Construction : Draw seg AF ⊥ seg BD.i. The circles with centres A and B touch each other at E. [Given]
∴ By theorem of touching circles,
A – E – B
∴ ∠ACD = ∠BDC = 90° [Tangent theorem]
∠AFD = 90° [Construction]
∴ ∠CAF = 90° [Remaining angle of AFDC]
∴ AFDC is a rectangle. [Each angle is of measure 900]
∴ AC = DF = 4 cm [Opposite sides of a rectangle]
Now, BD = BF + DF [B – F – C]
∴ 6 = BF + 4 BF = 2 cm
Also, AB = AE + EB
= 4 + 6 = 10 cm
[The distance between the centres of circles touching externally is equal to the sum of their radii]
ii. Now, in ∆AFB, ∠AFB = 90° [Construction]
∴ AB2 = AF2 + BF2 [Pythagoras theorem]
∴ 102 = AF2 + 22
∴ 100 = AF2 + 4
∴ AF2 = 96
∴ AF = √96 [Taking square root of both sides]
= √(16 x 6)
= 4√6 cm
But, CD = AF [Opposite sides of a rectangle]
∴ CD = 4√6 cm