Calculate the length of the chain ABCD? See picture below.

In this diagram both CD and AB are two direct common tangents of two circles of centers #O_1 and O_2#

Both the tangents are perpendicular at the points of contact with the radii of the circles.

If we draw a line #O_2E# parallel to CD from the point #O_2# and it intersects #O_1D# at E, then #DCO_2E# will be a rectangle.

Now in #DeltaO_1O_2E#

#O_1E=O_1D-DE=O_1D-O_2C=(8-3)cm=5cm#

#CD=O_2E=sqrt(O_1O_2^2-O_1E^2)#

#=sqrt(25^2-5^2)#

#=10sqrt6 cm#

So

#cos/_DO_1O_2 = cos/_EO_1O_2#

#=(O_1E)/(O_1O_2)=5/25=1/5#

#=>/_DO_1O_2 =cos^-1(1/5)#

Now

#/_CO_2B=/_DO_1A=2/_DO_1O_2#

#=2cos^-1(1/5)=2.74rad#

So
#"arcCB"/(O_2B)=2.74#

#=>arcCB=2.74xxO_2B=2.74xx3=8.22cm#

So length of the chain ABCD

#=AB+CD+arcBC#

#=2CD+arcBC#

#=2xx10sqrt6+8.22~~57.21cm#

If the full chain length in red is required then we are to measure the arc length opposite to reflex#/_DO_1A#

Now

reflex#/_DO_1A=2pi-2.74~~3.54rad#

So arc length opposite to reflex#/_DO_1A#
#=8xx3.54=28.32#

So total chain length #=57.21+28.32=85.53 cm#

Length of the chain #ABCD=DC+"minor arc "CB+BA# ......(1)

In the given picture both #DC and AB# are two common tangents of two circular cogs having respective centers #O_1 and O_2#

The tangents are perpendicular to respective radii of the circles at the points of contact.

Construction:
Let us we draw a line #O_2E# parallel to #DC# from the point #O_2# aso that it meets radius #O_1D# at #E#. As the figure #DCO_2E# has all internal angles #90^@#, it is a rectangle.

From construction
Opposite sides of rectangle are equal
#:.# Sides #DE=CO_2=3cm# ...(2)

In the right #Delta EO_1O_2#

#EO_1="Radius of bigger cog"-"Side DE"=DO_1-DE#
Using (3)
#EO_1=8-3=5cm#

Also side #DC=O_2E=sqrt((O_1O_2)^2-(O_1E)^2)#

#=>DC=sqrt(25^2-5^2)#

#=>DC=10sqrt6 cm#

From symmetry #DC=AB=10sqrt6 cm# ......(3)

To calculate #"minor arc "CB# we need to find minor #angleCO_2B#.
Which is #=#minor #angleDO_1A#

Again in the right #Delta EO_1O_2#

#cos/_EO_1O_2=(O_1E)/(O_1O_2)=5/25=1/5#

#=>/_EO_1O_2 =cos^-1(1/5)#

Now from symmetry

minor #/_CO_2B=# minor #/_DO_1A=2/_EO_1O_2#

#=2cos^-1(1/5)# .......(4)

We know that in a circle of radius #r# length of arc which subtends an angle #theta# at the center
Length of arc#=rtheta#
where #theta# is in radians.

#=>"minor arc "CB"=(O_2B)xx2cos^-1(1/5)#
#=>"minor arc "CB"=3xx2cos^-1(1/5)#
#=>"minor arc "CB"=6cos^-1(1/5)# ...... (5)

Inserting values from (4) and (5) in equation (1) we get

Length of the chain #ABCD=10sqrt6+6cos^-1(1/5)+10sqrt6#
#=>#Length of the chain #ABCD=20sqrt6+6cos^-1(1/5)#
#=>#Length of the chain #ABCD=48.99+8.22=57.2cm#, rounded to one decimal place.