SOLUTION: if there is an equilateral triangle of side 20 cm if 10 identical circle placed inside the triangle such that no two circle overlap or no circle cuts any side of the triangle. find

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Question 1041813: if there is an equilateral triangle of side 20 cm if 10 identical circle placed inside the triangle such that no two circle overlap or no circle cuts any side of the triangle. find the radius of each circle
Found 2 solutions by ikleyn, rothauserc:
Answer by ikleyn(52765)   (Show Source): You can put this solution on YOUR website!
.
I do not understand the question clearly.

Probably, you want to ask if such an equilateral triangle does exist.


Then the answer is "Yes, sure, if the radius of the circles is small enough".


How small?


Draw three parallel equidistant straight lines, all cutting the triangle and all parallel to one side.

Then draw three other equidistant parallel straight lines all parallel to the second side of the triangle.

How many small triangles did you get?
Count them. I believe the number of these triangles is 10 or more.

Now inscribe a circle into each small triangle.

It will give you what you want to get.

In any case, you got the idea.


Answer by rothauserc(4718)   (Show Source): You can put this solution on YOUR website!
We have an equilateral triangle with four circles on a side
:
draw a horizontal line connecting the four centers and draw a perpendicular line from the center of each circle on the ends(there are two) to the side.
:
label(AB) the distance from each end to the intersection point of the perpendicular line and the side
:
side length = 2AB + 6r, where r is the radius of a circle
:
note the angles in the triangles formed at each end are 30, 60, 90 with the AB's opposite the 60 degree angles
:
the sides in a 30-60-90 are in the ratio 1:square root(3):2, therefore
:
AB = r*square root(3) and we have
:
20 = 2r*square root(3) + 6r
:
10 = r * square root(3) + 3r
:
*****************************************
r = 10 / (square root(3) + 3) = 2.1132 cm
*****************************************
:
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