = 3*sqrt(3) * r.
and area = pi * r^2.
All the answer choices are relation between y and z and y already has a pi in it, so in the answer only C,D,E has pi multiplied for z.
Next step is simple, relation between z and y in terms of r and E is the only one that satisfies.
PS: Not sure this is the GRE method, please share if there is a shorter approach.
Let s be the side of the triangle and r be the radius of the circle, the relation between them is s = r * sqrt(3).
So z = 3 * s = 3*sqrt(3) * r.
and area = pi * r^2.
All the answer choices are relation between y and z and y already has a pi in it, so in the answer only C,D,E has pi multiplied for z.
Next step is simple, relation between z and y in terms of r and E is the only one that satisfies.
PS: Not sure this is the GRE method, please share if there is a shorter approach.
