Question

Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are opposite side of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.

Answer 1

Given : Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are opposite side of its centre and  the distance between AB and CD is 6 cm.

Let there is a circle having center O and let radius is b .

Draw ON perpendicular to AB and OM perpendicular to CD.

Now since  ON perpendicular to AB and OM perpendicular to CD and AB || CD.

So N, O,M are collinear.

Given distance between AB and CD is 6.

So MN = 6

Let ON = a, then OM = (6 - a)

Join OA and OC.

Then, OA = OC = b

Since we know that perpendicular from the centre to a chord of the circle bisects the chord.

and CM = MD = 11/2 = 5.5

AN = NB = 5/2 = 2.5

In ΔONA and ΔOMC  ,

OA² = ON² + AN²  

b² = a² + (2.5)².........(i)

and OC² = OM² + CM²

b² = (6 - a)² + (5.5)²......(ii)

From eq i and ii we get  :

a² + (2.5)² = (6 - a)² + (5.5)²

a²  + 6.25 = 36 + a² - 12a + 30.25

[(a - b)² = a² - 2ab + b²]

6.25 = -12a + 66.25

12a = 66.25 - 6.25

12a = 60

a = 60/12

a = 5

On putting a = 5 in eq i,

b² = 5² + (2.5)²  

b²  = 25 + 6.25

b² = 31.25

b = √31.25

b = √3125/100

b = √625/20

b = √125/4  

b = √(25 × 5)/4

b = 5√5/2 cm

RADIUS =  5√5/2 cm

Hence, the radius of the circle is 5√5/2 cm. HOPE THIS ANSWER WILL HELP YOU…..

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Answer 2

Let O be the centre of the given circle and let its radius be cm.

Draw ON ⊥ AB and OM⊥ CD since, ON ⊥ AB, OM ⊥ CD and AB || CD, therefore points N, O, M are collinear.

Let ON = a cm

∴ OM = (6 – a) cm

Join OA and OC.

Then, OA = OC = b c m

Since, the perpendicular from the centre to a chord of the circle bisects the chord.

Therefore, AN = NB= 2.5 cm and OM = MD = 5.5 cm

In ∆OAN and ∆OCM, we get

OA² = ON²+ AN²

OC² = OM² + CM²

⇒ b² = a² + (2.5)²

and, b² = (6-a)² + (5.5)² …(i)

So, a² + (2.5)² = (6 – a)² + (5.5)²

⇒ a² + 6.25= 36-12a + a² + 30.25

⇒ 12a = 60

⇒ a = 5

On putting a = 5 in Eq. (i), we get

b² = (5)² + (2.5)²

= 25 + 6.25 = 31.25

So, r = √31.25 = 5.6cm (Approx.)