Find the distance from the center of a circle to a chord 30 m long if...?

Question

...the diameter of the circle is 34 m?

Please explain!!!

Answer 1

Call it Chord AB and center of circle is C. Also call midpoint of AB M.

AB = 30, so AM = 15
CA = 17 (it's a radius)

M is not only the midpoint of Chord AB, but CM is perpendicular to AB. You can show this in many ways geometrically, but there is even a convenient little theorem in most geometry books that says "if a segment contains the center of a circle and bisects a chord, then it's perpendicular to the chord." The converse of this theorem also exists, by the way. So you have a right triangle containing CM, the distance from center to chord. 15² + (CM)² = 17² --> CM = 8

Answer 2

Draw a picture (can't do that in Yahoo Answers). Draw a circl with a chord. At the 2 points the chord intersects the circle, draw the radii. We now have an isoceles triangle with equal legs of 17 (the radii) and base 30. Draw the altitude from the center to the chord, which bisects the chord. Now we have 2 right triangles with one leg 15 and hypotenuse 17. By Pythagorean Theorem, the altitude = SQUAREROOT(17^2 - 15^2) = 8

Answer 3

If you draw a perpendicular from the center of the circle to the chord it will meet it at the midpoint of the chord. So now we have two right triangles. Let's look at one of them.

r = hypotenuse and also radius
c = one leg (half the chord length)
x = second leg (distance from center of circle)

x² = r² - c² = (34/2)² - (30/2)² = 17² - 15² = 289 - 225

x² = 64

x = 8

The distance from the center of the circle to the chord is 8 m.

Answer 4

Draw a perpendicular from center to chord.
Radius is 17, 1/2 chord = 15. Use pythagorean theorem.
17^2 = 15^2 + x^2

x = 8

Answer 5

I'll give you a hint:
diameter = 34, which implies the radius = 17
you now have an isoceles triangle with base 30 (the chord).
You can then split this into two right angle triangles, now use Pythagoras...