Extend the line BK such that you can draw a line that is perpendicular to it from H.

Name the foot of the perpendicular C(or any letter you like.

You now have a rectangle HABC.
(Note that ∠ HAB and ∠ KBA = 90° (tangent perpendicular to radius)

AB = HC, HA = CB

Since ∠ HCK = 90° (corner or a rectangle),

Triangle HCK is right-angled.

Then,

HK² = HC² + CK² (Pythagoras' Theorem)

30² = HC² + (CB - BK)²

30² = HC² + (HA - BK)² = HC² + (12 - 5)²

HC² = 30² - 7² = 900 - 49 = 851

HC = √851 ≈ 29.1719043 ≈ 29.2 (3s.f)

AB = HC ≈ 29.2 m

Looking at triangle HCK,

cos∠HKC = adj/hyp = CK/HK = 7/30

∠HKC = cos-¹ (7/30) ≈ 76.5066012° ≈ 76.5°(1 d.p)

∠AHK = ∠HKC = 76.5° (alternate angles, AH // CK)

Mark a point D on the circumference of the bigger circle such that ∠AHK = ∠DHK. Draw the line HD. Like HA, HD is also a radius of the bigger circle.

Mark a point E on the circumference of the smaller circle such that , ∠HKB = ∠HKE. Draw the line KE. Like BK, KE is also a radius of the smaller circle.

DE is also a common tangent to both wheels and DE = AB


∠AHD = 2∠AHK = 2 x 76.5066012°
= 153.013202°

Length of arc AD = (360°- ∠AHD)/360° x 2πr

= (360° - 153.013202°)/360° x 2π(12)

≈ 43.3512136

Realise that ∠BKE = ∠AHD
(BK // AH, KE // HD)
(You may want to extend the line HK to see how this is so)

So length of arc KE = 153.013202°/360° x 2π(5)

≈ 13.3529209

Length of chain

= arc AD + arc KE + AB + DE
= arc AD + arc KE + 2AB


= 43.3512136 + 13.3529209 + 2(29.1719043)

≈ 115.0479431

≈ 115 (3s.f)