AB and CD are two chords of a circle such that AB = 10 cm, CD = 24 cm and AB || CD. The distance between AB and CD is 17 cm. Find the radius of the circle.

Given chords AB = 10 cm, CD = 24 cm

Distance between AB and CD = 17 cm

From the figure,

OE ⊥ AB and OF ⊥CD.

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

⇒ AE = EB = 1/2 AB = 5 cm

⇒ CF = FD = 1/2 CD = 12 cm

Let OF = y cm, OE = 17 - y cm and OB = OD = x cm = Radius.

Consider ΔOEB,

By Pythagoras Theorem,

⇒ OB² = OE² + EB²

⇒x² = (17 – y)² + 5²

⇒x² = 289 + y² – 34y + 25

⇒x² = y² – 34y + 314 … (1)

Consider ΔOFD,

By Pythagoras Theorem,

⇒ OD² = OF² + FD²

⇒x² = y² + 12²

⇒x² = y² + 144 … (2)

From (1) and (2),

⇒ y² – 34y + 314 = y² + 144

⇒ -34y + 314 = 144

⇒ 34y = 314 – 144

⇒ 34y = 170

⇒ y = 170/ 34

∴ y = 5

Substituting y value in (2),

⇒x² = 5² + 144

⇒x² = 25 + 144

⇒x² = 169

∴ Radius = x = 13 cm