Question

Two chords ab and cd of lengths 5 cm and 11cm respectively of a circle are parallel. if the distance between the two chords is 11 cm, find the radius of the circle.

Answer 1

❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️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