Geometry circle circumference problem please!!!?

Question

Find, to the nearest cm, the circumference of a circle in which an 80-cm cord is 9 cm from the center.

here is the diagram:::

http://i65.photobucket.com/albums/h239/beast912/weirdpendalum.jpg

please and thanks

Answer 1

You do know you put this question in Geography, and not Geometry, right?

Answer 2

by pythagorean theorem:

the chord is 80 if we divide it by 2  = 40 cm (base) line from the center creating perpendicular which is 9cm (height)

and hypotenuse would be the radius:

r =sq. rt. (40^2 +9^2) = sq. rt (1681) = 41

circumference C = 2(pi)(r) = 2 (3.1416)(41)

C = 257.61 cm

Answer 3

Use Pythagoras:

80mm chord divided by 2 = 40 mm for segment A.

(which would be the left portion of the chord)

9mm equals segment B.

(which would be the distance from the chord to the center)

segment C = sqrt(A^2 + B^2) = sqrt(1681) = 41mm radius.

Then, to find circumference, it is 2*pi*radius, which gives you 257.6mm

Answer 4

Let the center of the circle be O, let the chord be AB, and let a perpendicular from O to AB intersect AB at P.  Then OPA is a right triangle in which OP is 9 cm and PA is half the length of the chord, or 40 cm.  Therefore by the Pythagorean Theorem,

OA = √((9 cm)^2 + (40 cm)^2) = √(1681 cm^2) = 41 cm.

OA is equal to the radius of the circle, so the circumference of the circle is 2π(41 cm) = 82π cm, or to the nearest centimeter, 258 cm.