Given: △abc is equilateral. the radius of each circle is r. find: ab

2r (1 + √3)

Step-by-step explanation:

Circle O₁ is tangent to AB.  Let's call the point of intersection point D.  If we draw a radius from the center O₁ to D, we know this forms a right angle.

△ABC is an equilateral triangle, so we know m∠A = 60°.  If we draw a line from A to O₁, we know that bisects the angle, so m∠DAO₁ = 30°.

So △DAO₁ is a 30-60-90 triangle.  We can find the length AD:

AD = r √3

Now on the other side, circle O₃ is tangent to AB.  Let's call the point of intersection point E.  We know it's the same triangle we found earlier, so:

EB = r √3

And finally, we can draw a rectangle connecting O₁, O₃, E, and D.  The distance between O₁ and O₃ is 2r, so:

DE = 2r.

Therefore:

AB = r√3 + 2r + r√3

AB = 2r√3 + 2r

AB = 2r (1 + √3)

Here's a graph showing the steps.  Hopefully this helps, let me know if you have questions!

desmos.com/calculator/hgaonfzxsm

answer: 140

step-by-step explanation:

so just add all then it will be your answer but i am not pretty sur eif it is but hope this i am here to people