Areas of Squares - March 4-8, 1996
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You have two squares and a circle. One square has the diameter of the circle as an edge. The other square is inscribed in the circle. (Draw yourself a good picture.) What's the area of the small square compared to the area of the big one?
Solutions
This is a neat problem that can be hard or easy, depending on how you draw your picture and what you "see" when you look at it. When I first did this problem myself a while back, I used a very complicated method because I didn't see something. You guys don't seem to have that problem, as everyone pretty much used the 'easier' ways of solving it.
There was a great range in the explanations. Brian Gordon provided a nice proof, using the fact that the area of a rhombus is half the product of the diagonals - since the square is a rhombus, this avoids all those square roots, which some people find comforting. Several folks changed the picture (or at least my vision of the picture) to put the vertices of the small square at the midpoints of the edges of the large square. This makes it really obvious that it's one half.
A number of folks did the problem in Sketchpad and then just stated what Sketchpad told them. While this gives you a good idea of what's going on, I think you should then look for a way to prove it 'algebraically' so that you know it's always true in the general case. Sketchpad provides a good jumping off point and an excellent way to explore the problem, but challenge yourself to figure it out on your own as well.
The same thing goes for those who figured out a specific case, for example making one of the lengths 4, figuring out the areas, and seeing that the ratio is 1:2. Once you've done this, then use the general case, using x for the length and figuring out the areas again. It is always a good idea to be as general as you can, though it never hurts to start with a specific example.
Following are highlights. The names of all the people who submitted correct solutions and most of the solutions are available, and we welcome 10 new schools, including firsts from Idaho, Indiana, and Newfoundland.
Sean Mostafavi
Grade: 10
School: Smoky Hill High School
Answer: The ratio of the large square over the small square is 2.
Large Triangle
-------------- = 2.00
Small Triangle
I used the Geometer's Sketchpad. I constructed a square inscribing
a circle. Then I drew a chord (or line) from one of the corners of
the square to the other (diagonal). Then I used this diagonal to
construct another square using a sample script in Sketchpad. The
larger square is 2 times bigger than the small one.
The ratio of "pi" will always remain the same, so in any circle
that you inscribe a square in, the diagonal of that square will
make another square exactly two times bigger than the first one.
The definition of a square states that all four sides must be
congruent, so you make four sides using that diagonal (or chord of
circle) to make the second square.
Scott Bridger, Mark Overton, Bryon Joel Grade: 9 School: Smoky Hill High School The ratio of the large square to the small square is 1:2. By drawing diameters as diagonals of the small square you create four congruent 45-45-90 triangles with a hypotenuse that is a radius of the circle. The same triangles when formed from the large square: the hypotenuse is the diameter of the circle, making the triangle similar in a ratio of 1:2.
Thomas S. Kuo
Grade: 7
School: Murray Junior High School, Ridgecrest, California
The ratio of the area of the small square to the area of the
big one is 1:2.
Let r be the radius of the circle and l be the length of the edge
of the small square.
Since the big square has the diameter of the circle, 2r, as an
edge, its area is (2r)^2 = 4r^2
A B
* * * * * * *
* * * *
* * * *
* * C *
* * * *
* * * *
* * * * * * *
Let the above square be the small square inscribed in the circle.
Point C is also the center of the circle. Point A and B are on
the circle. Angle ABC is 90 degrees. I also know that AC = BC = r.
Then I can solve the length of the edge of the small square, l, by
using Pythagorean Theorem.
l^2 = AB^2 = AC^2 + BC^2 = r^2 + r^2 = 2 r^2
Then the area of the small square is l^2 = 2 r^2.
Put them into ratio form and simplify it:
area of small square / area of big square
= 2 r^2 / 4 r^2 = 1/2
Gail Watling / Carolyn Watling Grades: 6 / 9 School: Vineyard E.H.S./ Granada High School If a square has the diameter of a circle as the length of its edge, and another square is inscribed within that circle, then the smaller square that is within the circle has its four corners at the midpoints of the sides of the larger square. This creates four right triangles which are on the outside of the smaller square, whose hypotenuses are the sides of the smaller square. The area of one of these triangles is, given side 'S', (1/2 S) squared divided by 2. The area of all four right triangles on the outside of the small square is 4 times that amount, or 4(1/2 S)squared divided by 2. When that is reduced, it is equal to 1/2 S squared. If the large square ,side length S, has an area of S squared, and the four triangles cut off the large square have a combined area of 1/2 S squared, then the remaining area, which is the smaller inscribed square, must have an area of 1/2 S squared. The area of the small square compared to the larger is one-half. This is very easy to see using tangrams, as you can just flip the small right triangles over to cover the small square. Drawing it on graph paper made it easy to count the area squares too.
Nathan Roberts
Grade 11
School: College Park High School, Pleasant Hill, California
Well, it took me about one minute to draw a diagram and see for
myself what the correct answer was, two minutes to algebraically
prove it to myself, and now for the hard part: describing it in
words.
The diagram I drew was a circle with D as the diameter, and a
square with D as the length of the sides, so the area of the
larger square is D^2 (D squared). The circle was inside the
square, so the circle touched the larger square at exactly four
points - the midpoints of the sides.
Then I drew the smaller square inside the circle. I could
"rotate" the smaller square inside the circle to wherever I wanted
and the square would be congruent, so I made the 4 points of the
smaller square touch the 4 points I mentioned above - the
midpoints of the larger square. Now, I already know that if the
four points of a square touch the midpoints of a larger square,
then the smaller square is exactly half the size, but here is how
I proved it algebraically:
The diagonal of the smaller square goes from the top midpoint of
the larger square to the bottom, so the diagonal of this smaller
square is also D (the diameter of the circle and the length of the
larger square's sides). I want to find the length of the sides of
the smaller square (to find the area with). Now I have a right
triangle. The legs are two sides of the smaller square (which I'll
call "X"), and the hypotenuse is the diagonal of the smaller
square. Now I use the Pythagorean theorem:
X^2 + X^2 = D^2
or
2X^2 = D^2
Now I solve for X.
sqrt2X = D
X = D/sqrt2 (Divide both sides by sqrt2. Now I have the length
of the sides of the smaller square)
Now I know the length of the sides of both squares: D for the
larger, D/sqrt2 for the smaller. Now I square them both to get
the areas:
D^2 Larger square
(D/sqrt2)^2 = D^2/2 Smaller square
Now you can see, through substitution, that the area of the smaller
square is exactly half of the larger.
Wow, explaining that turned out to be a lot easier than I thought.
Brian Gordon
Dartmouth '92
This one IS easier... The big square is twice the area of the
small square.
The area of a square is either s^2, or .5 * d^2 (since a square
is a rhombus and the diagonals are equal).
Well, the diameter of the circle is s for the big square and it's
d for the small square.
Officially, if we let x=the diameter of the circle:
Big square area: x^2
Small squarea: .5x^2
So it's a 2-to-1 ratio.
--bri
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