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A regular hexagon and an equilateral triangle are both inscribed in the
same circle so that the hexagon and the triangle share three vertices. The
radius of the circle is 10 units. What is the area of the region between
the two polygons?
Make sure you explain your answer - answers that include just a number will not be considered correct! I know the answer already - what I want to know is how you figured it out. SolutionsMost of the people this week figured out what to do, but made mistakes doing it, particularly finding the area of the hexagon or the triangles. 70 people got it right, and 37 got it wrong. When you are finding the area of an equilateral triangle, make sure you draw a very good picture and carefully add the lines that you need. A number of people assumed that the edgelengths of the triangles they used were nice numbers - like 10, 8, and 6 - and ended up with the wrong area. There are a lot of ways to find the area of a triangle - Chris Shaw of Middletown High School North used Heron's Formula. He didn't provide much of an explanation, so you won't learn what Heron's Formula is from his solution, but I've included it anyway. A few people got the problem wrong because they drew an "accurate" picture of things and then measured the lengths with a ruler. Almost never will you be able to get a good answer this way. Your drawing, no matter how careful you are, will never be as accurate as the facts and formulas you know about triangles. There were a lot of different ways of figuring this problem out. A number of people found the area of the hexagon and the triangle and then subtracted. That works, but it's a lot less work to notice that the area we're looking for is equal to the area of the big equilateral triangle, or equal to half the hexagon. Then you only need to find one thing. Alison Falkenhagen of Highland Park Senior High School noticed that she only had to find the area of some of the parts, and she did it two different ways, so I've included her answer. Jen Ramirez of Germantown Academy gave a very thorough explanation which you can also read below. To find the area of the hexagon you can break it up into equilateral triangles, which is what I always do since I can't remember any of the fancy formulas for the area of a hexagon, or even the area of an equilateral triangle. But some of you are smart enough to look it up and find those formulas and you save yourself a little bit of work. Several people noticed that you only have to find half the area of the hexagon, which is a trapezoid, and found the area of the trapezoid. Alison Miller, who is homeschooled, did it this way, and I've highlighted her answer. This was the sort of problem that produces fairly short answers and fairly long ones. That's okay - read your solution carefully and see if you have said enough, and that's as long as it has to be. It is sometimes hard to know how much explanation you should give, but at the very least you should probably explain where all of your numbers came from. One more thought: simplify your answers! Ideally in a problem like this I would like to see the answer as "75 sqrt3, which is equal to 129.9." That tells me a little bit about how you figured the answer out, and also gives me a number to think about. I don't want to see "15 sqrt75" - you know better than that. A list of all the people who got this problem right follows the highlighted solutions below, and most of the solutions are also available.
From: Chris Shaw
Grade: 10
School: Middletown High School North, Middletown, New Jersey
I found the area of the hexagon using the radius of the circle
as the length of one side of an equilateral triangle. I found
the area of the triangle using Heron's formula and multiplied
my answer by six because there are six equilateral triangles in
a regular hexagon. The area of the hexagon is appr. 259.8.
Then, I noticed that the area of the inscribed triangle is
equal to that of the left over hexagon. Therefore, the answer
is half of the area of the hexagon.
My answer is: 129.9 units^2
Chris Shaw
From: Alison Falkenhagen
Grade: 9
School: Highland Park Senior High School, St. Paul, Minnesota
Subject: Oct 13-17 POW
Alison Falkenhagen Grade 9, Geometry IB
Highland Park Senior High School, (612) 293-8940
www.stpaul.k12.mn.us/hpsh/highland.html
POW, October 13 - 17, 1997
From: Jen Ramirez
Grade: 10
School: Germantown Academy, Fort Washington, Pennsylvania
Subject: Problem of the Week, Oct. 13-17
Jen Ramirez, Grade 10
Germantown Academy
Fort Washington, PA
Problem of the Week, October 13-17
I first drew the circle, then inscribed a hexagon by plotting six points on
the circle each a distance apart that is equal to the radius (10 units, in
this case). By connecting these points, a hexagon is formed. As seen in the
attached diagram (Geometer's Sketch Pad), I labeled the points clockwise from
the top A, B, C, D, E, F, and the center of the circle G.
From: Alison Miller
Grade: 6
School: homeschooled, Niskayuna, New York
Subject: October 13-17 Geometry POW
Dear Annie,
In my picture, the hexagon is ABCDEF, and the triangle is BDF.
The following students submitted correct solutions this week:Patrick Tartar, Grade 8, Odle Middle School, Bellevue, WashingtonJulia Le, Grade 11, Minnechaug Regional High School, Wilbraham, Massachusetts James Tong, Grade 11, Klein High School, Klein, Texas Gordon Bockus Jr., Grade Freshman, Eastern Oklahoma State College, Wilburton, Oklahoma Bill Braund, Grade alum, Lindblom Technical High School, Chicago, Illinois Josh Richard, Grade 8, Albright Middle School, Alief, Texas Joseph Pacold, Grade 7, Homeschooled, Tarrytown, New York Jennie Doss and Denise Dotson, Grade 9, Franklin County High School, Rocky Mount, Virginia Victor Munos and Melinda Whiteman, Grade 11 & 10, Granada High School, Livermore, California J. Chaja, Grade 9, Chetek High School, Chetek, Wisconsin Steven Abt, Grade 10, William Penn Charter School, Philadelphia, Pennsylvania Lev Navarre, Grade 6, Odle Middle School, Bellevue, Washington Arielle Cohen, Grade 10, Akiba Hebrew Academy, Merion, Pennsylvania David Nguyen, Grade 8, Albright Middle School, Alief, Texas Katie Madden, Grade 9, Mount Saint Joseph Academy, Flourtown, Pennsylvania Colleen Kelly, Grade 10, Mount Saint Joseph Academy, Flourtown, Pennsylvania Jane Milton and Sara Fitzsimmons, Grade 10, Mount Saint Joseph Academy, Flourtown, Pennsylvania Michael McCloskey, Grade 9, Calistoga High School, Calistoga, California Holly Black, Grade 7, Odle Middle School, Bellevue, Washington michelle cheng, Grade 12, University of Toronto Schools, Toronto, Canada Chris Shaw, Grade 10, Middletown High School North, Middletown, New Jersey Sam Van Rooy, Grade 8, Oak Hill School, Eugene, Oregon John Heckel, Grade 7, Spring Hill Christian Academy Frannie Laks, Grade 10, Akiba Hebrew Academy, Merion, Pennsylvania Greg Schoppe, Grade 6, homeschooled Lunenburg, Vermont Jon Gantman, Grade 10, Akiba Hebrew Academy, Merion, Pennsylvania Sorin Ionescu, Grade 8, Ecole Secondaire Dorval, Quebec, Canada Danny McKenna, Grade 10, William Penn Charter School, Philadelphia, Pennsylvania Jenny Kaplan, Grade 7, Castilleja Middle School, Palo Alto, California Julia Fischer, Grade 10, Granada High School, Livermore, California Jennifer Liang, Grade 8, Odle Middle School, Bellevue, Washington Anna Wu, Grade 10, Monte Sant' Angelo Mercy College, Sydney, Australia Thomas Kuo, Grade 10, Burroughs High School, Ridgecrest, California Tiffanie Lam, Grade 8, Sequoia Middle School, Pleasant Hill, California Laura Roos, Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania Rob Johnson, Grade 10, Redmond High School, Redmond, Oregon Clayton Dillaway, Grade 8, Odle Middle School, Bellevue, Washington Katie Beranek, Grade , Shaler Area High School, Pittsburgh, Pennsylvania Ben Yeckel, Grade , Shaler Area High School, Pittsburgh, Pennsylvania Megan Morgan, Grade , Redmond High School, Redmond, Oregon Deanna Smith, Grade , Redmond High School, Redmond, Oregon Andy Lane, Grade , Remond High School, Redmond, Oregon William Christianson, Grade 9, Highland Park Senior High School, St. Paul, Minnesota Kelsey Long, Grade 9, Highland Park Senior High School, St. Paul, Minnesota Emily Buzicky, Grade 9, Highland Park Senior High School, St. Paul, Minnesota Brandon Gilchrist, Grade 9, Highland Park Senior High School, St. Paul, Minnesota Alison Falkenhagen, Grade 9, Highland Park Senior High School, St. Paul, Minnesota Gina Guss, Grade , Redmond High School, Redmond, Oregon John Martin, Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania Brian Vita, Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania Alex Chen, Grade , Odle Middle School, Bellevue, Washington Aaron Cho, Grade , Calistoga Junior/Senior High School, Calistoga, California Jen Ramirez, Grade 10, Germantown Academy, Fort Washington, Pennsylvania Alex Chernyavsky, Grade , Akiba Hebrew Academy, Merion, Pennsylvania Zimran Douglas, Grade 11, Wingate High School, Brooklyn, New York Alison Miller, Grade 6, homeschooled, Niskayuna, New York Anna Warszawa, Grade 9, Germantown Academy, Fort Washington, Pennsylvania Ylenia Adornato, Grade , Istituto Comprensivo, Novi de Modena, Italy Scott Brown, Grade , Germantown Academy, Fort Washington, Pennsylvania Ken Rockot, Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania Kasey Jones, Grade , Redmond High School, Redmond, Oregon Danielle, Grade , Redmond High School, Redmond, Oregon Greg Always, Grade , Redmond High School, Redmond, Oregon Candace Murray, Grade , Redmond High School, Redmond, Oregon Ikechekwu Njoku, Grade 12, Wingate High School, Brooklyn, New York Lisa Oakland, Grade , Redmond High School, Redmond, Oregon Kristin Kirkman, Grade , Redmond High School, Redmond, Oregon Jamie Larson, Grade , Redmond High School, Redmond, Oregon Kerianne Schubert, Grade , Redmond High School, Redmond, Oregon Shameica Edwards, Grade 12, Wingate High School, Brooklyn, New York |
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We are supposed to find the area of the yellow part when the radius is 10.
It is clear from the picture that the big equilateral triangle is half of the
regular hexagon. The big equilateral triangle has three congruent isosceles
triangles and the regular hexagon has six. The yellow part that we are trying to
find the area of also has three congruent isosceles triangles, so it is also
half of the regular hexagon.
In the Polygon chapter in our geometry book there is a formula for finding the
area of a regular hexagon: A = (1/2)ap, where a is the apothem and p is the
perimeter. The perimeter of the regular hexagon is 60 because each side is the
same as the radius, or 10. The apothem is the distance from the center to the
midpoint of a side. In the picture it is labeled a, and we can find how long it
is by useing the Pythagorean Theorem.
a^2 + 5^2 = 10^2
a^2 = 75
a = radical 75 = 5(radical 3)
The area of the regular hexagon is (1/2)(5 radical 3)(60) = 150(radical 3), and
the shaded part is half of that, or 75(radical 3).
There is another way to solve this problem that is even easier.
Each yellow triangle is isosceles and has two congruent right triangles in it.
The small gray equilateral triangle also has two congruent right triangles in
it. The yellow part is three times the area of one isosceles triangle, and it is
also three times the area of one gray equilateral triangle.
In the Polygon chapter in our geometry book there is a formula for finding the
area of an equilateral triangle: A = s^2(radical 3)/4, where s is the side of
the equilateral triangle. In our problem, the side is the same as the radius, or
10. So the area of the equilateral triangle is (100)(radical 3)/(4) = 25(radical
3), and the shaded part is three times that, or 75(radical 3).
I then inscribed a triangle by drawing segments from E to C, A to E, and A to
C, forming the pink equilateral triangle. I then connected all the points to
the center of the triangle by constructing segments. This made six congruent
equilateral triangles, and each side of these triangles was 10 units long.
Also, each of the triangles was divided in two. The area of the space
between the polygons is in 3 different sections, each section having the same
area as the triangles.
Thus, the way to solve this problem is to find the area of one equilateral
triangle, then multiply it by three to find the total area of the space
between the two polygons.
So I used the Pythagorean Theorem to find the height of the equilateral
triangles:
a`2 ("a-squared"- sorry, my computer refuses to superscript) + b`2=c`2. We
know the hypotenuse (C) is 10 units, and one side is 5 units (half of C).
5`2 + b`2 = 10`2
25 + b`2 = 100
b`2=75...
Therefore, b= 5 times the square root of 3.
So the height of the triangle is 5 root 3. The area of a triangle is 1/2 the
base times the height. So the area of each yellow area is 5 root 3 x 5,
which equals 25 root 3. Multiply this by 3, yielding the total area of the
space between the two polygons: 75 times the square root of 3, or about
129.9 square units.
The End. :)
The area of the triangle is half of the area of the hexagon, because you can
divide the hexagon into three rhombi (rhombi is the plural of rhombus) like
BCDG. You can divide the big triangle into three smaller triangles, which are
not equilateral, like BGD, and you can do it so that each triangle is half of
a rhombus (my illustration shows that).
The trapezoid EFAB is also half of the hexagon. So, it has the same area as
the triangle BDF. So, the trapezoid BCDE has the same area as the area of the
region between the hexagon and the triangle.
Now, if you want to find the area of a trapezoid, you multiply the average of
the lengths of the bases by the altitude. The average of the lengths of the
bases is (10 + 20)/2. That is equal to 15.
Now to find the altitude of the trapezoid, you split the trapezoid into three
little equilateral triangles. The altitude of one of those triangles is equal
to the altitude of the trapezoid. You can find that altitude by splitting the
little triangle into two 30-60-90 triangles. Now, in a 30-60-90 right
triangle, the length of the hypotenuse is equal to twice the length of the
shorter leg. Also, in a 30-60-90 triangle, the length of the longer leg is
equal to sqrt(3) times the length of the shorter leg. In this 30-60-90
triangle the hypotenuse is 10 (because that's the radius of the circle, and a
radius of the circle is a side of the little triangles). So, the length of
the shorter leg is 5. That also means the length of the longer leg is
5*sqrt(3). That is also the altitude of the small triangle, which means it is
also the altitude of the trapezoid. Now we are ready to calculate the area of
the trapezoid.
If you remember, you do that by multiplying the average of the bases by the
altitude. We now know both numbers. So, the area of the trapezoid is
15*5*sqrt(3), or 75*sqrt(3) square units. If you remember, we showed that that
is also equal to the area of the region between the two polygons. So, our
answer is 75*sqrt(3), or approximately 129.9 square units.
Alison Miller, Grade 6
Homeschooled, Niskayuna, NY