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.
<pre><b>
{{{drawing(400,400,-1.5,1.5,-1.5,1.5,

line(0,-1/2,0,1),line(0,0,-sqrt(3)/2,-1/2),
locate(-.41,-.5,15), locate(.39,-.5,15),locate(-.56,.33,30),
locate(.01,-.2,d),locate(-.45,-.12,r), locate(.018,.33,r),
locate(-sqrt(3)/2,-1/2,M),locate(0,-1/2,N),locate(.01,1.1,P), locate(.01,.07,O),   
triangle(-sqrt(3)/2,-1/2,0,1,sqrt(3)/2,-1/2), circle(0,0,1))}}}

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, 

MN<sup>2</sup> + NP<sup>2</sup> = MP<sup>2</sup>

And for right triangle MNO,

MN<sup>2</sup> + NO<sup>2</sup> = MO<sup>2</sup>

Translating those in terms of d and r:

{{{system(15^2+(r+d)^2=30^2,15^2+d^2=r^2)}}}

{{{system(225+(r+d)^2=900,225+d^2=r^2)}}}

{{{system((r+d)^2=675,225=r^2-d^2)}}}

{{{system(r+d=sqrt(675),225=(r-d)(r+d))}}}

{{{r+d=sqrt(675)=sqrt(225*3)=15sqrt(3)}}}

{{{r+d=15sqrt(3)}}}

Substituting {{{15sqrt(3)}}} for r+d in the second
equation of the system:

{{{225=(r-d)(15sqrt(3))}}}
{{{225/(15sqrt(3))=r-d}}}
{{{15/(sqrt(3))=r-d}}}
{{{15sqrt(3)/3=r-d}}}
{{{5sqrt(3)=r-d}}}

So we have the system:

{{{system(r+d=15sqrt(3),r-d=5sqrt(3))}}}

Adding the two equations we get

{{{2r=20sqrt(3)}}}
{{{r=10sqrt(3)}}} cm

Subtracting the two equations, we get

{{{2d=10sqrt(3)}}}
{{{d=5sqrt(3)}}} cm

Edwin</pre></b>