Question

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

Answer 1

`~~57.21cm`

Explanation:

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`

Answer 2

Explanatory steps to solution already posted by @dk_ch

Explanation:

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.