If an equilateral triangle of side 'a' is inscribed inside a circle, then area of circle = Pi (a^2) / 3 (or)
radius = a / Sqrt(3)
Given radius = 4, area = 16Pi
16Pi = Pi (a^2) / 3
a^2 = 16* 3 -> a = 4Sqrt(3)
Perimeter = 3a = 12Sqrt(3) D IMO
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difficult geometry
Source: Beat The GMAT — Problem Solving |
if a equilateral triangle is inscribed in a circle: the radius of a circle is R= 2/3 X height
=>height of the triangle = 4*3/2 = 6
If height of equilateral triangle is given as 6;
as a^2= 6^2 + (a/2)^2
you can calculate the side (a) to be 4V3
Thus perimeter is 3*a => 12V3
Clearly D
=>height of the triangle = 4*3/2 = 6
If height of equilateral triangle is given as 6;
as a^2= 6^2 + (a/2)^2
you can calculate the side (a) to be 4V3
Thus perimeter is 3*a => 12V3
Clearly D
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neelgandham
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Let O be the centre of the circle and r be the radius of the circle. As triangle ABC is a equilateral triangle the lines joining the centre and the vertices bisect the angles as shown in the diagram.
Angle DBO = Angle OBA = 60/2 (As the angle made by two sides of a equilateral triangle is 60 degrees).
we also know that the line joining the centre of the circle and any chord is perpendicular to and bisects the chord . To find the area of a equilateral triangle, one should know the value of the side or the height of the triangle.
Height of the triangle = AO + OD = r + OD
From Right angled triangle ODB, OD = r*Sin 30 = r/2
Height of the equilateral triangle = r + OD = r + (r/2) = (3/2)* r
Height of an equilateral triangle = ((Square root(3))/2)*a where a is the length of the side of the triangle.
So, ((Square root(3))/2)*a = (3/2)* r
a = (Square root(3)) * 4 (r= 4)
Perimeter of the triangle = 3a = 3*(Square root(3)) * 4 = 12*(Square root(3))
IMO Option D
Angle DBO = Angle OBA = 60/2 (As the angle made by two sides of a equilateral triangle is 60 degrees).
we also know that the line joining the centre of the circle and any chord is perpendicular to and bisects the chord . To find the area of a equilateral triangle, one should know the value of the side or the height of the triangle.
Height of the triangle = AO + OD = r + OD
From Right angled triangle ODB, OD = r*Sin 30 = r/2
Height of the equilateral triangle = r + OD = r + (r/2) = (3/2)* r
Height of an equilateral triangle = ((Square root(3))/2)*a where a is the length of the side of the triangle.
So, ((Square root(3))/2)*a = (3/2)* r
a = (Square root(3)) * 4 (r= 4)
Perimeter of the triangle = 3a = 3*(Square root(3)) * 4 = 12*(Square root(3))
IMO Option D
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Anil Gandham
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The sides of a 30-60-90 triangle are proportioned s : s√3: 2s.shikh wrote:If the circle above has a radius of 4, what is the perimeter of the inscribed equilateral triangle?
A. 6v2
B. 6v3
C.12v2
D.12v3
E.24
The figure below illustrates that when an equilateral triangle is inscribed in a circle, s = r√3:
In the problem above, s = 4√3, so p = 12√3.
The correct answer is D.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
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For more information, please email me (Mitch Hunt) at [email protected].
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In an equilateral triangle, centroid & circumcenter both are same. and centroid divides the altitude in 2:1 ratio. In equilateral triangle altitude is nothing but perpendicular bisector. Using this info you can solve the problem.
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