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

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

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

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 sides of a 30-60-90 triangle are proportioned s : s√3: 2s.
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.

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.

user123321

thank you so much.. i got it nw