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.

Found 2 solutions by jim_thompson5910, solver91311:
Answer by jim_thompson5910(35256) About Me  (Show Source):
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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.

Answer by solver91311(24713) About Me  (Show Source):
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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

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