SOLUTION: Equilateral triangle ABC has sides of 22cm. A circle of radius 5 cm inside it is tangent to sides AB & AC. Find the distance from the circle’s center to side BC, in cm.

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Question 1148427: Equilateral triangle ABC has sides of 22cm. A circle of radius 5 cm inside it is tangent to sides AB & AC. Find the distance from the circle’s center to side BC, in cm.
Answer by ikleyn(52786) About Me  (Show Source):
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The altitude of the triangle ABC from vertex A, drawn to the side BC  is  

    h = 22%2A%28sqrt%283%29%2F2%29 cm.


Let the center of the circle be the point O at this altitude.


And let the points D and E are the tangent points at the sides AB and AC.


Then the triangle ODA is a right angled triangle with the shorter leg of 5 cm and the acute angle OAD of 30 degrees.


Hence, its hypotenuse OA is twice the radius. i.e. 10 cm.


Thus the distance from the center O of the circle to the side BC is

    d = 22%2A%28sqrt%283%29%2F2%29 - 10 cm = 19.053 - 10 = 9.053 cm.


ANSWER.  The distance from the center O of the circle to the side BC is  d = 22%2A%28sqrt%283%29%2F2%29 - 10 cm = 9.053 cm.

Solved.