Question

plsss send me the answer fast

Answer 1

Answer:

Step-by-step explanation:

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.

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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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Hope this will help you....

Answer 2

The answer of this question is in the image.