SOLUTION: An equilateral triangle, each side of which is 30cm, is inscribed in a circle. Find [a] the distance from the center of the circle to each side. [b] the radius of the circle.

Algebra ->  Surface-area -> SOLUTION: An equilateral triangle, each side of which is 30cm, is inscribed in a circle. Find [a] the distance from the center of the circle to each side. [b] the radius of the circle.      Log On


   

Question 1067832: An equilateral triangle, each side of which is 30cm, is
inscribed in a circle. Find [a] the distance from the
center of the circle to each side. [b] the radius of the
circle.
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
An equilateral triangle, each side of which is 30cm, is
inscribed in a circle. Find [a] the distance from the
center of the circle to each side. [b] the radius of the
circle.



We want to find d and r.

I could use trigonometry, but maybe you haven't had that,
so I'll only use the Pythagorean theorem and algebra:

For right triangle MNP, 

MN2 + NP2 = MP2

And for right triangle MNO,

MN2 + NO2 = MO2

Translating those in terms of d and r:

system%2815%5E2%2B%28r%2Bd%29%5E2=30%5E2%2C15%5E2%2Bd%5E2=r%5E2%29

system%28225%2B%28r%2Bd%29%5E2=900%2C225%2Bd%5E2=r%5E2%29

system%28%28r%2Bd%29%5E2=675%2C225=r%5E2-d%5E2%29

system%28r%2Bd=sqrt%28675%29%2C225=%28r-d%29%28r%2Bd%29%29

r%2Bd=sqrt%28675%29=sqrt%28225%2A3%29=15sqrt%283%29

r%2Bd=15sqrt%283%29

Substituting 15sqrt%283%29 for r+d in the second
equation of the system:

225=%28r-d%29%2815sqrt%283%29%29
225%2F%2815sqrt%283%29%29=r-d
15%2F%28sqrt%283%29%29=r-d
15sqrt%283%29%2F3=r-d
5sqrt%283%29=r-d

So we have the system:

system%28r%2Bd=15sqrt%283%29%2Cr-d=5sqrt%283%29%29

Adding the two equations we get

2r=20sqrt%283%29
r=10sqrt%283%29 cm

Subtracting the two equations, we get

2d=10sqrt%283%29
d=5sqrt%283%29 cm

Edwin