Question

2.
Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel
to each other and are on opposite sides of its centre. If the distance between AB and
CD is 6 cm, find the radius of the circle.​

Answer 1

Answer:

5.6 cm

Step-by-step explanation:

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).

#Hope my answer will help you!