Question 1101450: Equilateral triangle ABC has sides of 16 cm. A circle of radius 4 cm inside it is tangent to sides AB and AC. Find the distance from the circle's centre to side BC, in cm.
Found 2 solutions by greenestamps, josgarithmetic:
Answer by greenestamps(13195)   (Show Source):
You can put this solution on YOUR website!
(Sorry -- I labeled the vertices of the triangle differently than in the statement of the problem. The solution is still valid....)
Call the equilateral triangle ABC, with O the center of base AB; the circle with center F is tangent to sides AC and BC at D and E, respectively.
The radii DF and EF are perpendicular to the respective sides of the triangle.
That means triangles CEF and COB are similar; and since O is the midpoint of AB, the ratio of similarity is 1:2.
That means CF is 8.
CO (the altitude of the equilateral triangle) is 8*sqrt(3); CF is 8, FG is 4.
The length we are asked to find is CO-CF-FG = 8*sqrt(3)-12.
PS -- The solution by josgarithmetic is not valid; the circle is not tangent to all three sides of the triangle. (If it were, its radius would not be 4.)
In response to your question about putting the diagram in my answer....
Can you see, when you look at my response, a place you can click on with your mouse that says "show source"? As a tutor, I can see that when I am looking at the solution that any tutor has written to a problem; I'm hoping you can see that too.
If you can see that, then clicking on it will show you what I typed in my response to build the figure.
I am having a lot of "fun" trying to figure out what kinds of things I can do in terms of drawing figures; the figure in this response uses a couple of types of things I hadn't tried before. So it will be a learning curve for you to try to learn how to include figures....
Answer by josgarithmetic(39613)  (Show Source):
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