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Topic: POW Solution, Dec. 20-24
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Problem of the Week

Posts: 292
Registered: 12/3/04
POW Solution, Dec. 20-24
Posted: Jan 3, 1994 11:48 AM
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Problem of the Week, December 20-24

A chord of a circle is the hypotenuse of an isosceles right triangle whose
legs are radii of the circle. The radius of the circle is 4 times the
square root of 2. What is the length of the minor arc subtended by the
chord?

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Correct responses were received from:

Erin O'Connor Grade 8, St. Michael Elementary School, Olympia, WA
Dave Eisenreich Grade 12, Shaler Area Senior High School, PA
Bipin Mujumdar Grade ??, Shaler Area Senior High School, PA
Jake Petrasch Grade 10, Dover-Sherborn High School, Dover, MA
Wendy Brian Grade 10, Dover-Sherborn High School, Dover, MA
Gina Metrakas Grade 10, Dover-Sherborn High School, Dover, MA
Lindsay Madeira Grade 10, Dover-Sherborn High School, Dover, MA
Alie Curran Grade 10, Dover-Sherborn High School, Dover, MA
Donny Wright Grade 10, Dover-Sherborn High School, Dover, MA
Jill Marden Grade 10, Dover-Sherborn High School, Dover, MA
Megan McAfoose Grade 10, Dover-Sherborn High School, Dover, MA

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From: "Dick O'Connor" <djo7613@u.washington.edu>

This is my first post, and my Dad is doing the typing, so I hope he gets
it right!

The length of the minor arc is 2 time pi times the square root of 2.

Method: The circumference of a circle is defined as 2 times pi times the
radius. The ratio between the length of a subtended minor arc and the
circumference of the entire circle is equal to the ratio between the
angle formed by the radii which touch each end of the chord and 360 degrees.

(I got that last part from my geometry book. I can't prove it yet.)

The radii here are the legs of a right triangle, so the angle in question
is 90 degrees and the ratio of 90 to 360 is 1:4. Thus, the ratio of
the length of the arc to the circle circumference is 1:4in this case.

Circumference is 2 times pi times (4 times the square root of 2)
One-fourth of that is the length of the minor arc, or 2 times pi times root 2.

Erin O'Connor
Grade 8
St. Michael Elementary School
Olympia, WA

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From: dave eisenreich <eisenrei@one.sasd.k12.pa.us>

Hi! My name is Dave Eisenreich, I am a senior at Shaler Area
Senior HighSchool.

First, I drew a picture to illustrate the problem. Realizing the two
legs were congruent, and it was a right angle triangle, I used the fact
that THE ratio of the sides are 1:1:sqr. rt. of 2. I found the chord to
be 8 un its long. Then using the equation of the length of an arc,

(measure of the angle) times (Radius),
4 times rt. of 2 times pi/2= 2pi(root of 2)

My final answer was 2pi(root of 2).

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From: bipin mujumdar <mujumdar@one.sasd.k12.pa.us>

To solve the problem I first drew a picture. I then drew the right
isosceles triangle and put in the chord. Then I labeled the two equal
sides. Next, I drew in the rest of the triangles and discovered that
the finished result was a sqaure. Since each side of the square was
equal, I knew that the arc subtended by each hypotenuse had to be
equal. Since there are 4 sides to a square I divided the circumfrence
of the circle by four. So I took 2 times pi times (4xsqr 2)-radius.
After solving it I found the final answer to be 8.8857.using pi to
the 9th place.

Bipin Mujumdar Shaler Area Senior Highschool





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