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

AB = (2+2√3)r

Step-by-step explanation:

All three sides of an equilateral triangle equals 60° each.

Given that the circles are equal and are inscribed in a triangle, the angle bisectors pass right through the center of the circle present in front of that angle.

For example a figure has been attached with the answer, where angle bisectors make a triangle with center of the circle and a perpendicular projection of the center on side AB.

Finding AB:

Let us divide the side AB into three parts. One is the line joining the center of the two circles which is = 2

Then we have two equal parts, each joining one vertices with the center of the circle.

Let us assume that there is a point P on the side AB which forms a line segment PO₁ ⊥ AB.

We have the right angled triangle APO₁. Angle A = 30°    PO₁ = r

let the base AP = x

We know that tan 30° = perp/base

1/√3 = r/x

=> x = √3  r

Hence Side AB = √3  r + 2r + √3  r

AB = (2+2√3)r

squares, rhombus, rectangles

a 270° rotation about point p says on e d g e n u it y (had to space it out because it says it's a misspelling so i couldn't submit it