# Two chords AB and Given chords AB = 6 cm and CD = 12 cm,

Draw OE AB intersecting CD at F,

As, AB||CD

OF CD [Corresponding angles]

Distance between AB and CD, EF= 3 cm

Now, we know that perpendicular from center to chord bisects the chord.

Then CF = FD = 1/2 CD = 6 cm

AE = EB = 1/2 AB = 3 cm

Let OF = y cm, OE = 3 + y cm and OD = OB = x cm = Radius.

Consider ΔOFD,

By Pythagoras Theorem,

OD2 = OF2 + FD2

x2 = y2 + 62

x2 = y2 + 36 … (1)

Consider ΔOEB,

By Pythagoras Theorem,

OB2 = OE2 + EB2

x2 = (3 + y)2 + 32

x2 = 9 + y2 + 6y + 9

x2 = y2 + 6y + 18 … (2)

From (1) and (2),

y2 + 36 = y2 + 6y + 18

36 = 6y + 18

6y = 36 – 18

6y = 18

y = 18/ 6

y = 3

Substituting y value in (1),

x2 = y2 + 36

x2 = 32 + 36

x2 = 9 + 36

x2 = 45

[45 = 3 × 3 × 5]

Radius = x = 3√5 cm

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