Saturday 31 July 2010

The radius of a circle is 17 cm. A chord (XY) lies 9 cm from the center and divides the circle into two segments. Find the perimeter of the minor...

We are given
a circle with radius 17 cm. There is a chord 9 cm from the center. We are asked to find the
perimeter of the minor segment formed by the chord.

Let the center of the
circle be O, and the endpoints of the chord be A and B. We can draw a radius perpendicular to
the chord; label the intersection of this radius M.

The perimeter of the
minor segment is the length of minor arc AB plus the length of the chord AB.


First, find the length of the chord. Note that OB = 17 (it is a radius.) OM = 9 since
the definition of the distance of a chord from the center of the circle is the length of the
segment drawn from the center perpendicular to the chord.

Using the
Pythagorean theorem we find that

`MB = 4sqrt(13) => AB
= 8sqrt(13)`

If we measure the central angle `angle AOB`
in radians, the arc length s of the minor arc from A to B is given by `s = r theta` where r is
the radius (17) and `theta` is the measure of the central angle.

Using
right-triangle trigonometry, we find that `m angle BOM = cos^(-1)(9/17)`, so the measure of the
central angle is `m angle AOB = 2cos^(-1)(9/17)` .

Then `s =
17*[2cos^(-1)(9/17)]=34cos^(-1)(9/17)` .

If we measure in degrees, the length
of the arc is `s = (m angle AOB)/360*(2 pi r) = (2cos^(-1)(9/17))/360 * (2 pi 17)`


The perimeter of the minor segment is then:


`p = 8sqrt(13)+34cos^(-1)(9/17)` (where the arccos, or inverse
cosine, is measured in radians), which is approximately 63.28cm.


If we measure in degrees, it is:

`p =
8sqrt(13)+(17pi)/90 cos^(-1)(9/17)`

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