Two chords AB, CD of lengths 5cm and 11cm respectively of a circle are parallel. if the distance between AB and CD is 6 cm, find the radius of the circle​

Given: AB = 11cm

CD = 5cm

AB ║ CD

Draw OP ⊥ AB and OQ ⊥ CD

PQ = 3cm

To Find: Radius (r) of the circle.

Sol: In triangle AOP

AP = 1/2 AB (The perpendicular from the centre of the circle bisects the chord.)

AP = 1/2 x11 = 5.5cm

Let OP = x cm

OA² = AP² + x²

r² = (5.5)² + x²

r² - (5.5)² = x²

In ΔCOQ,

CQ = 1/2 CD   (theorem)

CQ = 1/2 x 5 = 2.5

OC² = CQ² + OQ²

r² = (2.5)² + (x+3)²

Now,

(5.5)² + x² = (2.5)² + (x+3)²

(5.5)² - (2.5)² = (x+3)² (x)²

8x3 = (2x+3) (3)

∴2x + 3 = 8

x = 5/2 = 2.5 cm

Substituting x = 2.5

r² = (5.5)² + (2.5)²

r² = 36.5

r = √36.5 = 6.04 cm

*Hope it helps*

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Answer

4.4/5

11

Given: AB = 11cm

CD = 5cm

AB ║ CD

Draw OP ⊥ AB and OQ ⊥ CD

PQ = 3cm

To Find: Radius (r) of the circle.

Sol: In triangle AOP

AP = 1/2 AB (The perpendicular from the centre of the circle bisects the chord.)

AP = 1/2 x11 = 5.5cm

Let OP = x cm

OA² = AP² + x²

r² = (5.5)² + x²

r² - (5.5)² = x²

In ΔCOQ,

CQ = 1/2 CD   (theorem)

CQ = 1/2 x 5 = 2.5

OC² = CQ² + OQ²

r² = (2.5)² + (x+3)²

Now,

(5.5)² + x² = (2.5)² + (x+3)²

(5.5)² - (2.5)² = (x+3)² (x)²

8x3 = (2x+3) (3)

∴2x + 3 = 8

x = 5/2 = 2.5 cm

Substituting x = 2.5

r² = (5.5)² + (2.5)²

r² = 36.5

r = √36.5 = 6.04 cm

*Hope it helps*

eddibear3a and 105 more users found this answer helpful

THANKS  

49

4.3

(56 votes)

Answer

4.4/5

11

Step-by-step explanation:

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