What is the relationship between the diameter of a circle and a side of an equilateral triangle inscribed insi

Question

If an equilateral triangle is inscribed inside a circle, how many times in length will be one of its sides compared to the diameter of the circle?

Answer 1

If a is the length of the side of the inscribed equilateral triangle and d is the diameter of the circle then

a = one half of (square root of three) times d

i.e. a = 0.866....d

Answer 2

Since that is an equilateral triangle, each angle is equal to 60 degree.

Since it is also inscribed equilateral triangle, you can draw a bisector line (perpendicular bisectors) from each vertex to the opposite side.

All 3 lines will intersect at the center to create 6 equal divisions and each angle is 60 degree at the center of the circle. Now you also find each vertex of the triangle is bisected by 30 degree each.

Now take only one piece of right triangle among 6.

One side is half of the side of the equilateral triangle which is s/2.
the hypotenuse is radius of the circle, means half of the diameter = d/2

The right triangle is 30, 60, 90 special right triangle.
Using trigonometry, cos 30 =adjacent side /hypotenuse

that means 0.866 = s/2 / d/2

When you simplify 0.866 = s/d
(by removing the similar denominators)

So the relation between the diameter and the side is

s = 0.866 d
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Since that is 30, 60, 90 special triangle,
You can use the formula in ratio format
1 x : Square root 3 x: 2x
we need opposite of 60 degree side.
s = square root of 3 (d/2)
s = 1.732 (d/2)
s = 0.866 d
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Answer 3

First draw three lines (radius length) from the corners of the triangle to the center of the circle.

Form an obtuse triangle from two of the radi, and one of sides of the equilateral triangle. Use the law of cosines to determine the side of the obtuse triangle. The obtuse angle of the triangle must be 120 degrees since it divides the circle evenly into three sections 360 / 3 = 120.
s^2 = r^2 + r^2 - 2(r)(r)(cos 120)
s^2 = (2)r^2 - 2r^2(-0.5)
s^2 = (2)r^2 + r^2
s^2 = (3)r^2
s = sqrt(3)(r)
s = sqrt(3)r
r = d/2
s = sqrt(3)d/2
s = 0.866025403(d)

Answer 4

for any triangle inscribed in a circle of radius R
a/sinA=b/sinB=c/sinC=2R=>D
where a,b,c are the sides and A,B,C are the corresponding angles and R is the radius of the circumcircle
and D the diameter
since in an equilateral triangle a=b=c and SinA=sinB=sinC=
root3/2 we have 2a/root3=2R or D or a=(root 3/2) times D

Answer 5

d = diameter
s = side of eq triangle inscribed

d = 2r

Let O = center of circle
ABC = triangle(with vertices)

Draw radii OA, OB, OC.
r = OA = OB = OC
(radii of the same circle are congruent)

AB = BC = CA = s
(def'n of equilateral triangle)

Thus,
triangle AOB = triangle BOC = triangle COA
by SSS Postulate

therefore
angle AOB = angle BOC = angle COA
by CPCTC

angle AOB + angle BOC + angle COA = 360
bec. the sum of all angles at O is 360

thus, by simplification
angle AOB = angle BOC = angle COA = 120

draw 3 alt. OD to BC, OE to AC, OF to AB.
Def'n of alt.

Thus, you form 6 congruent right triangles
BD = DC = CE = EA = AF = FB = 1/2 s

Using triangle-angle theorems, the ratios of these 6 sides to every radius are sqrt3/2.

therefore,
BD/OA = sqrt3/2
BD = sqrt3/2 OA

Since BD = 1/2 BC and OA = r,
1/2 BC = sqrt3/2 r
BC = sqrt3 r

since d = 2r, r = d/2 and BC = s

s = sqrt3/2 d
d = 2/sqrt3 s

Hope this helps^_^
^_^