Posted by Knights on Saturday, February 2, 2013 at 3:44pm.
circle A at bottom
circle B at upper right
circle C at upper left
side of triangle at upper right = x
so
(1/2) of the bottom is x/2, draw vertical altitude through top of triangle and center of circle A
NOW
draw line through centers of circles A and B extending beyond at both ends
It hits right side of triangle at right angle (tangent to circle perp to radius)
It hits our center altitude at 60 degree angle (30 at top of triangle so 60 at altitude intersection)
NOW
distance from intersection with altitude and right side of triangle = 2 radii, 3 + 3 + 3 = 9
SO
distance from top of triangle to that intersection = 9 sqrt 3
there is another radius length along the altitude to the center of the bottom of the triangle so the total height of the altitude = 3 + 9 sqrt 3
cos 30 = (3+9sqrt 3)/x = sqrt 3/2
x = (6+18 sqrt 3) / sqrt 3
simplify that and check my arithmetic
Joining all the centres will produce another equilateral triangle with sides 6
So there is no problem seeing that the distance from tangent contact point to tangent contact point on the bigger triangle is 6.
let's concentrate on one end of the figure
From one of the circles draw the radii to each of the tangents.
label the centre C, the contact points A and B and the intersection of the two tangents as D. Join D and C to form the right-angled triangle ACD
angle DCA = 60°
AC = 3
so tan 60° = AD/3
AD = 3tan60=3√3
At the end of each circle we have 2 of these lengths
so the total perimeter
= 6(3√3) + 3(6)
= 18√3 + 18 units
Thank you so much guys I really appreciate it!!!
kkkkkkkkkk