what is the radius of an inscribed circle inside a triangle that has a side length of 3.65 in?
Find the radius of the circle inscribed in an equilateral triangle with side length of 3.65in.
The center of the inscribed circle is the point of intersection of the angle bisectors of the triangle. Also, the triangle is tangent to the circle on all three sides.
(1)From the center of the circle draw a radius to the point of tangency of a side of the triangle. The radius drawn is perpendicular to the side of the triangle. The radius also bisects the side of the triangle. (The triangle is equilateral and thus isosceles -- the angle bisectors of a vertex of an isosceles triangle are also the medians and altitudes.)
(2)From the center of the circle, draw a segment to a vertex whose angle includes the point of tangency used in step (1). Thissegment bisects the angle at the vertex.
(3) Since the angles of an equilateral triangle are all `60^circ` , we have created a 30-60-90 right triangle. The radius of the circle can be found by:
`tan30^circ = r/(1.825)`
`r=1.825 tan30^circ ~~ 1.054`
So the radius of the incircle is approximately 1.05in.
(Another method is to find the length of an altitide of the triangle: `alt=1.825 tan 60^circ ~~3.161` and realize that the center of the incircle is not only the intersection of the angle bisectors, but in an equilateral triangle is also the intersection of the medians and altitudes. Then the length of the radius of the incircle is 2/3 of the altitude or 1.05in)