Take a circle and two squares. One square has the diameter of the circle as an edge. The other square is inscribed in the circle. What's the area of the small square compared to the big one?What if the small square were inscribed in the semicircle? Then how do the areas of the two squares compare? The big one still has the diameter of the circle as an edge.
Solutions aren't due for two weeks, so take your time and do a good job. I want good explanations! Simple numerical answers will not be acceptable.
Annie says: Solutions
There are obviously several ways to go about solving this problem. Depending on how you draw your picture for the first part, you will see different things. Tomomi Nakajima of Newport High School drew his squares so that both were centered at the center of the circle - this showed very clearly that the one inside was half the one outside (his solution is highlighted below).
Leif Linden from Germantown Academy, Amber Rau of Rogers High School, and Keith Ballantine of Lajes High School in the Azores all wrote very nice, clear explanations. You'll also notice that everyone drew pretty good pictures, no easy feat where circles are involved!
A slightly different approach comes from Dave Peterson. He used the idea of similar triangles to solve this, and his answer is very cool. I encourage you to read it carefully. It also gives a good explanation of how to best construct the inscribed square, which isn't easy.
Many people didn't bother to use the "d" that I gave in the problem. This doesn't really matter, of course, since we are finding ratios, not actually numbers, and I think that I probably shouldn't have given anything at all. Some people figured it out by assigning some number to the radius or diameter and going from there. This is fine, and will get you the right answer, but try doing it with a variable instead - it's good practice, and it's kind of neat how everything works out in the end.
To get the second part, you have to really know what "inscribed" means - a lot of people, I think, assumed that the square in the semicircle would be half the area of the one in the circle. This makes sense at first glance, since the semicircle is half the circle, but because of the way you must inscribe the square, it doesn't work out that way. However, it still works out to be a rational (nice) answer.
More than one person made the same mistake I did on the second part: when they (and I!) assigned x and 2x to the edges of the small square and then squared both of those, they got x^2 and 2x^2, forgetting to square the 2! I could not figure out why I wasn't getting the right answer! And yes, then I did kick myself, but it really pays to pay attention and use parentheses well. More than a few people made assumptions about the square inscribed in the semicircle that were not right. For example, they figured that the edge of the square was equal to the radius. This is a case where a well-drawn picture is a great help in figuring things out.
A list of all the people who got this problem right and most of the solutions are also available.
Tomomi Nakajima School: Newport High School, Bellevue, Washington For the first problem, I drew a square and inside that square, I drew a circle with the diameter equalling the edge of the square. Inside the circle, I drew a smaller square. The 4 points of the smaller square touch the same places as where the four points the circle touch the larger square. Those 4 points are the midpoints of the sides of the square. That tells me that the area of the smaller square is half the area of the larger square. For the second problem, I drew a square with a circle inside (the diameter being the same as the edge of the square). Then I drew 2 lines making the circle into a semicircle and cutting it up into 4 equal parts. And then, inside the semicircle, I drew a square that touches two points on the circle, one side being part of the line drawn earlier. _________________________ | | | | | | *one side of the larger square =2a | 2b | | a *a=2b |O-----------|P | *square OPRQ is divided in half at |b | | point U T |____2b______|U____a______| * OU=QU=a | | | * angle OUT=angle QUT |b | | * OT=QT=PU=RU=2b |Q-----------|R | * a^2=b^2+(2b)^2 | 2b | | a a^2=b^2=4b^2 | | | a^2=5b^2 |____________|____________| (I couldn't draw a circle inside the larger square. The circle touches the 4 midpoints in the larger square and point O and point Q touches the circle) How I solved to get the area of the smaller square: (2b)(2b)=4b^2 I simply multiplied two sides of the smaller square to get the area. How I solved to get the area of the larger square: (2a)(2a)=4a^2 I simply multiplied the two sides of the larger square to get the area. How I solved to get the comparison of the areas of the larger and smaller square: 4a^2=4(5b^2) = 20b^ 4a^2 is the area of the larger square and 5b^2 is equal to a^2 (solved earlier by the picture): 20b^2/4b^2 = 5 So the area of the smaller square is 5 times the area of the larger square.
Leif Linden School: Germantown Academy, Fort Washington, Pennsylvania In the first example, d is the diameter of the circle which is also one of the larger square's sides. The general equation for the area of a square is a=side squared, so in the case of the larger square, its area would be d squared. To find the area of the smaller square, you have to divide it into two right triangles. Let's label the legs of one right triangle x. They are both x because they make up two of the sides of the small square, and all sides of a square are the same length. The hypotenuse of the triangle is d, because it spans the diagonal of the inscribed square, making it the diameter of the circle. Using the Pythagorean theorem, we have x squared + x squared = d squared. If you solve for d, you get d = the square root of 2 * x. Then we can solve for x, which is d divided by the square root of 2. Now that we know two of the sides of the square (the two forming the legs of the triangle), we can figure that its area is x*x, or d squared over 2. So the big square has an area of d squared, and the small one an area of 1/2 d squared. The area of the smaller square appears always to be 1/2 of that of the bigger one. For the second part of the problem, the larger square still has an area of d squared, because it is the same as the one described in part 1. The smaller square is inscribed in the upper half of the origional circle. We will again attempt to use a triangle to find its total area. Let's take the midpoint of the base of the square (which is also the center of the circle). The distance from there to the upper right corner of the square is d/2 (as shown in the diagram) because the upper corner of the square is on the edge of the circle, and the line from there to the center is the radius of the circle. That line is the hypotenuse of the right triangle inside of the smaller square. Now, one of the legs of the triangle we will call x, because it is also one of the sides of the inscribed square. The bottom leg of the right triangle spans a distance from the midpoint of the base of the square to the edge of the square, making its length half of one of the sides of the square, or x/2. Next, the Pythagorean theorem comes into use when we find the value for x. We start with d/2 squared = x squared + x/2 squared, so x = d divided by the square root of 5. If we multiply the whole thing by the square root of 5 over the square root of 5 to simplify, and then multiply the answer by 5/5, we get d squared divided by 5. So: the area of the large square is d squared and the area of the small square is 1/5 d squared. I think we can safely say the area of the small square compared to the area of the larger square is 1:5.
Amber Rau Grade: 9 School: Rogers High School, Newport, Rhode Island Find the ratio of two squares As = Area of small square Ab = Area of big square Find As/Ab r = radius of circle Ab = (2r)^2 = 4r^2 1) Small square inscribed inside circle The hypotenuse is the radius of the circle that meets the corner of the square. The base and the leg are from the top right quarter of the square, in which the radius of the circle is cut in half leaving a right triangle. /| / | r/ |x / | /________| x side of square = 2x r^2 = x^2 + x^2 = 2x^2 x = r / 2(1/2) 2x = 2r/ 2(1/2) As = (2x)^2 = 4r^2 /2 As/Ab = 4r^2 /2 / 4r^2 = 1/2 *** Answer 2) Small square inscibed in semi-circle The hypotenuse is also the radius of the circle meeting the corner of the square. Side "a" is the height of the square and side "b" is half of the length of the square. /| / | / | r/ |a / | /_____| b side of square = a = 2b a^2 + b^2 = r^2 (2b)^2 + b^2 = r^2 5b^2 = r^2 b = r/ (5)^(1/2) As = a^2 = (2b)^2 = 4r^2 /5 As/Ab = 4r^2 /5 / 4r^2 = 1/5 *** Answer
Keith Ballantine School: Lajes High School, Azores Square 1: Side = 2r where r = radius of circle Area = side squared = (2r) squared = 4 r(squared) Square 2: inscribed in circle whose diameter is a side of square 1 Diagonal = 2r Separate out one triangle consisting of 2 sides and the diagonal. 45 45 90 triangle. Ratio of sides is 1, 1, sqrt(2) so side of second square is 2r/sqrt(2) Area of second square is side squared = 4 r(squared)/2 so ratio of areas is {[4 r(squared)]/2}/4 r(squared) = 1/2 to 1 or 1 to 2 Small square in semi-circle, big square, as before, has side 2r, Area 4 r(squared) Small Square: a radius of the circle could go from the midpoint of the flat side of the semi-circle to a corner of the square that touches the curved arc of the half-circle. This makes a triangle: The legs of this triangle consist of half of a side of the square and a side of the square, that is, 1:2 ratio. The radius of the circle, which goes from the midpoint of the side of the square to the corner of the square, is the hypotenuse of this triangle, and has length of sqrt(5) because: 1(squared) + 2(squared) = 5 Since the hypotenuse is also the radius, its length is r Length of half a side is r/sqrt(5), length of a whole side is 2r/sqrt(5) Area of small square is 4r(squared)/5 Area of large square was 4r(squared) [4r(squared)/5]/4r(squared) = 1/5 to 1, or 1 to 5
Dave Peterson Homeschool teacher Rochester, New York First, for the square inscribed in a circle, I drew the large square as circumscribing the circle, and the small square tilted so that it touches the sides of the large square. This way it is almost obvious that the ratio of areas is 1:2. (For proof, just fold the corners of the large square inward to cover the small one.) +----------*----------+ | * / \ * | | * / \ * | |* / \ *| | / \ | |/ \| * * |\ /| | \ / | |* \ / *| | * \ / * | | * \ / * | +----------*----------+ For the second part, I can draw the large square with one side on the diameter of the semicircle, and draw a segment from the center of that side to an opposite corner. If I imagine "deflating" the square while holding the center of the side fixed, the corner will follow this line, so that where it intersects the circle will be the corner of the inscribed square. (Formally, I would use similar triangles to prove this.) Using similar triangles, taking the side of the large square to be 2 so that the radius of the circle is 1 and the diagonal segment is sqrt(5) by Pythagoras, I get x/2 = 1/sqrt(5), so x = 2/sqrt(5) and x^2 = 4/5. The ratio of areas is then 4/5:4 or 1:5. +---------------------+ | /| | / | | sqrt(5) / | | / | | * / | 2 | *---------* | | * | /| * | | * | 1 / | * | |* | / |x *| | | / | | *-----+----*----+-----* 1 1
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