Question

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Answer 1

Hey mate!

The Answer to your question is :

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Two chords AB and CD of length 5 cm and 11 cm respectively of a circle are parallel to each other and are on the same side of the centre.If the distance between AB andCD is 3 cm, find the radius of the circle.

The perpendicular from the centre of a circle to a chord bisects the chord.

Radius is the line segment joining the centre and any point on the circle is called the radius of the circle. All radius have a same length in a circle.

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

From ΔONA and ΔOMC

OA² =ON² +AN²

b²=a² + (2.5)².........(i)

and OC² = OM²+CM²

b²= (6-x)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

6.25 = -12a+ 66.25

12a = 66.25 - 6.25

12a = 60

a= 60/12

a= 5

Put a = 5 in eq i,

b²= 5²+ (2.5)²

b²= 25 + 6.25

b² = 31.25

b= √31.25

b= 5.6 (approx)

RADIUS=b= 5.6cm (approx)

Hence, radius of the circle is 5.6 cm (approx).

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