an equilateral triangle is incribed in a circle of radius 4cm. Find its side.
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The centre of circle is O.
If the ∆ is ABC then radius = OA
We know in an equilateral ∆, Median AD bisects the side and is perpendicular also. (property 1)
Also in an equilateral ∆, circumvented and centroid coincide.
So, O is circumcentre and centroid.
We know centroid divides AD in 2:1.
AO = 4 (radius)
So,
AD /AO= 3/2
AD = 3/2*(4) = 6
In ∆ ADC
AC = x (let)
DC = x/2 (by property 1)
AD = 6
So by pythgoras
6^2 + (x/2)^2 = x^2
We need to find x
Solve above equation to get
36 = 3x^2/4
X = √48
X = 4√3 ans
If the ∆ is ABC then radius = OA
We know in an equilateral ∆, Median AD bisects the side and is perpendicular also. (property 1)
Also in an equilateral ∆, circumvented and centroid coincide.
So, O is circumcentre and centroid.
We know centroid divides AD in 2:1.
AO = 4 (radius)
So,
AD /AO= 3/2
AD = 3/2*(4) = 6
In ∆ ADC
AC = x (let)
DC = x/2 (by property 1)
AD = 6
So by pythgoras
6^2 + (x/2)^2 = x^2
We need to find x
Solve above equation to get
36 = 3x^2/4
X = √48
X = 4√3 ans
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