An equilateral triangle of side length 2 units is inscribed in a circle. Find the length of a chord of this circle which passes through the midpoints of two sides of the triangle.
The height of the larger equilateral triangle is sqrt ( 3)
So....the height of the smaller equilateral triangle formed by the chord is sqrt (3)/ 2 = (1/2)sqrt (3)
The radius of the cirlcle can be found as
2^2 = 2R^2 - 2R^2 cos 120°
4 = 2R^2 + R^2
4 = 3R^2
R^2 = 4/3
R = 2/sqrt (3) = (2/3)sqrt (3)
So the distance from the center of the circle to the chord is (2/3)sqrt (3) -(1/2) sqrt(3) = (1/6)sqrt (3)= sqrt (3)/6
Using the Pythagorean Theorem, 1/2 the chord length can be found as
sqrt [ ( 4/3) - ( sqrt (3)/6 )^2 ] = sqrt [ 48/36 - 3/36 ] = (1/6) sqrt (45) = (1/6) sqrt (9 * 5) =
(3/6) sqrt (5) = sqrt ( 5)/2
So....the chord length is twice this = sqrt (5)
An equilateral triangle of side length 2 units is inscribed in a circle. Find the length of a chord of this circle which passes through the midpoints of two sides of the triangle.
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Find the diameter of a circle D = 2(1 / cos30º) = 2.309401076
Height of a triangle h = √3 = 1.732050808
Subtract height from diameter M = D - h = 0.577350268
The length of a chord L = sqrt( D2 - M2 ) = √5