A Math Forum Project

Geometry Forum - Problem of the Week Solutions - Length of the Hypotenuse, Feb. 13-17, 1995 Annie says: Nice job this week. Generally the explanations were very good - there's a bunch of stuff going on here that's difficult to explain. In many cases, it wouldn't hurt to give things some labels, as Ruth Carver's class and David and Sarah McMenomy did. Then we can talk about "the leg of the new triangle which is only half of one of the legs of the original triangle" by just saying "AE" or whatever. Much neater, eh? Also, note the differences in length in the solutions. They're all very different. Read them for yourself and see which ones you think give enough information/too much information. What parts can you gloss over? What parts are very important? Aron Freidenreich POW Feb 13, 1995 You know that the medians with the lengths given bisect the legs of the triangle. Therefore, call each half of one leg y and each half of the other leg x. That makes one leg of the triangle 2y and the other 2x. The original triangle can be turned into two separate triangles, each with a hypotenuse being one of the given medians. In this way you have a measure for all sides of two small right triangles. Applying the Pythagorean Theorem to both triangles, one gets a system, which then can be solved for x and y which then can be substituted into the original equation to give the measures of the two legs and then by again applying the Pythagorean Theorem, you get that the hypotenuse of the original triangle is the sq. root of 52. Aron Freidenreich 10th Grade Akiba Hebrew Academy 10th Grade Geometry Class at Mt. St. Joseph's Hi Annie, Here are Mount St. Joseph Academy's solutions for the Problem of the Week (2/13-17, 1995). The following students submitted solutions: Ashley Hall (10th), Kristin Biermann (10th), Kourtney Beahan (10th), Bridget Coyle (10th), Siobhan O'Brien (10th), Kathy Diamond (10th), Sarah McLaughlin (10th), Karen Lannutti (10th). All the students took the same approach. If BCA is a right triangle with right angle at C, draw the median from acute angle A to side BC and label the midpoint D. Draw the median from acute angle B to side CA and label the midpoint E. Let DC=x, then BC=2x. Let CE=y, then CA=2y. Given: AD=5, BE=sq root of 40 In right triangle DCA, (DC)^2 +(AC)^2 = (AD)^2 x^2 + (2y)^2 = 5^2 or x^2 = 25-4y^2 In right triangle BCE, (BC)^2 + (CE)^2 = (BE)^2 (2x)^2 + y^2 = (sq root 40)^2 4x^2 + y^2 = 40 Substituting for x^2: 4(25-4y^2) + y^2 = 40 100- 16y^2 + y^2 =40 -15y^2 = -60 y^2 = 4, so y=2 Substituting for y yields : x=3 Therefore, the legs of right triangle BCA are 4 and 6. Using the Pythagorean theorem, the hypotenuse of the original triangle is the square root of 52 or 2 times the square root of 13. Thanks for the challenge! David and Sarah McMenomy We hope that we are sending this solution to the right address. It is a little unclear on the forum page where exactly answers to the problem of the week should be sent. This is the solution produced by David and Sarah McMenomy (home- schooled) ages 12 and 10, resp., (grades 7 and 4) with a trivial amount of coaching from their dad. They are both currently working through Jurgensen, Brown, and Jurgensen, _Geometry_ as part of their home curriculum. The answer we came up with was 2*sqrt(13), or sqrt(52): the former if answers are required in simplest form, the latter if consistent with the terms of the problem, which contained the unreduced sqrt(40). Should we include explanations? --- Construct triangle ABC, w/ right angle at C. Median AD (D the midpoint of seg BC) and median BE (E the midpoint of seg AC). AD = 5; BE = sqrt(40). Construct ED. --- AC^2 + CD^2 = AD^2 = 25 BC^2 + EC^2 = BE^2 = 40 AC^2 + BC^2 = AB^2 (25 - CD^2) + (40 - EC^2) = AB^2 65 - (CD^2 + EC^2) = AB^2 CD^2 + EC^2 = ED^2 65 - ED^2 = AB^2 ED = AB/2 (similar triangles) 65 - (AB/2)^2 = AB^2 65 - (AB^2)/4 = AB^2 65 * 4 = 5 * AB^2 260/5 = AB^2 52 = AB^2 AB = sqrt(52) = 2* sqrt(13) ----- There is an alternative method working from the fact that tri ADC is a 3-4-5 triangle, but we'll let that go for now. Daniel Gabriel Problem of the Week February 13-17 Daniel Gabriel 10th Grade Akiba Hebrew Academy The length of the hypotenuse of the right triangle is the square root of 52. (I cannot e-mail square root signs. Sorry.) The first thing to do in solving this problem is to draw a right triangle and then draw in the two medians which are involved in this problem. Then, one needs to think of the whole triangle as two separate triangles, one, having a hypotenuse of 5 and the other having a hypotenuse of the square root of 40. It may be useful to then redraw the one triangle as two. Since it is given that we have two medians to begin with, we know that the legs are divided into two equal parts by the medians. Then let the equal halves of the sides both equal x or y. (Both halves of one leg =x) Therefore, two times a number (x or y) will give us the length of a side. Let us assume that one leg therefore is equal to 2y and the other side of the original triangle is 2x. Now we shift back to our two smaller triangles. Each of these triangles has one entire leg of the original triangle as one of its legs and half of the other original leg as its other leg. (i.e. one of the two small triangles has legs 2y, x and 5, and the other small triangle has legs 2x, y, and sq. root of 40.) This being the case, we now have two right triangles for which we have all of the sides. We can apply the Pythagorean Theorem (a^2 + b^2 = c^2) twice (once on each triangle) to give us a system as follows: (NOTE: a carrot ^ indicates a power) (2y)^2 + x^2 = 5^2 y^2 + (2x) ^2 = sq. root of 40 ^2 Once we have this system we can, by using algebra, determine that y^2 = 4 and that x^2 = 9. With this information, we can go back to our original triangle and fill in 3 for the two X's and 2 for the Y's. We then, having the measures for the legs of the original triangle, can again apply the Pythagorean Theorem giving us the equation: (2 + 2) ^2 + (3 + 3) ^2 = c^2 4^2 + 6^2 = c^2 16 + 36 = c^2 52 = c^2 c = sq. root of 52 Geometry Class at Holland Hall School We are a geometry class in Holland Hall school in Tulsa OK. Our teacher is Libby Jones. The solution: We knew the median cut the sides of the larger triangle in half so we set up a substitution problem. Let 2x be the length of one side and 2y be the length of the other. We found two equations using the Pythagorean theorem. One was: (2x)^2 + (y)^2 =5^2 and the other was: (2y)^2 + x^2 = (sqrt (40))^2. We had 2 equations and 2 unknowns so we solved using algebra. 4x^2 + y^2 = 25 x^2 + 4y^2 = 40 ---------------- Multiply the bottom row by -4. Then add the bottom row to the top row and solve for y. Y will then equal 3. Put 3 back in for y and solve for x. X will then yield 2. Then we put the numbers back into the figure so the length of one side was (2)3 and the length of the other side was (2)2. We then used the Pythagorean theorem to solve for the Hypotenuse. 4^2 + 6^2 = C^2. The answer was the sqrt(52). So we now are looking forward to the next challenge. Libby Jones Ariel Collis Sorry, I had trouble attaching a file, then I realized I have no idea how to send a pict file from the Geometer's Sketchpad or a binary file in this system. Good for me to practice these things. The diagram I'll try to send is a right triangle with hypotenuse Z, vertical side broken into x and x, horizontal divided into y and y. The median of the triangle divide the triangle into two right triangles. Since medians are given, the sides opposite the acute angles are divided equally into two lengths. Using the Pythagorean theorem we can make the following equation. x2+(2y)2=52 and y2+(2x)2=the square root of 40 squared. We then say that X2=25-4y2 then we substitute it in the other equation, which we could complete to be y2+4x2=40. So the substitution would be y2+ 4(25-4y2)=40, which we can do out to y2+100-16y2=40 so -15y2=-60 so y2=4 and y=2. Substitute that into the original equation of y2+4x2=40 and you get 4+4x2=40 so 4x2=36 and 4x2=36 and x2=9 so x=3. Again using the Pythagorean theorem, now for the bigger triangle, we can set up the equation (2x)2+(2y)2=z2 simplified that is 16+ 36=z2 so z= the square root of 52 or 2 root 13. - Ariel Collis, Akiba Hebrew Academy Previous page || Next problem || Previous problem || Table of Contents || Forum Home Page