SOLUTION: Suppose the diameter of a circle is 26 ft. long and a chord is 10 ft. long. Find the distance from the center of the circle to the chord. Show your work.
Algebra ->
Circles
-> SOLUTION: Suppose the diameter of a circle is 26 ft. long and a chord is 10 ft. long. Find the distance from the center of the circle to the chord. Show your work.
Log On
Question 613306: Suppose the diameter of a circle is 26 ft. long and a chord is 10 ft. long. Find the distance from the center of the circle to the chord. Show your work.
Note: the diameter is 26, so the radius is 26/2 = 13. Also, the chord is bisected by the diameter, which allows us to cut that chord of 10 in half to get 2 pieces of 5.
So we can see we have a right triangle. Let's use the pythagorean theorem to solve for x
a^2+b^2 = c^2
5^2+x^2 = 13^2
25+x^2 = 169
x^2 = 169-25
x^2 = 144
x = sqrt(144)
x = 12
So the distance from the center to the chord is 12 feet.
Let represent the center of the circle. Let and be the endpoints of the 10ft chord. Construct a radius that is the perpendicular bisector of the chord. Name the point of intersection of the radius and the chord . Construct another radius through point .
Since the radius through is a bisector of , . Since , is a right triangle. Since , which is the hypotenuse of the right triangle, is a radius, .
Now that you have the hypotenuse and one leg of a right triangle use Pythagoras to calculate the measure of the other leg.
John
My calculator said it, I believe it, that settles it