A Math Forum Project

Geometry Forum - Problem of the Week Solutions - Two Squares and a Circle, March 6-10, 1995 Annie says: This was a deceptive problem. Many students had problems with the ratio idea, and couldn't seem to state a conclusion after doing a bit of work. The solutions included below are exceptions. Given that there were no measurements included, it's a good exercise in talking in general terms, using x as a length. Several people had problems finding the right circles - the problem must be read very carefully to make sure the right things are being compared (especially since there aren't any pictures). The second part of the problem was neat - I certainly didn't expect it to come out in such a nice ratio! Maybe the students did. : - ) I also didn't use the method that Daniel, Avi, and Charles and Indira did - they turned the square inscribed in the semicircle into a rectangle inscribed in the circle. Much easier! I used the same method that Kim and Cecilia did. Sure, it works, but it's not nearly as cool as the rectangle way. The solutions included also show that you can get through a problem like this in many different lengths! Remember, longer doesn't mean more right. The more concisely you can explain your answer, the easier the solution will be to understand (usually - as long as short doesn't mean you left out a lot of details). Daniel Gabriel Problem of the Week March 6-10, 1995 Daniel Gabriel Akiba Hebrew Academy 10th Grade ** I apologize for the length of this, however, it is very difficult to explain without a picture. This week's problem really had two separate parts and it is very useful to think of them as two entirely separate problems. First, solve the first part and then begin entirely again to solve the second part. The best place to start is in the picture. Simply draw what is given to exist. You now have a circle and the diameter of the circle is an edge of another square as well as being a diagonal of the inscribed square. The diameter, as the diagonal of the inscribed square divides the inscribed square into two right triangle, both isosceles because all the sides of the square are equal. Since both of these are isosceles, you know that the base angles of each triangle formed by the diameter are each 45 degrees. Now you have two 45- 90-45 right triangles. Let the diameter of the circle (also the diagonal of the square and the edge of square B) equal 2x. Since it, being the hypotenuse of the right triangles created from the inscribed square by the diagonal, is root 2 times either leg, you can divide it by root 2 to find each leg. 2x / rt.2 = x rt. 2 Now you have the length of the side of the square. You can then determine the area of the square by multiplying length times width, or x rt.2 * x rt. 2. You now the area of the inscribed square is 2x^2. Remember now, that you know the length of the diameter of the circle which also is an edge of the second square (Square B). Since all sides of a square are equal, simply multiply length times width again (2x * 2x = 4x^2) giving you the area of the uninscribed square which is 4x^2. The area of the inscribed square is 2x^2 and that of the uninscribed square is 2x^2. Therefore, the ratio of the inscribed square to square B is 1 to 2. In other words, square B is twice as big as the inscribed square. Now we can move on to the second part of the problem. Again, begin by drawing. Draw a semi-circle with a square inscribed in it. Then, complete the circle and draw another square in the other semi-circle. Call each side of these squares X. (Different X from above) Both of these squares are now equal. Since they share one side (at the bottom of the semi-circle) they can be thought of as one larger shape. If you ignore this shared side, you have a rectangle which is X by 2X. Draw a diameter of the circle which is also a diagonal of the rectangle which was created by fusing the two squares. Remember, this diagonal was really also an edge of the other square involved in this problem. This diagonal divides the rectangle into two right triangles. Each right triangle has one leg being X and the other leg being 2X. The diagonal of the rectangle (also the diameter of the circle) can be found by the Pythagorean Theorem. Add the squares of the legs (x^2 + 4x^2 = 5x^2) Now take the square root of this to find the length of the hypotenuse ( diameter, diagonal, and edge of the second square involved in this problem). Sq. Root of 5x^2 = x rt. 5. Now going all the way back to the sides of the little squares inscribed in the semi-circle. Recall that the length of one of their sides was X. Find the area of the little square. Length X width = area (X* X = X^2) The area of the square inscribed in the semi-circle is X^2. Now you also have that the length of the diagonal of the rectangle is x rt. 5. You know that this diagonal is also an edge of the second ( the big square) square in this problem, so you now know that each side of the second square is x rt. 5. Again, multiplying length by width, you derive that the area of the big square is 5x^2. You now know that the area of the little square inscribed in the semi-circle is X^2 and the area of the big square is 5X^2. Now you know that the ratio between the area of the small inscribed square to the big square is 1 to 5, or you could say that the big square is 5 times bigger than the little one. Avi Klein Avi Klein Akiba Hebrew Academy 10th Grade Once one has constructed the picture one gives the length of the diameters as 2X, therefore the area of the square with a side as the diameter is 4x^2, because the diameter of the square is a 45 -45 -90 right triangle, we know that the diameter is rt. 2 as large as either side. Therefore, each side is x rt. 2, and when the area is calculated you get 2x^2. The comparison between the two squares is a 1 to 2 ratio. For the second half let us give one side of the square a length of just x. Then let us make a mirror image of the semicircle & square so we have a rectangle with the lengths X and 2X, because we have a rectangle we know we have a right angle. Therefore, we gave a right triangle, which we will call triangle ABC with C= hypotenuse. Using the Pythagorean theorem, X^2 + 4X^2 = C^2, C^2= 5X^2, the square root of which is x rt. 5. So the sides of the second square = x rt. 5. Therefore the area = X rt. 5 * X rt. 5 which is 5x^2. We also know that the area of the first square is X^2, since each side was X. The two compare in the ratio of 1 to 5. Charles Floyd and Indira Sarma Solution to pow, mar 6-11 In the first problem, assume that the length of one side of the inscribed square is X. Because the diagonal of a square inscribed in a circle is equal to the diameter of the circle, the length of one side of the external square, which is equal to the diameter of the circle, would be equal to the diagonal of the inscribed square. The length of a diagonal of a square is the square root of two multiplied by the length of any side. Thus, the area of the inscribed square would be X squared, while the area of the external square would be 2X squared. Consequently, the ratio of the inscribed square to the external one is 1:2. In the second problem, when the semicircle and the square are both reflected in the diameter of the circle, the resulting figures added to the preexisting ones create a rectangle and a circle. Again assuming that the length of any side of the original square was X, the longer length of the newly formed rectangle would be 2X and the shorter side would remain X. Using the Pythagorean theorem, the diagonal of the rectangle, which is equal to the diameter of the circle, is the square root of 5 times X. Thus the area of the external square would now be 5X squared while the inscribed square would be X squared. Consequently, the ratio of the inscribed square to the external one is 1:5. Charles Floyd and Indira Sarma Woodward Academy Kim Carbone Kim Carbone, grade 9, Fairfield HS Question #1 If a square is drawn with one of its sides the diameter of a circle and another square is inscribed in the circle, then the ratio of the inscribed square to the other square is 1:2. This is because the diameter of the circle is a side of the larger square and also the diagonal of the inscribed square. The diagonal, therefore, cuts the inscribed square into two congruent isosceles right triangles or 45-45-90 triangles. Therefore, if one of the sides of the inscribed square is a, then the hypotenuse of the 45-45-90 triangle is a*sqrt(2). This means that the area of the inscribed square is a^2, and since the hypotenuse of the 45-45-90 triangle is also a side to the larger square, then the area of the larger square is (a*sqrt(2))^2 or 2a^2. Therefore, the ratio of the smaller square to the larger square is a^2:2a^2 or 1:2. Question #2 If the same square is drawn from the diameter of the circle but another square is inscribed in the semicircle (in the larger square), first say that the diameter is a. If you divide the large square in half (right to the circle's center) with a segment, then the segment will be a perpendicular bisector to both sides that it intersects. It actually divides both squares in half. Draw a radius from the center of the circle to a corner of the small square. It is the hypotenuse for a triangle formed by a side of the small square and 1/2 of an adjacent side. The hypotenuse (radius of the semicircle) = 1/2 a, since the diameter is a. Then make a side of the inscribed square x, so the adjacent half a side (ending at the center) that makes up the triangle is 1/2 x. Therefore, using the Pythagorean Theorem, x^2 + (.5x)^2 = (.5a)^2 or 5x^2 = a^2. That tells you that, since the area of the square inscribed in the semicircle has an area of x^2, and the large square with a diameter as a side has an area of a^2, that the area of the large square is 5 times the area of the small square. Therefore, the ratio of the square inscribed in the semicircle to the square with a diameter as a side is 1:5. Cecilia Merediz Cecilia Merediz, grade 9, Fairfield HS The small square is 1/2 the size of the big square. If the 2 legs of the triangle formed by the diagonal of the small square are given the value of x, then the diagonal of the small square (which is also a side of the big square) has a measure of x*sqrt(2). This is because in a 45-45-90 triangle, the hypotenuse is the leg times sqrt(2). Since each side of the small square is x, then the area is x^2. Since the sides of the big square are x*sqrt(2), the area is 2x^2 or two times the area of the smaller square. Therefore, the ratio of the smaller square to the larger square is 1:2. The ratio of the small square to the big square is 1:5. The center of the circle is also the midpoint of the lower side of the small square. I named each half of this side 'a'. So each side of the small square is 2a, leaving the area to be 4a^2. I connected the center of the circle to the upper corners of the small square and labeled these two radii c. I came up with the equation c^2 = (2a)^2 + a^2 by the Pythagorean Theorem. So c^2 = 5a^2 or c = a*sqrt(5). Since c is a radius and two radii make up a diameter, 2c = d, where d is the diameter and a side of the large square. d = 2a*sqrt(5). So d^2 will be the area of the large square and (2a*sqrt(5))^2 = 20a^2. This is 5 times the area of the small square. So the ratio of the small square to the big square is 1:5. Ruth Carver's Geometry Class Annie, Here is my students' solution for this week's problem. We are also attaching a sketch done on Sketchpad. Mount St. Joseph Academy POW 3/10/95 The Problem: A square has the diameter of the circle as an edge. Another square is inscribed in the circle and another square is inscribed in the semicircle. How do their areas compare? Solution: Let y be a side of the inscribed square. Then the area of the inscribed square is y^2 and the length of the diagonal of this square is y*square root of 2. The diagonal of the inscribed square is the diameter of the circle and thus a side of the other square. The area of the square with diameter y*square root of 2 is 2y^2. The square inscribed in the semicircle has side of length y/2 and area of y^2/4. The ratio of the area of the square whose side is a diameter to the area of the inscribed square to the area of the square inscribed in the semicircle is 2y^2 : y^2 : y^2/4 or 8 : 4 : 1 Mr. Gehrett's 7th Period Geometry Class Mr. Gehrett's 7th Period Geometry Class Ridgeview High School, Bakersfield, CA This is our solution. Problem #1 The diameter of the circle is x. The diameter is the same as the diagonal of the small square. Then, the diagonal splits the small square into two 45-45-90 right triangles. Therefore, if we use the 45-45-90 right triangle theorem, the side length of the small square is x divided by the square root of 2. The area of the big square is x squared and the area of the small square is x squared divided by 2. If we divide the areas, then we get 2. Therefore, the area of the big square is double the small square. Problem #2 The area of the big square is still x squared. The degree measure of the horizontal diameter is 180 degrees. Cut the semicircle into 3 equal parts by using radii. Therefore, the 180 degrees is divided by 3 and make 3 equal angles of 60 degrees. Split the small square vertically in half to create two 30 degree angles. Then, you have 4 30-60-90 right triangles. The hypotenuse of the right triangles is half the diameter. Therefore, using the 30-60-90 right triangle theorem, the length of the short leg is x divided by 4. Therefore, we double this length to get x divided by 2. The area of the small square is x squared divided by 4. Therefore, if we divide the areas, we get 4. The area of the big square is 4 times the area of the small square. Mr. Gehrett's 2nd Period Geometry Class Mr. Gehrett's 2nd Period Geometry Class Ridgeview High School, Bakersfield, CA Problem #1 - The diameter of the circle is x. Then, the side length of the big square is x. Then, the diagonal of the small square is x long which cuts the small square into two 45- 45-90 right triangles. Because of the 45-45-90 theorem, the sides of the small square are x divided by the square root of 2. Then, if you multiply out the areas each square you have: big square = x squared small square = x squared divided by 2 If you then divide the area of the big square by the area of the small square, you get 2. Therefore, the area of the big square is twice as large as the area of the small square. Problem #2 The area of the big square is still x squared. But the area of the small square has changed. If you double the vertical sides of the small square and extend them across the circle, you create two equal arcs that are 1/3 of the circle. If the arc is 1/3 of the circle, then the central angles are 120 degrees. The bottom horizontal side of the small square cuts the angles in half making 60 degree angles. Therefore, the right triangles formed in the small square must all be 30-60-90 degree triangles. If you apply the 30-60-90 right triangle theorem to these triangles, you find a side of the small square has to be 1/2 x. Therefore, the area of the small square is 1/4 x squared. The area of the big square is 4 times the area of the small square. Previous page || Next problem || Previous problem || Table of Contents || Forum Home Page