# AB and CD are two parallel chords of a circle such that AB = 10cm and CD = 24. The chords are on opposite sides of the centre and distance between them is17cm. Find the radius of the circle Given AB = 10 cm, CD = 24 cm and distance between their center is 17 cm.

Let OQ = y cm and radius of circle = OB = OD = x cm

Therefore, OP = (17 - y) cm

OP ⊥ AB, (Perpendicular from the centre of the circle to a chord bisects the chord)

OQ ⊥ CD (Perpendicular from the centre of the circle to a chord bisects the chord)

In right ∆OPB,

OB2 = OP2 + PB2 (Pythagoras Theorem)

⇒ x2 = (17 - y)2 + (5)2  .... (1)

Similarly,

In right ∆OQD,

OD2 = OQ2 + QD2 (Pythagoras Theorem)

x2 = y2 + 122  ... (2)

On equating (1) and (2), we get

(17 - y)2 + (5)2 = y2 + 12

⇒ 289 + y2 -34y + 25 = y2 + 144

-34y + 314 = 144

-34y = 144 - 314

-34y = -170

y = 5

On putting the value of y in (2), we get

x2 = 52 + 122

x2 = 25 + 144

x2 = 169

x = 13 cm

Thus, radius of the circle is 13 cm.

• 92

Let the distance between( the perpendicular) be MN,

AB=10 CD=24 and MN=17

Then, AM=5, CN=12

Let the distance from centre O to M be x, then the distance from centre O to N will be (17-x)

and let the radius be r.

Therefore, using the pythagoras theorem,

52 + x2

• -25

Let the distance between( the perpendicular) be MN,

AB=10 CD=24 and MN=17

Then, AM=5, CN=12

Let the distance from centre O to M be x, then the distance from centre O to N will be (17-x)

and let the radius be r.

Therefore, using the pythagoras theorem,

52 + x2= r2 and 122 + (17-x)2= r2

therefore, 52 + x2 = 122 + (17-x)2

25 + x2= 144+ 172 + x2 - 2(17*x)

25 = 144 + 289 -34x

34x= 408

x= 408/34

x=12

then using the pythagoras theorem,

52 + 122 = r2

25+ 144= r2

r2 = 156

therefore, r= 13cm

• 1
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