Suppose the diameter of a circle is 30 centimeters long and a chord is 24 centimeters long. How do you find the distance between the chord and the center of the circle?

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Suppose the diameter of a circle is 30 centimeters long and a chord is 24 centimeters long. How do you find the distance between the chord and the center of the circle?

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Answer 1 · 2016-04-09T03:52:00

9 cm.

Explanation:

If AB is the chord, M is its midpoint and C is the center of the circle,
the distance between the chord and the center is $C M = \sqrt{{(C A)}^{2} - {(A M)}^{2}} = \sqrt{15^{2} - 12^{2}} = \sqrt{81} = 9$ $CM = sqrt((CA)^2-(AM)^2)=sqrt(15^2-12^2)=sqrt81=9$
CA = radius of the circle = 15 cm and AM = (length of the chord) $/$ $/$ 2 = 12 cm

Answer 2 · 2016-04-10T14:30:17

$9 c m$ $9cm$

Explanation:

Consider the image

enter image source here

Let the distance between the chord and centre of the circle be $x$ $x$

We need to find $x$ $x$

For that we need to recreate this image

enter image source here

Now we have formed a right angle triangle

Now the problem has become easy!

Use Pythagoras theorem

$a^{2} + b^{2} = c^{2}$ $color(blue)(a^2+b^2=c^2$

Where,

$a and b$ $aand b$ are the right containing sides ( $x and 12$ $xand12$ )

$c$ $c$ is the Hypotenuse (longest side- $15$ $15$ )

$\to x^{2} + 12^{2} = 15^{2}$ $rarrx^2+12^2=15^2$

$\to x^{2} + 144 = 225$ $rarrx^2+144=225$

$\to x^{2} = 225 - 144$ $rarrx^2=225-144$

$\to x^{2} = 81$ $rarrx^2=81$

$\Rightarrow x = \sqrt{81} = 9$ $color(green)(rArrx=sqrt81=9$

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Suppose the diameter of a circle is 30 centimeters long and a chord is 24 centimeters long. How do you find the distance between the chord and the center of the circle?

2 Answers

Explanation:

Explanation:

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