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Topic: Geometry POW Solution, November 15
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Geometry Problem of the Week

Posts: 159
Registered: 12/6/04
Geometry POW Solution, November 15
Posted: Dec 21, 1996 11:48 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

******************************
Geometry Problem of the Week, November 11-15

You might have to read this a couple of times and draw a picture before it
makes any sense!

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 6 times the
square root of 2. What is the length of the minor arc subtended by the
chord?

What if it were an equilateral triangle instead of just isosceles?

What if I told you that the non-base angle (so the angle at the vertex at
the center of the circle) was x degrees? Then what would the answer be?

This problem has lots of parts, but I think if you really understand the
first part, the second two won't be too bad. Make sure you really explain
your answer!

******************************

A pretty good job this week! 80 correct solutions were submitted, and 35
incorrect. The majority of the incorrect solutions were missing the third
part - it is sometimes hard to generalize something even though you used
it correctly to solve two other parts.

One common mistake was to get the third part backwards. That is, dividing
by x/360 instead of multiplying by it. This is after doing exactly the
first thing in the first two parts. You need to look very carefully at
what you did in the specific cases.

Several people used a formula for arc length that requires the length of
the chord and the angle. It works okay for the first two parts, but makes
the
third part harder, I think.

For accuracy, leave everything in terms of pi and sqrt2 until the very
end. You might also find that a lot of stuff cancels out before you're
done, and
to me, the less numbers, the better! In fact, with a problem like this, I
like to see the answers in terms of pi and sqrt2 _and_ as just numbers. It
makes the answer easier to follow and connects it with the circle and the
original radius.

This problem is a good time to work on writing good equations. Explain
carefully what you are going to do, and where the numbers come from, but
when it comes to presenting the numbers, write some equations.

I've highlighted two solutions this week. Thomas Bereknyei from Hillview
Junior High gave a simple explanation. Greg Moore and Shawn Phelen
from Shaler Area High School pretty much explained it perfectly. There are
also a number of good solutions in the full list of solutions, so be sure
you read them over.

The following students submitted correct solutions this week. A few of the
solutions are highlighted below. Most of the rest of the solutions are
also available.

Zack Subin Grade 10, Atlantic City HS, Atlantic City, New Jersey
Josh Grochow Grade 8, Georgetown Day School, Washington, DC
Katherine Walther Grade 9, John Glenn Junior High
Clayton Smith Grade 11, Lambton-Kent Composite, Chatam, Ontario, Canada
Thomas Bereknyei Grade 8, Hillview Junior High, Pittsburgh, California
--
Brian Gordon Grade 1992, Dartmouth
Justin Lam Grade 8, Sequoia Middle School, San Francisco, California
Jason Yeung Grade 10, Iolani School, Honolulu, Hawaii
Chris Collora Grade 9, Roselle Park High School, Roselle Park, New Jersey
Alicia Carbone and Kate McTernan
Grade 9, Roselle Park High School, Roselle Park, New Jersey
--
Carmen Castillo and Huma Safdar
Grade 9, Roselle Park High School, Roselle Park, New Jersey
Ayan Bose Grade 9, Roselle Park High School, Roselle Park, New Jersey
Jenny Collins Grade 9, Roselle Park High School, Roselle Park, New Jersey
Melanie Griffin Grade 9, Roselle Park High School, Roselle Park, New Jersey
Bryan Lee and Maxine Kwan and Ben Ball and Hayden Edwards
Grade 9, The Science Academy at LBJ HS, Austin, Texas
--
Candice Diaz Grade 9, Roselle Park High School, Roselle Park, New Jersey
Allie Hess and Laura Merkel
Grade 9, Roselle Park High School, Roselle Park, New Jersey
Kathy Ryan and Katie Shanley
Grade 9, Roselle Park High School, Roselle Park, New Jersey
Joshua Goldstein Grade 9, Roselle Park High School, Roselle Park, New Jersey
Michael Jose Grade 10, Roselle Park HS, Roselle Park, New Jersey
--
Rich Deo Grade 10, Roselle Park HS, Roselle Park, New Jersey
Christine Joerge Grade 10, Roselle Park HS, Roselle Park, New Jersey
Karen McKinney Grade 10, Roselle Park HS, Roselle Park, New Jersey
Sarah Hartley & Jill Matthews
Grade 10, Roselle Park HS, Roselle Park, New Jersey
Arnette Wheeler Grade 10, Roselle Park HS, Roselle Park, New Jersey
--
Dimple Patel Grade 10, Roselle Park HS, Roselle Park, New Jersey
Liz Miller Grade 10, Roselle Park HS, Roselle Park, New Jersey
Craig Frost and Shawn McCoy and Jerry Elliott
Grade 10, Roselle Park HS, Roselle Park, New Jersey
Monish Desai Grade 10, Roselle Park HS, Roselle Park, New Jersey
Lorena Reyes and Soraya Jallad
Grade 10, Roselle Park HS, Roselle Park, New Jersey
--
Paul Johnson and David Modero
Grade 10, Roselle Park HS, Roselle Park, New Jersey
Rich Petz and Pat O'Connell
Grade 10, Roselle Park HS, Roselle Park, New Jersey
Tammy Heskeyahu and Purvi Patel and Maria Pineiro
Grade 10, Roselle Park HS, Roselle Park, New Jersey
Mick Lorusso Grade 9, Ignacio High School, Ignacio, Colorado
Bob Young and Christian Paul
Grade 10, Shaler Area High School, Pittsburgh, Pennsyvania
--
Amy Bruecken and Alexis Sauter
Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania
Natalie Navarro and Jackie Rose
Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania
Becky Spinella and Erin May
Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania
Adam Yacono and Gretchen Schneider
Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania
Kristen Pirozzi and Jessica Thompson
Grade 9, Roselle Park High School, Roselle Park, New Jersey
--
Kyle Haire and Trevor Glasgow
Grade 10, South High School, Bakersfield, California
Nicole Forostoski Grade 10, Martin County High School, Stuart, Florida
Jessica Grade 10, South High School, Bakersfield, California
Jen Baer and Brett Bernardo
Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania
Gretchen Ross and Pranav Shetty
Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania
--
Melissa Antoszewski and Lynn DeLuca
Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania
Greg Moore and Shawn Phelen
Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania
Andrew Miller and Sarah Kost
Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania
Katie Behling and Chris Vendilli
Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania
Lisa Jasneski and Richard Erb
Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania
--
Keith Dougall and Jenny Booth
Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania
Kurt Davies Grade 10, Smoky Hill High School, Aurora, Colorado
Mike Rogers Grade 9, Smoky Hill High School, Aurora, Colorado
Michael Shulman Grade 11, North Hollywood HS, Hollywood, California
Cassie Gorish Grade 9, Burroughs High School, Ridgecrest, California
--
Roger Mong Grade 8, Zion Heights JHS, Richmond Hill, Ontario, Canada
Sheridan Waller Grade 10, Smoky Hill High School, Aurora, Colorado
Jennifer Kaplan Grade 6, Castilleja Middle School, Palo Alto, California
Bob Jackson Grade ,
Gavin Calkins Grade 8, Waluga Junior High School, Portland, Oregon
--
Terry Haslam Grade 11, Southern Trinity HS, Mad River, California
Becky Dunlap and Lauren Kupersmith and Swathi Bala
Grade 9, Germantown Academy, Fort Washington, Pennsylvania
Adam Hobson Grade , Brookstone School, Columbus, Georgia
YA   Grade , Smoky Hill High School, Aurora, Colorado
Daniel Ornstein Grade 8, Georgetown Day School, Washington, DC
--
Hector Mercedes Grade 12, George Wingate HS, Brooklyn, New York
Thomas Kuo Grade 9, Burroughs High School, Ridgecrest, California
Katie Quinn-Kerins Grade 10, Germantown Academy, Fort Washington, Pennsylvania
Giscard Pongnon Grade 12, George Wingate HS, Brooklyn, New York
Renee Grade 9, Smoky Hill High School, Aurora, Colorado
--
Kirk Vissat Grade 10, Smoky Hill High School, Aurora, Colorado
Leif Linden Grade , Germantown Academy, Fort Washington, Pennsylvania
Ryan Baker Grade 9, Newport High School, Bellevue, Washington
Jonathan Biderman Grade 8, Georgetown Day School, Washington, DC
Jon Cooper Grade 8, Georgetown Day School, Washington, DC
--
Oluseyi Ojeifo Grade 8, Georgetown Day School, Washington, DC
Zoe Durner-Feiler Grade 6, Sand Lake Elementary School, Anchorage, Alaska

***************************************

From: ZSUBIN@AOL.COM

From: Zack Subin
Grade: 10
School: Atlantic City HS, Atlantic City, New Jersey

If the legs are the radii of the circle, then the hypotenuse must be the chord,
and the central angle must be 90 degrees.
The minor arc between two radii is equal to the same fraction of the
circumfrence as the fraction created by the angle over 360 degrees.
Therefore, if the radius is equal to 6 times the square root of 2, the
circumfrence is equal to 12pi times the square root of 2 (2*pi*R).
The central angle (90 degrees) over 360 degrees equals one fourth, which,
multiplied by the circumfrence, equals 3pi times the square root of 2.
If the triangle was equilateral, then all three angles, including the central
angle, would be 60 degrees. 60/360 times the circumfrence equals 2pi square
root of 2.
Finally, if the angle was x degrees, then the arc would equal (x / 360)(12pi
times the square root of 2).
However, if the angle was greater than 180 degrees, the minor arc would be (360
- x)/360 * 12pi*square root of 2.


***********************************************

From: joshg2099@aol.com

From: Josh Grochow
Grade: 8
School: Georgetown Day School, Washington, DC

We shall call the point at the center of the circle point A, the
circumference of the circle shall be C and the radius shall be R.
R = 6 times the square root of 2. C=2 times pi times R. So,
C=12 times the square root of 2 times pi. The total number of
degrees in a circle is 360, and since the two legs of the
triangle are the radii of the circle and angle A is 90 degrees,
the minor arc subtended by the hypotenuse of the triangle (also
the chord) is one quarter of C, which is 3 times root 2 times pi.
If the triangle were equilateral, then it would also be
equiangular, so angle A = 60 degrees. Since 60 is 1/6 of 360,
the minor arc subtended by the chord would be 1/6 of C, which is
2 times root 2 times pi.
If angle A = x, then the minor arc subtended by the chord would
be equal to (12 times root 2 times pi) divided by 360/x.



Josh Grochow
Grade 8
Paul Nass
Georgetown Day School
November 11-15, 1996


***********************************************

From: NRGM22A@prodigy.com

From: Katherine Walther
Grade: 9
School: John Glenn Junior High

To figure this problem out, all you need to know is the measure of the
vertex
angle of the triangle and the radius of
the circle.
The measure of the minor arc created by the chord is equal to the measure
of the
vertex angle over 360 times the circumference of the circle.

minor arc = 90/360 (6.28x 6 times the square root of 2)
minor arc = 1/4 (53.2876)

The first answer is 13.3219

Using this formula, we can easily find the second answer, because we know that
all the angles in an equilateral triangle
are equal to 60.

minor arc = 60/360 (circumference from problem 1)
minor arc = 1/6 (53.2876)
The minor arc of the equilateral triangle would be 8.8813.

The same can be done with X if it is the vertex angle.

minor arc= X/360(2Pi x radius)
The length of the minor arc would be X/360(53.2876)


***********************************************

From: keither@ciaccess.com

From: Clayton Smith
Grade: 11
School: Lambton-Kent Composite School, Chatam, Ontario, Canada

Since the triangle in the first part of the problem is a right
triangle, the angle of the arc subtended by the chord is 90
degrees or pi/2 radians. Since the length of the arc is 6 times
the square root of 2, the length of the arc is 3 times the
square root of two times pi. If the triangle were isosceles,
the angle subtended would be 60 degrees or pi/3 radians.
Therefore, the length of the arc would be 2 times the square
root of two times pi. Were the angle x degrees, the subtended
arc would have a length of the square root of two times x times
pi over 30. These calculations all are based on the fact that
the length of an arc is equal to the radius of the circle times
the represented by the arc.


***********************************************

From: imre_bereknyei@ccgate.apl.com

From: Thomas Bereknyei
Grade: 8
School: Hillview Junior High, Pittsburgh, California

Subject: Problem of The Week Solution from Thomas B

1. The answer to the first question is:13.32
Here is why:
Because the non-base angle is 90 degrees (right triangle) the arc would
be 1/(360/90)=1/4 of the circle's circumference.
The cicrcle's circumference = 2r Pi = 2 * 6* SQRT(2) * 3.14= 53.28
The length of the minor arc is: 53.28 / 4 = 13.32

2. Instead of dividing by 4, now we divide by 360/60=6
The length then of the minor arc is: 53.28 / 6 = 8.88

3. We already know the circumference is 53.28.
360 / x is what you would divide from 53.28 so the length of the minor
arc is:
53.28 * x
------------ in terms of x which is the angle
360

Solution by Thomas Bereknyei 11 years old 8th grade student at Hillview
Jr. High School
Addr: 2061. Biscay Dr. Pittsburg CA 94565
e-mail addr: imre_bereknyei@ccgate.apl.com


***********************************************

From: bmgordon@ntplx.net

From: Brian Gordon
Grade: 1992
School: Dartmouth

Part one is a quarter of a circle, since you have a 90 degree central
angle. So
the arc is 1/4 timesthe circumference of the circle:

1/4 * 2 * pi * 6 * sqrt(2) = 3sqrt(2) * pi

For an equilateral triangle, the angle is 60 degrees, so we have 1/6 of a
circle. The arc is

1/6 *2 * pi * 6 * sqrt(2) = 2sqrt(2) * pi

For an angle of x degrees, we have x/360 of a circle, so it's

x/360 * 2 * pi * 6sqrt(2) = x*sqrt(2)*(1/30)*pi.

--bri


***********************************************

From: quan.lam@ucop.edu

From: Justin Lam
Grade: 8
School: Sequoia Middle School, San Francisco, California

The circumference of the circle with radius of 6SQRT(2) is 2(PI)(6SQRT(2))=
12SQRT(2)PI.

(PART 1) Since 90 degree is 1/4 of 360 degree, the lentgth
of the minor arc is (1/4)(12SQRT(2)PI) = 3SQRT(2)PI.

(PART 2) If the triangle is equilateral, then the vertex angle
is 60 degree which is 1/6 of 360 degree. Therefore,
the length of the minor arc is (1/6)(12SQRT(2)PI) = 2SQRT(2)PI.

(PART 3) If the vertex angle is x degree which is (x/360) of 360 degree.
Therefore, the length of the minor arc is
(x/360)(12SQRT(2)PI) = xSQRT(2)PI/30.


***********************************************

From: arthur@iolani.honolulu.hi.us

From: Jason Yeung
Grade: 10
School: Iolani School, Honolulu, Hawaii

Answer: 3 (sqrt 2) * pi
2 (sqrt 2) * pi
x (sqrt 2) * pi / 30

The isosceles right triangle's right legs are the radii of the
circle, thus the angle is a central angle, which implies if the
angle is 90 degrees, then the angle of the arc is also 90 degrees.
The circum. of the circle is 12 sqrt 2 pi. The arc is 90 degrees
over 360 degrees (the whole circle), or 1/4 the circum. of the
circle. Therefore, the length of the minor arc is 3 (sqrt 2) * pi.

If the triangle is equilateral instead, then the central angle
would be 60 degrees, and the arc would be 1/6 of the total circle.
Therefore, the length of that minor arc is 2 (sqrt 2) * pi.

If the non base angle is x degrees, then the central angle would
be x degrees, and the arc would be x/360 of the total circle.
Therefore, the length of that arc is x (sqrt 2) * pi / 30.


***********************************************

From: 74620.2745@compuserve.com

From: Chris Collora
Grade: 9
School: Roselle Park High School, Roselle Park, New Jersey

1. First I found that 8.5 was the radii. Then using the central angle, I
found
the ratio of it to the circumfenrce was 90/360=1/4. Next I found the
circumference by using pi*D, 53.4, then multiplied it by the 1/4 ratio to get
the arc. The arc was 13.35.

2. I followed all the same steps in the second one that I used in the first.
However the central angle was 60 so the ratio was 60/360=1/6. The arc =8.9.

3. Because the central angle is x, it could be any size. The ratio would be x/
360 and the arc would be x/360*53.4.


***********************************************

From: 74620.2745@compuserve.com

From: Alicia Carbone and Kate McTernan
Grade: 9
School: Roselle Park High School, Roselle Park, New Jersey

A circle has 360. The radius of this one rounds off to 8.5. An isosceles
right
triangle has 2 = sides. One < is 90. 360 divided by 90 is 4 so the isosceles
triangle is 1/4 of the circle. The we got the circumference of the whoe circle
= 53.38. Since the triangle takes up 1/4 of the circle, we difided 53.38 by 4
and got 13.345, so the minor arc = 13.345.

An equialteral triangle has 180 and each < is 60. We divided 360 by 60 to see
how many triangles would fit in the circle. We got 6. We got c=53.4 and
divided by 6. This arc = 8.9.

For the last part we would divide 360 by x since for the 1s one we divided 360
by 90 and the 2nd one we divided by 60 then we divide the 53.38 by 360/x.


***********************************************

From: 74620.2745@compuserve.com

From: Carmen Castillo and Huma Safdar
Grade: 9
School: Roselle Park High School, Roselle Park, New Jersey

A circle is 360. The radius here is 8.5. Divide the 360 by 90=4. So the
triangle is 1/4 of the circle. C=53.38. Since the triangle takes up 1/4
of the
circle divide 53.38 by 4 = 13.3, so the minor arc is 13.3.

We divided 360 by 60 for the equilateral triangle and we got 6. We got the
circumference to be 53.4. Then we divided by 6 and got 8.9 to be the answer.

Fopr the last part, we would divide 360 by x and divide c by the answer.


***********************************************

From: 74620.2745@compuserve.com

From: Ayan Bose
Grade: 9
School: Roselle Park High School, Roselle Park, New Jersey

The length of the arc is 13.32864881. I found that by using the formula for
circumference; c=2*pi*r=53.3 and multiplying that by .25 because that arc
is 1/4
of the circ.e

If it was an equilateral triangle, the arc would be /6 * 2*pi*r = 8.885176.

If the non-base angle was x, the answer would be x/360*2*pi*8.481281374.


***********************************************

From: 74620.2745@compuserve.com

From: Jenny Collins
Grade: 9
School: Roselle Park High School, Roselle Park, New Jersey

If the angle of an isosceles triangle fitted into circle = 90 and circle = 360
then the angle is a quarter of the circle. In relation, the subtended arc is a
quarter of the circumf=1/4(53.3 = 13.326.

If the angle of an equilateral triangle, fitted into circle=60 and the circle
equals 360 then the < is a sixth of the circle. In relation, the subtended are
is a sixth of the circumference = 1/6(53.3)=8.883

If a non-base angle = x then x/360* circumference = measure of the
subtended arc
= x/360(53.3)


***********************************************

From: 74620.2745@compuserve.com

From: Melanie Griffin
Grade: 9
School: Roselle Park High School, Roselle Park, New Jersey

The non-base < in which the hyp. is a cord of the circle, measures 90. Since
there are 360 in a circle, this angle is 1/4 of the circle. Therefore, the
subtended arc is 1/4 of the circumference. C=2*pi*r. 4=8.48. Therefore c=
53.28745703. 1/4 of the circumf. = 13.32189176

In the circle with 60 central angle. This is 1/6 of the circle.
Therefore, the
measure of the subtended arc is 1/6 of the circumference. The circumf. is the
same, so 1/6(53.28756703)=8.881261172.

If the measure of the angle is x, then the subtended arc is x/360*circumference
of the circle.


***********************************************

From: bryan@thepentagon.com

From: Bryan Lee and Maxine Kwan and Ben Ball and Hayden Edwards
Grade: 9
School: The Science Academy at LBJ HS, Austin, Texas

The length of the minor arc subtended by the chord of an isosceles
in a circle with a leg of 6 times the square root of 2 is 3 pi times
the square root of 2, because the circumfrence of the circle is 12
pi times the square root of 2, or 2 times the radius (6 times the
square root of 2) times pi. Since the non-base angle of the
isosceles triangle is 90, then the length of the arc subtended by
the hypotenuse is one fourth the entire circumfrence of the circle,
because 90 degrees is one fourth of 360, the total degrees in a
circle.
The length of the minor arc subtended by the hypotenuse of an equilateral
in a circle with a side of 6 times the square root of 2 is 2 pi times
the square root of 2. Since the non-base angle of the equilateral
triangle is 60 (each angle is 60 degrees in an equilateral triangle)
then the length of the arc subtended by the hypotenuse of the
equilateral triangle is one sixth of the entire circumfrence of the
circle because 60 degrees is one sixth of 360.
The length of the minor arc subtended by the chord of the circle
is X times pi times the square root of 2, the quantity over 30, where
X is the measure of the angle at the vertex
at the center of the circle, because the circumfrence needs only
to be divided by the times the angle goes into 360, so the circumfrence
of the circle divided by the quantity of 360 divided by the angle
at the center of the circle.


***********************************************

From: 74620.2745@compuserve.com

From: Candice Diaz
Grade: 9
School: Roselle Park High School, Roselle Park, New Jersey

Radius=8.485281374
Circumference = 2Pi r = 53.31459526.

90 Angle represents 1/4 of a 360 degree circle. Subtended arc = 1/4 of
circumference = 13.32189176

60 angle represents 1/6 of a 360 degree circle. Subtended are=1/6 of
circumference = 8.885765877

If the vertex angle = x at the center of the circle, the measure of the arc
would be x/360 (circumference)


***********************************************

From: 74620.2745@compuserve.com

From: Allie Hess and Laura Merkel
Grade: 9
School: Roselle Park High School, Roselle Park, New Jersey

r = 8.49 c = 53.34

A. Right triangle: subtended arc = 53.34/8.49 = 13.34

B. Equilateral traingle: subtended arc = 53.34 / 8.49 = 8.89

C. Triangle with x for vertex angle: subtended arc = (x/360)*53.34


***********************************************

From: 74620.2745@compuserve.com

From: Joshua Goldstein
Grade: 9
School: Roselle Park High School, Roselle Park, New Jersey

Since an isoscele triangle has 2=sides and 2=angles and the central angle =90.
Since the arc is 90 it is 1/4 of the circumference of the circle. The
length of
the arc = (2Pi*4/x)=13.3. x=360/90 = 4.

When an equilateral triangle is plugged in instead, the 3=angles are 60 each so
the x=6 and the arc = 8.9.

If the central angle is x then the formula for the arc would =
2Pi*4/y y=360/x


***********************************************

From: 74620.2745@compuserve.com

From: Rich Deo
Grade: 10
School: Roselle Park High School, Roselle Park, New Jersey

C=53.4 r=8.5 A=287.75

A. 90/360 = 1/4 to find what % of the circle is taken.
arc = 1/4circumference = 1/4(53.4)=13.35

B. 60/360 = 1/6 of the circle
arc = 1/6 circumference = 1/6(53.4)=8.9

C. x/360 to find what %age is taken
are=x/360 circumference = x/360(53.4)


***********************************************

From: 74620.2745@compuserve.com

From: Karen McKinney
Grade: 10
School: Roselle Park High School, Roselle Park, New Jersey

C=53.4 C=2Pi*r circle=360

8.5=radius

a) 90/360=1/4 arc=(1/4)(c)=13.35

b) 60/360=1/6 arc=(1/6)(C)= 8.9

c) x/360 arc=(x/360)(C)=(x/360)(53.4)


***********************************************

From: 74620.2745@compuserve.com

From: Sarah Hartley & Jill Matthews
Grade: 10
School: Roselle Park High School, Roselle Park, New Jersey

r=8.5
C=2Pi*r=2(3.14)(8.5)=53

90 triangle: Divide by 4 because the triangle is 1/4 of the circle
because 4 90
angles make a circle. 53/4=13

60 triangle: Divide by 6 because the triangle is 1/6 of the circle. 6
triangles fit in it. C=53. 53/6=8.9=minor arc

last part arc=53/y 360/x=y


***********************************************

From: 74620.2745@compuserve.com

From: Arnette Wheeler
Grade: 10
School: Roselle Park High School, Roselle Park, New Jersey

r=8.5 C=2Pi*r = 2(Pi*8.5) = 53

360/90=4 so the arc=53/4 = 13

360/60=6 so the arc=53/6 = 8.9

360/x=y so the arc = 53/y


***********************************************

From: 74620.2745@compuserve.com

From: Dimple Patel
Grade: 10
School: Roselle Park High School, Roselle Park, New Jersey

r-8.5
c=2Pi*r
c=2(3.14)(8.5)
c=53.38

arc 1 = 53.38/4 4 because 360/90 = 4

arc 2 = 53.38/6 6 because 360/60 = 6

arc 3 = 53.38/y y because 360/x = y


***********************************************

From: 74620.2745@compuserve.com

From: Lorena Reyes and Soraya Jallad
Grade: 10
School: Roselle Park High School, Roselle Park, New Jersey

\c=2Pi*r c=52.8 r=8.4

The 1st arc = 1/4 of the circumference of the circle = 13.2.

If it were an equilateral triangle it would be 1/6 of the circumference of 52.8
so it = 8.8

The last arc = 52.8/(360/x)


***********************************************

From: 74620.2745@compuserve.com

From: Tammy Heskeyahu and Purvi Patel and Maria Pineiro
Grade: 10
School: Roselle Park High School, Roselle Park, New Jersey

C=2Pi*r r=6(square root of 2)=8.48=8.5
C=2(3.14)8.5=53.38

arc=53.38/4=13.345 divide by 4 because 90/360=4

arc=53.38/6=8.896 divide by 6 because 60/360=6

arc=53.38/y divide by y because x/360=y


***********************************************

From: mickl@earthlink.net

From: Mick Lorusso
Grade: 9
School: Ignacio High School, Ignacio, Colorado

The minor arc subtended by the hypotenuse of the right isosceles
triangle has a length of about 13.33. The minor arc subtended by
the chord in the equilateral triangle has a length of about 8.886,
while the central angle formed by the legs of the right triangle
is equal to 90 degrees (x=90 degrees).

I obtained this information by first diagraming the problems.
Since the right anle of the triangle is the central angle of the
circle, I divided 90 degrees by 360 degrees (There are 360 degrees
in a circle). This gave me a fraction of 1/4. Thus, the central angle
occupies 1/4 of the circle. Corespondingly, the minor arc subtended
by this trianle has a length equal to 1/4 of the circumference of
the circle.
Thus: 1/4*C=M, where M= the minor arc subtended by the chord
C= the circumference of the
circle.

Since the circumference of any circle is equal to the radius
multiplied by 2 and pi (about 3.1416), the total length of the minor
arc can be expressed in the folowing formula:

1/4*(3.1416*2*6*(the square root of 2))=M

When this is reduced one gets: 3.1416*3*(the square root of 2)=M

The answer becomes about 13.33.

I applied the same method to the equilateral triangle.
Since the sum of any trianle's angles must be equal to 180, the
equilateral triangle must be composed of three 60 degree angles
(180/3= 60). 60 degrees divided by 360 degrees is equal to 1/6,
so the central angle must comprise 1/6 of the whole circle.
Corespondingly, the minor arc must also have a length equal to 1/6
of the circle's circumference.

Thus: M=1/6*(3.1416*2*6*(the square root of 2))
Reduced: M= 3.1416*2(the square root of 2))

Punched into a calculator this equals about 8.886.

For the last part here's what I got:

x/360degrees multiplied by the circumference=length of arc.

This formula will give you the length of the arch formed by the
radii of any central angle. I determined this by noting that any circle has
360 degrees, and any central angle in the circle forms some portion of the
circle. Since this portion of the circle can be expressed by x/360, then
one can multiply x/360 by the circumference to determine the lenght of the arc.


***********************************************

From: Lishack@sasd.k12.pa.us

From: Bob Young and Christian Paul
Grade: 10
School: Shaler Area High School

1. If the radius of the circle is 6 square-roots of 2 then, the circumference
of the circle is 53.31. In the first part this should be divided by 4, since
four triangles can fit into this circle. This would be 13.33.
2. Since the circumference is the same, 53.31, then all we had to do was
divide
this by 6 instead of four, since 6 triangles fit into the circle. The
answer is
8.89.
3. If you don't know what degree the triangle is, use this formula for your
answer: 2*Pi*R(x/360).


***********************************************

From: Lishack@SASD.K12.PA.US

From: Amy Bruecken and Alexis Sauter
Grade: 10
School: Shaler Area High School, Pittsburgh, Pennsylvania

To find our answers we used this formula:Degrees/180*Pi*r
For the isoceles triangle the degrees was 90, so we put it into
the formula and got 13.3.
For the equilateral triangle the degrees was 60, so we put it
into the formula and got 8.89.
Then it asked to find degrees which was x, so we got 26.7x/180.


***********************************************

From: Lishack@SASD.K12.pa.us

From: Natalie Navarro and Jackie Rose
Grade: 10
School: Shaler Area High School, Pittsburgh, Pennsylvania

First we defined all of the unknown terms. Then we came to conclu
usions on how to draw the drawing. We inscribed an isosceles
right triangle into a circle. Since we know the legs of the
triangle are the radii of the circle, they are both 6*the
square root of two, which is 8.49. To find the length of the
minor arc, we used the formula (q/180)*pi*r, where q is equal
to the degree of the triangle. Since the first triangle is a
right triangle it is 90 degrees. The final answer is 13.34
rounded to the nearest hundreth. A equilateral triangle
is 60 degrees, we put it into the same formula to find the minor
arc which was 8.89. When the angle is "x", using the same formula
(q/180)*pi*r), the answer in terms of x was (26.67x/180).



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