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What is the measure of the radius of the circle inscribed in a triangle whose sides measure $8$, $15$ and $17$ units? I can easily understand that it is a right angle triangle because of the given edges. but I don't find any easy formula to find the radius of the circle.
I would suggest that you read a book on telescope mirror making. The telescope maker's approximate formula for the radius of curvature of a mirror surface is s = r^2/2R, where r is half the mirror diameter, s is the radius of curvature of the surface (half the focal length), and R is the saggita (depth of the curve).
Middle school geometry problem about a triangle inscribed in a circle with very particular properties Rational divisors on Calabi–Yau threefolds What do border officials do with my passport when I tell them that I'm entering as the spouse of an EU national?
Get the equation of a circle through the points $(1,1), (2,4), (5,3) $. I can solve this by simply drawing it, but is there a way of solving it (easily) without having to draw?
The standard circle is drawn with the 0 degree starting point at the intersection of the circle and the x-axis with a positive angle going in the counter-clockwise direction. Thus, the standard textbook parameterization is: x=cos t y=sin t. In your drawing you have a different scenario.
Intuitively, "an equation" cuts down the dimension by one, but to get a circle in space you have to lower the dimension by two. (A small piece of a circle looks like a line, so a circle is "one dimensional" for present purposes.) Here are a few alternative descriptions that you may find helpful or interesting.
Yes. Draw a picture : From the circle's center draw a radius to a vertex and a line to the midpoint of a side with that vertex at one extreme. Then you get a $\;30-60-90\;$ little triangle and thus the line from center to the midpoint is opposite to the angle of $\;30^\circ\;$ and is thus half the hypotenuse which is the radius.
The radius of the circle is not the side of this equilateral triangle; it is the height of the equilateral triangle. $\endgroup$ – fleablood Commented May 1, 2016 at 15:41
$\begingroup$ It should be noted that this equation of the circle is simply the Law of Cosines with appropriate variables. $\endgroup$ – eh9 Commented Jun 7, 2015 at 14:29
the integral of r(d(theta)) from 0 to 2π is 2πr (the circumference of a circle with radius r), now integrate 2πr(dr) from 0 to r and the answer is πr^2 (the area of a circle of radius r). This is my derivation.