Geometry POW Solution, December 13
Geometry Problem of the Week
******************************
Geometry Problem of the Week, December 9-13
Two congruent 10cm x 10cm squares overlap. A vertex of one square is at
the center of the other square. What is the largest possible value for the
area where they overlap? (The one square is movable, as long as the vertex
remains in the center.)
******************************
I love this problem. Many of you liked it too - 77 correct solutions, and
only 10 wrong. Most of the ones that were wrong were actually right, in
that they stated the maximum area was 25, but didn't offer ANY
explanation. Like NONE.
It's true that the "maximum" area covered is in fact the ONLY area that
can be covered. A number of people pointed this out, but only a few
bothered to explain why. When you say something like that, something
convinced you that you were true - explain that! What made you think you
were right? And can you write an explanation that would convince _anyone_
that you are right? Better yet, can you draw a picture?
When I stated the problem, I didn't say to explain everything, but to just
say what the maximum area is. You must include _some_ explanation, as
usual, and it's a good idea that if you say that "25 cm^2 is the maximum",
you should say something about why you think it can't be bigger. This is
a problem where I think most people _think_ a lot more than they _write_.
Remember that you are always writing solutions for readers who might
not be familiar with the problem or the solution. That will help you to be
very thorough.
People used two ways to show that the area is a constant. The easiest to
see, and the least work (and consequently my favorite, of course) is to
show that since the square takes up 90 degrees at the center, you can put
three more squares around the center and they'll cover four congruent
areas. This was the method used in two of the highlighted solutions, Brent
Tworetsky of JP Taravella High School, and Thomas Kuo of Burroughs
High School. The other way to do it was used by a bunch of folks. Start
with the squares in "square overlap" position. As you rotate the one
square, you can show that the new area you cover is equal (and congruent)
to the area you uncover. Shameica Edwards of Wingate High School
used this method.
Brian Gordon, one of our "adult" contingent, decided to be a little
extreme with his answer, and figured he could get away with folding the
moveable
square. Since I didn't say you couldn't, I gave him credit.
And this week I did something I've never done before, but I got a little
carried away. I love this problem so much that I have a sketch that
illustrates how it works. Since no one drew "the perfect picture", I've
included a picture of my sketch, which is linked to the sketch itself for
those of you who have Sketchpad. Criticism is welcome :-).
There was an interesting question raised by a couple of people this week.
Justin Lam asked about other shapes - will this work for triangles?
Regular polygons? Any irregular ones? Robert Hughey thinks it will work
for rectangles and rhombii. What do you think? (Personally, I think it
sounds like a good Project of the Month!)
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.
Megan Bean and Alice Tran
Grade 9, Smoky Hill High School, Aurora, Colorado
Justin Lam Grade 8, Sequoia Middle School, Pleasant Hill, California
Gavino Walker Grade 10, Sammamish High School, Bellevue, Washington
Brian Gordon Grade 1992, Dartmouth
Brent Tworetzky Grade 10, JP Taravella High School, Coral Springs, Florida
--
Rick Bullock Grade 10, Golden Plains High School, Rexford, Kansas
Marc A. Tuler Grade 9, Smoky Hill High School, Aurora, Colorado
Erin Muehlenkamp Grade 9, Smoky Hill High School, Aurora, Colorado
Ken Duisenberg Hewlett Packard Engineer, Stanford University '91
Kenneth Yan Grade 10, Sammamish High School, Bellevue, Washington
--
Ryan Salvas Grade 8, Old Saybrook Middle School
Michael Martin Grade 9, Sammamish High School, Bellevue, Washington
Nicole Forostoski Grade 10, Martin County High School, Stuart, Florida
Christine Chung Grade 10, Smoky Hill High School, Aurora, Colorado
Roger Mong Grade 8, Zion Heights JHS, Richmond Hill, Ontario, Canada
--
Andrea Wahlen Grade 9, Smoky Hill High School, Aurora, Colorado
Jeremy Reedy Grade 11, Redbank Valley, Pennsylvania
Jess Gilburne and Eric Faden and Jared Joiner
Grade 8, Georgetown Day School, Washington, DC
Davia Bell Grade 8, Murray Middle School, Ridgecrest, California
Stephanie Parsons Grade 9, Smoky Hill High School, Aurora, Colorado
--
Jason Howerton Grade 10, Granada High School, Livermore, California
Kurt Davies Grade 10, Smoky Hill High School, Aurora, Colorado
Erin McCloskey Grade 9, Smoky Hill High School, Aurora, Colorado
Nick Davis Grade , South High School, Bakersfield, California
Heidi Taylor Grade 10, Smoky Hill High School, Aurora, Colorado
--
Laura Mannino and Candice Diaz
Grade 9, Roselle Park High School, Roselle Park, New Jersey
Lyndsey Hayes and Gwen Dillow
Grade 10, Sammamish High School, Bellevue, Washington
Sandra Wai Grade 9, Smoky Hill High School, Aurora, Colorado
Thierry Cote Grade 7, College Marie de France, Montreal, Quebec, Canada
Jenny Kaplan Grade 6, Castilleja Middle School, Palo Alto, California
--
Kirk Vissat Grade 10, Smoky Hill High School, Aurora, Colorado
Preston Sherrell Grade 9, Young Junior High School, Arlington, Texas
Michael Chin Grade 10, Smoky Hill High School, Aurora, Colorado
Ralph Boleslavsky Grade 12, Council Rock High School, Newtown, Pennsylvania
Katherine Walther Grade 9, John Glenn Junior High, San Angelo, Texas
--
Jonathan Shing Grade 9, Newport High School, Bellevue, Washington
Libbie Skelton Grade 9, Martin County High School, Stuart, Florida
Hector Mercedes Grade 12, George Wingate High School, Brookly, New York
Shameica Edwards Grade 11, George Wingate High School, Brooklyn, New York
Paul Chun Grade 9, Newport High School, Bellevue, Washington
--
Simon Park Grade 9, Newport High School, Bellevue,
Washington Megan LaChance Grade 10, Cheshire High School,
Cheshire, Connecticut
Robert Hughey Grade 10, Cheshire High School, Cheshire, Connecticut
Ashley Turner Grade 10, Cheshire High School, Cheshire, Connecticut
Ally Schulthess Grade 10, Cheshire High School, Cheshire, Connecticut
--
Peter Suchy Grade 10, Cheshire High School, Cheshire, Connecticut
Diana Bojka Grade 10, Cheshire High School, Cheshire, Connectictu
Matt Mansur Grade 9, Cheshire High School, Cheshire, Connecticut
Leah Rosen Grade ,
Thomas Kuo Grade 9, Burroughs High School, Ridgecrest, California
--
Andrea Marks Grade 8,
Jackie Roth Grade , Germantown Academy, Fort Washington, Pennsylvania
Jackie Wong Grade , Newport High School, Bellevue, Washington
Katie Quinn-Kerins Grade 10, Germantown Academy, Fort Washington, Pennsylvania
Kate Martin Grade 10, Cheshire High School, Cheshire, Connecticut
--
Jessica LaSalle Grade 10, Cheshire High School, Cheshire, Connecticut
Suzanne Davis Grade 10, Newport High School, Bellevue, Washington
Giscard Pongnon Grade 12, George Wingate High School, Brooklyn, New York
Ginni Wentworth Grade , College Park High School, Pleasant Hill, California
Joya Guha Grade , College Park High School, Pleasant Hill, California
--
Juri Miyamae Grade , College Park High School, Pleasant Hill, California
Marisa Mannari Grade 9, Newport High School, Bellevue, Washington
Krissy Wall Grade , College Park High School, Pleasant Hill, California
Laurel Kaminski Grade 9, Newport High School, Bellevue, Washington
Joe Fantini and Greg Mellor
Grade 9, Germantown Academy, Fort Washington, Pennsylvania
--
Adam Peyman Grade 10, Cheshire High School, Cheshire, Connecticut
Tim Pawlush Grade 10, Cheshire High School, Cheshire, Connecticut
Scott Gorski Grade 10, Cheshire High School, Cheshire, Connecticut
Hope Langan Grade 9, Smoky Hill High School, Aurora, Colorado
Jackie Roth Grade , Germantown Academy, Fort Washington, Pennsylvania
--
Rachel Collins Grade 9, Newport High School, Bellevue, Washington
Brenda Hoskinson Grade 9, Newport High School, Bellevue, Washington
Chris Roe Grade 9, Newport High School, Bellevue, Washington
Heather Hook Grade 9, Newport High School, Bellevue, Washington
Meghan Bowen Grade , Newport High School, Bellevue, Washington
--
Lauren Hillman Grade , Germantown Academy, Fort Washington, Pennsylvania
Renee Grade 9, Smoky Hill High School, Aurora, Colorado
***********************************************
From: Monaco...@ossinc.net
From: Megan Bean and Alice Tran
Grade: 9
School: Smoky Hill High School, Aurora, Colorado
The answer to this POW is 25cm squared. No matter what position the movable
square is in. The way we figured this out is by cutting two 10cm X 10cm
squares
out and then put a brass plated fastener at the vertex of one square and the
center of the other square. Then we moved the squares in all possible positions
and measured the areas and we found that all the areas of the overlapped part
was 25 cm squared.
***********************************************
From: quan...@ucop.edu
From: Justin Lam
Grade: 8
School: Sequoia Middle School, Pleasant Hill, California
E *
A X * * Let ABCD and OEFG be two identical squares
***********B * with side = 10 cm. Let O be the center of Square ABCD.
* * * * Want to prove that no matter how you rotate Square
OEFG
* * * *about O, the area that is created from the overlapping of
* O* * *F Square ABCD and Square OEFG is always the same. In
another
* * * * word, there is no one overlapping area that is maximum
* **Y * or minimum - they are all the same.
*********** *
C D G
Let X be the intersection of OE and AB.
Let Y be the intersection of OG and BD.
Claim: the Triangle OXB is congruent to Triangle OYD.
OB = OD They both are half of the diagonal of Triangle ABCD
angle OBX = angle ODY They are 45 degree angles because OB and OD are
part of the diagonals
angle XOB = angle DOY angle XOB+ angle BOY=angle XOY=90 degrees
(vertex angle of Square OEFG) and
angle DOY+angle BOY=angle DOB=90 degrees (Diagonals COB
and AOD intersect at a right angle)
ASA implies that the two triangles are congruent.
The overlapped area OXBY = Area of Triangle OXB + Area of Triangle OBY
= Area of Triangle OYD + Area of Triangle OBY
= Area of Triangle ODB
= 1/4*(Area of Square ABCD)
= 1/4*(10*10)
= 25 square cm.
***********************************************
From: shg...@oasis.bellevue.k12.wa.us
From: Gavino Walker
Grade: 10
School: Sammamish High School, Bellevue, Washington
What I did is I took 2 equal squares, put the vertex of one in the middle
of the other, like you suggested, spun around the outside square until
No matter how you rotate the square on the midpoint
the area is always going to be 25cm-squared.
No matter how you do it the result is the same.
good luck responding to all the rest, I know how you feel
gavino
***********************************************
From: bmgo...@ntplx.net
From: Brian Gordon
Grade: 1992
School: Dartmouth
Well Annie, since this problem is a rerun, you will not be
receiving the standard answer of 25 cm^2 from me this week.
My answer this week is 50 square centimeters if the movable square
must remain a square, and 87.5 if not. The easiest number to
show is the 87.5. This is obtained by folding the square back over
across the points of intersection with the fixed square:
okay, here we go. the original diagram should look like this:
------------------------
| |
| |
| |
| |
| -----------B----------
| | | |
| | | |
| | | |
| | | |
| | | |
------------A----------D |
| |
| |
| |
| |
----------------------
Let A and B be the points of intersection of the two squares.
Fold the lower square over itself across segment AB. It lands right on top
of the first square. You will note the only area where there is no
overlap is triangle ABD, which has area 1/2 * 5 * 5=12.5, so the overlap is
100 - 12.5 = 87.5. The line AB is a line of symmetry for the picture.
Of course, if you insist that this remain a square, simply
repeat the folding on all four corners. The square would then
have area 50 cm^2, or half the original square.
You didn't say we couldn't move the square out of the plane.
*phtphtbt*
see ya!
--bri
***********************************************
From: bla...@ix.netcom.com
From: Brent Tworetzky
Grade: 10
School: JP Taravella High School, Coral Springs, Florida
in this problem, we are placing the vertex
of a square at the center of another. this means we are
creating a right angle at the center of the square.
seeing as 90 degrees is a quarter of 360, the area
of the overlapping squares is a constant 25%. to
further prove this, extend the sides of the square
with its vertex in the center of the other until the lines
touch the sides of the square. the result is four congruent
shapes (wow!) this can be done at any angle.
* * * * * * * * * *
* * * * * * *
* * * * * * * *
* * * * * * *
* * * * * * * * * *
any angle you connect the two squares, four congruent
shapes are formed. the overlapping is one of these squares;
1 out of 4, 25%
thus, the answer is 25% of 100, being 25
***********************************************
From: ber...@ruraltel.net
From: Rick Bullock
Grade: 10
School: Golden Plains High School, Rexford, Kansas
It would be 25 units. If the vertex is right in the middle of the
square then all you would have to do is find the area of a triangle.
And since the vertex is in the center the height would be 5,because
it is a 10 unit long square, and the bottom rail would be ten.
So taking that into the triangle formula I got the solution of
25 units.
***********************************************
From: mtu...@shhs1.smoky.org
From: Marc A. Tuler
Grade: 9
School: Smoky Hill High School, Aurora, Colorado
The answer I got to this weeks problem is 25cm squared.
I got this answer because I knew that both squares were
congruent. So the sides of the one rectangle must be bisected in order to
put the vertex of the one square onto the midpoint of the other. So now
each side becomes 5cm.
5cm x 5cm= 25 cm squared
***********************************************
From: asan...@shhs1.ccsd.k12.co.us.
From: Erin Muehlenkamp
Grade: 9
School: Smoky Hill High School, Aurora, Colorado
I made two 10 cm. squares out of notebook paper. I put one vertex
of a square in the middle of the other square, and I moved it
around, keeping the vertex in the center of the square. I came to
the conclusion that it can cover the area of 100 cm. as it is
rotated around. It can only cover a total area of 25 cm. at a
time however.
***********************************************
From: ke...@ecst.csuchico.edu
From: Ken Duisenberg
Grade: Hewlett Packard Engineer
School: Stanford University '91
Answer: The overlap is always 25 sq. cm.
Solution: Place four 10x10 squares together at a corner and a fifth
square centered on that common corner. In this way, it is obvious that
the last square is evenly divided into the other four squares no matter
how it is rotated. Each of the four overlaps with one quarter of the
rotating square, or 25 sq. cm.
***********************************************
From: K61...@aol.com
From: Kenneth Yan
Grade: 10
School: Sammamish High School, Bellevue, Washington
The half of the edges of the square is 5cm, and the edges of the
overlap square is 5cm. So, 4 sides are 5cm, then the largest
possible value for the area is (5X5)sq. cm= 25sq. cm
***********************************************
From: bpac...@connix.com
From: Ryan Salvas
Grade: 8
School: Old Saybrook Middle School
The largest possible value for the area would be 25 square centimeters. No
matter how you rotate the overlapping square, the area is always 25 square
centimeters.
***********************************************
From: mart...@aol.com
From: Michael Martin
Grade: 9
School: Sammamish High School, Bellevue, Washington
The overlap of the two squares always remains the same. It remains constant as
long as the vertex of one of the squares is the
center of the other square.
Let's put it this way...
____
Take a four-sided figure. | |
| |
----
If you move two opposite points in tandem, as long as two points in opposite
corners stay in the same place, the area of the figure
will always stay the same.
Uhh, never mind... This is hard to draw with text.
Anyways, with the overlap of the two squares, the two moving points are the
points where the sides of the squares meet.
The two nonmoving points are the corners of the squares that form the overlap.
The two moving points move in tandem,
so the area of the overlap stays the same.
Well, that's pretty much it, so I'll just send you this now.
See ya' later!
Michael Martin
***********************************************
From: foros...@worldnet.att.net
From: Nicole Forostoski
Grade: 10
School: Martin County High School, Stuart, Florida
Problem of the week November 18-22
Nicole Forostoski
Martin County High School
Stuart, Florida
Mrs. Summers Per. 6
Dear Annie,
łTwo congruent 10cm x 10cm squares overlap. A vertex of one
square is at the center of the other square. What is the
largest possible value for the area where they over lap?˛
The area where they overlap is 1/4 or 25% of either 10cm x 10cm
square. 1/4 of 100cm squared is 25cm squared. as illstrated
below.
10cm
x x x x x x x x x x
x x
x x
x x
10cmx 100cm2 x
x x x x x x x x x x x
x x x x
x x 25cm2 x x
x x x x
x x x x x x x x x x x
x x 10cm
x 100cm2 x
x x
x x
x x x x x x x x x x
10cm
***********************************************
From: monaco...@ossinc.net
From: Christine Chung
Grade: 10
School: Smoky Hill High School, Aurora, Colorado
The largest possible area value for the area where they overlap is 25 cm sq.
Since the square is 10 cm by 10 cm, that means that the sides are 10 cm by 10
cm.
So the area is 100 cm sq.
Since the vertex of one square is at the center of the other square that means
that
the square overlaps the other square by a quarter. A quarter of 100 is 25.
So the answer is 25 cm sq.
***********************************************
From: mon...@iponline.com
From: Roger Mong
Grade: 8
School: Zion Heights Junior High School, Richmond Hill, Ontario, Canada
Assumming there's two lines cuting the square into four
sections, If they meets at the centre of the square and is
perpendicular to each other, four equal sections are formed,
since the square(including the cuts) has a rotation
symmetry of four.
Let's say one of the four sections is part of a square same
size as the original one, the section where the squares overlap
is always a quarter of one of the square, and the vertex of one
square is the same point as the centre of the other.
Therefore, the area that overlaps is a quarter of the area of
one of the square, no matter how it is turned.
So in this case: area of square: 100cm2
quarter of the area of square: 25cm2
The two area where the two squares overlap is always a25cm2.
***********************************************
From: CWa...@aol.com
From: Andrea Wahlen
Grade: 9
School: Smoky Hill High School, Aurora, Colorado
For this weeks problem I found the answer to be 25cm squared.
To get this answer I started by drawing a square. Then I drew a
nother square on top of it with one of the seccond squares
virticies on the center of the first square.This gave me a small
square where the two squares overlapped that had an area of 25
square centemeters. I moved the top square to where there was a
triangle where the two squares overlapped and I got the same area
of the overlapping section. I repeated this process a few more
times and I came up with the conclution that nomatter where you
put the square as long as the virticy is touching the center the
area of the overlapping part will always be 25 square centemeters.
To figure out that the first overlapping square's area was 25cm
squared I figured out that the length of one side was 5cm long
because a line from the middle to the edge would be half as long
as the whole side of the big square. Since it was a square all of
the sides are equal and the area is 5*5=25. When I moved it I
made a triangle. This time the highth of the triangle was 5cm and
the base was 10cm. To find the area of the triangle I took 1/2bh
and got 25.
This was how I did the problem and my final answer was 25cm squared.
***********************************************
From: she...@rbv.iu6.k12.pa.us
From: Jeremy Reedy
Grade: 11
School: Redbank Valley, Pennsylvania
when the square in the center is at a right angle it is 1/4 the other
sqaure but
this is always true no matter how
you turn the sqaure. the answer is 1/4 the square wich is 2.5 cm.
***********************************************
From: gds...@worldnet.att.net
From: Jess Gilburne and Eric Faden and Jared Joiner
Grade: 8
School: Georgetown Day School, Washington, DC
Subject: from Eric,Jess,Jared at GDS
Jess Gilburne,Eric Faden, and Jared
Joiner
Georgetown Day School
Grade 8
Paul Nass
We have found in answer to this weeks problem that the maximum area that the
two squares can overlap, is 25cm squared.
We have found two ways to do this. In both cases the colored area is 1/4 of
the total area of the square.
In Figure 1, the small square is 5x5 and has a total area of 25cm squared.
In figure 2, the total area of the triangle is 25cm squared(1/2 base x height).
[sketch with both square and isosceles triangle solution]
***********************************************
From: ssu...@ridgecrest.ca.us
From: Davia Bell
Grade: 8
School: Murray Middle School, Ridgecrest, California
Subject: POW Dec 9-13
Solution- 25 square cm
Explanation- I used two pieces of graph paper, each with 100 square units.
I folded both papers into fourths and I drew a dot in the center of one of
the pieces. I took the paper without the dot in the center, and placed the
corner of it to the dot. I spun the top paper to fit into one of the
folded squares of the marked paper. This square is equivalent to a forth
of the paper which is 25 square cm. Even if you spin it not to fit into a
square of the marked paper, you would get 25 square cm. Trace the outside
of the unmarked paper on the marked paper. It should cover two squares.
There should be a small space in between the folding line and the traced
line. The area of the traced figure should fit into the second square,
making it a full square, there for making it a fourth of 25 square cm.
Davia Ball, Grade 8
Murray Middle School
Ridgecrest, CA
***********************************************
From: Pars...@ix.netcom.com
From: Stephanie Parsons
Grade: 9
School: Smoky Hill High School, Aurora, Colorado
Subject: answer to the pow
My name is Stephanie Parsons and I'm a 9th grader at Smoky Hill High
School.
The answer that I got for the problem of the week is 25cm squared. I got
that by graghing out a 10cmx10cm square. I then found the center of the
square by making sure I had the same number of 1x1cm squares above the
center, below it and on each side of it. However when doing that, you
discover that not 1 but 4 1cmx1cm squares makes up the center. I also know
that the center of the 10x10 square is where the 4 1cmx1cm squares
intersects. Then, starting at the middle of the 10x10 square I go up 10 cm,
and over 10 cm drawing in the lines. That is my second square but I find
that the first 5 cm up and the first 5 cm to the sides are share with the
first square. If I have 5square cm up and 5 to the side, I have a 5cmx5cm
square. Therefor, the area of the overlapped square is 5cm times 5cm = 25cm
squared.
***********************************************
From: Kurt_...@shhs1.ccsd.k12.co.us
From: Kurt Davies
Grade: 10
School: Smoky Hill High School, Aurora, Colorado
The largest possible area where they overlap is 25 sauare
centimeters. This is because the largest area of the overlaping
space is the triangle from two consecutive corners of the square
being overlaped to the center of the square. The height of the
triangle would be five because the length from the center of the
square to the edge is half f it's total lenght. The area would
be 1/2*5*10=25.
***********************************************
From: Beau...@aol.com
From: Erin McCloskey
Grade: 9
School: Smoky Hill High School, Aurora, Colorado
The answer I got is 25 is the greatest possible value. No matter how you turn
the square, the value of the area is always 25. If it is a square it is
5x5, if
it's a triangle, the 1/2 (10)(5) and always end up with 25.
***********************************************
From: nda...@khsd.k12.ca.us
From: Nick Davis
Grade:
School: South High School, Bakersfield, California
I think that there is only one possible area. I think that it is 25 square
units. I think this is because no matter which angle you rotate the overlapping
square at, you will always cover one-fourth of thoe bottom square.
***********************************************
From: Heidi_...@Shhs1.ccsd.K12.co.US
From: Heidi Taylor
Grade: 10
School: Smoky Hill High School, Aurora, Colorado
Answer: 25cm
Given that the two squares are
congruent and given that the
one vertex always has to be in the center of the other square, this
means that the overlaping portion will always have the area of 1/4
of the area of the whole square. Therefore since the area of the whole square
is
100 that means the overlaping section at its largest is 25.
***********************************************
From: 76420...@compuserve.com
From: Laura Mannino and Candice Diaz
Grade: 9
School: Roselle Park High School, Roselle Park, New Jersey
We drew one 10 x 10 square on the computer by measuring one point 10 cm
from another until we had four points that, when connected, formed a 10 cm X 10
cm square. Then by drawing a line connecting each pair of opposite corners of
the square, we found the center. We then measured 10 cm over and drew a point.
We used this as the 10 cm radius of a circle. Next, we used this distance to
measure another 10 cm X 10 cm square. Using the computer, we could rotate the
square around the circle with the mouse. We told the computer to measure area
and perimeter of the two squares' overlapping area. The area remained
constant--
25.00 cm. The perimeter changed with every movement of the mouse. Thus, 25
cm is
the highest and only area of the two squares' overlapping portion.
***********************************************
From: geom...@oasis.bellevue.k12.wa.us
From: Lyndsey Hayes and Gwen Dillow
Grade: 10
School: Sammamish High School, Bellevue, Washington
Dear Ms. Fetter,
Our answer is the largest area that the section could be is
25 centimeters squared. We found our answer by drawing a picture
of the two squares. We saw that only 1/4 of the area of the
square could be covered so we figured out the area of both of the
squares is 100 centimeters squared. We divided 100 by 4 and
got 25. That is our answer.
***********************************************
From: sw...@shhs1.smoky.org
From: Sandra Wai
Grade: 9
School: Smoky Hill High School, Aurora, Colorado
The largest possible value for the area where they overlap is
25 square cm.
The center of a square is the midpoint of the sides, so if I
overlap to a square, the lenght of the side would be 5. So I used
the formula to find the area of the square which is (side x side),
(5cm x 5cm)=25 square cm. Also I did try to overlap to a triangle
which base is 5 and the height is 5, too. And I use the formula to
find the area of the triangle: (base x height)/2, (5 x 5)/2 which
is 12.5 square cm. Since I try to overlap to different shape and
finally I found out the largest possible area is 25 square cm.
***********************************************
From: a...@pacbell.net
From: Jenny Kaplan
Grade: 6
School: Castilleja Middle School, Palo Alto, California
1. First I drew a square, and thought of two obvious
possibilities,
1)That the square whose vertex was at the center of the other
square took up a corner,
then the area would be 25 (it is a 5 x 5 square because
the vertex is at the center, and the line is going
straight out splitting the side (10)in half.
2)That the square whose vertex was at the center of the other
square took up a side,
then the area would also be 25 (the side is 10, and
the height is 5).
2. Then I thought of another possiblity,
That the square took up some side and some corner to find
this figure's area this is what I did,
1, I drew the square who took up a side (2)) over
over this one.
2, It had 3 parts (two small triangles, and a big one).
I found that the two smaller triangles were congruent. This is
because they both had a 45 degree angle. And, at the center where
the two 90 degree
angles overlapped the two outside angles when added to the middle
angle
equaled 90 degrees so the two outside angles are equal. Also two
sides are equal
because they are half diagonals.
3, Also the square that took up a side had a small
triangle and a big one, same with the one that
had part of a side and a corner, therefore
no matter where you put the center square the area
will always be 25!!!!!
***********************************************
From: fvi...@aol.com
From: Kirk Vissat
Grade: 10
School: Smoky Hill High School, Aurora, Colorado
The area where the two squares overlap is 25 cm^2.
First, I drew a 10 X 10 square. Then I drew the second square
with its bottom left vertex in the center of the first square.
The sides of the overlapped square was 5 X 5. Finaly I
multiplied the length by the height to find the area.
**Note: rotating the square about its vertex will not change
the area that is overlapped.**