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# circle

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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.

Jan 4, 2021

#1
+114171
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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)

Jan 4, 2021
edited by CPhill  Jan 4, 2021
#2
+836
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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.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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

Jan 4, 2021