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Geometry POW Solution, November 15
Posted:
Dec 21, 1996 11:48 AM


****************************** Geometry Problem of the Week, November 1115
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 nonbase 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!
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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, LambtonKent 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 QuinnKerins 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 DurnerFeiler Grade 6, Sand Lake Elementary School, Anchorage, Alaska
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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.
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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 1115, 1996
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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)
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From: keither@ciaccess.com
From: Clayton Smith Grade: 11 School: LambtonKent 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.
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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 nonbase 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 email addr: imre_bereknyei@ccgate.apl.com
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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
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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.
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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.
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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.
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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.
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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.
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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 nonbase angle was x, the answer would be x/360*2*pi*8.481281374.
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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 nonbase angle = x then x/360* circumference = measure of the subtended arc = x/360(53.3)
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From: 74620.2745@compuserve.com
From: Melanie Griffin Grade: 9 School: Roselle Park High School, Roselle Park, New Jersey
The nonbase < 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.
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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 nonbase 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 nonbase 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.
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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)
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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
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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
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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)
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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)
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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
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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
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From: 74620.2745@compuserve.com
From: Dimple Patel Grade: 10 School: Roselle Park High School, Roselle Park, New Jersey
r8.5 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
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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)
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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
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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.
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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 squareroots 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).
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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.
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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).



