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 March 8th, 2017, 02:25 AM #1 Newbie   Joined: Mar 2017 From: Australia Posts: 2 Thanks: 0 Mystery Hypotenuse (I'm only a Year 7, so please explain clearly how you found the solution.) In the right-angled triangle ABC, a point M on the hypotenuse BC is such that AM is perpendicular to BC. Also, MC is 8 cm longer than BM, and the ratio AB:AC=3:5. How many centimetres is the hypotenuse? I started by calling the length of BM y, and MC y+8 and then using similar triangle ratios created fractions AM/MC = BM/AM which cancelled out to make BM/MC which substitutes into y/y+8 but then I got stuck. Can anyone help me? By the way, the answer is 17. Last edited by skipjack; March 8th, 2017 at 06:41 AM.
 March 8th, 2017, 03:27 AM #2 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,094 Thanks: 701 Math Focus: Physics, mathematical modelling, numerical and computational solutions Hint: BM and MC can also be obtained by considering: 1. the aspect ratio of the triangle, which means that one of the angles of the triangle is fixed; and 2. Pythagoras' theorem. Once you get a second evaluation for the length BC in terms of y, you can solve it to get a specific value of y and hence the value of BC. I personally set the length AB to a length 3x and the length AC to a length 5x and then solved for x and hence the length of the hypotenuse, but it should still be possible doing it in terms of BM instead.
 March 8th, 2017, 05:09 AM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,861 Thanks: 768 You'd think the ratio should be given as 3:4, not 3:5, so based on 3-4-5 right triangle... Havva looksee here: How to Solve Problems with the Altitude-0n-Hypotenuse Theorem - dummies
 March 8th, 2017, 06:30 AM #4 Global Moderator   Joined: Dec 2006 Posts: 18,718 Thanks: 1533 As BM/MC = (BM/AM)(AM/MC) = (AB/BC)² = 9/25, BM = (9/25)MC, but MC = BM + 8cm, so (16/25)MC = 8cm, which implies that MC = 12.5cm and BM = 4.5cm.
 March 10th, 2017, 10:27 AM #5 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,861 Thanks: 768 As a point of interest: smallest primitive Pythagorean triangle where BM, CM and AM are integers: 65-156-169 ; BM=25, CM=144, AM=60. Smallest non-primitive: 15-20-25 ; BM=9, CM=16, AM=12.
 March 12th, 2017, 09:14 AM #6 Newbie   Joined: Feb 2017 From: Omaha, Nebraska Posts: 8 Thanks: 0 Math Focus: Algebra and Trigonometry Not sure, but... let hypotenuse line segment MC = x and, let hypotenuse line segment BM = x - 8: x + x - 8 = 2x - 8 = x = 4 = MC x + 8 = BM = 12 hypotenuse line BC = BM + MC = 16 hypotenuse slope = rise/run = AC/AB = 5/3 = 1.6667 rise is 2/3 greater than run. assuming 60 degree slope, then we have a 30-60-90 triangle: BC = C = 16 = 2b ==> b = 8 AC = B = (sqrt 3) b = 13.856406 AB = A = b = 8 C^2 = A^2 + B^2 16^2 = 8^2 + 13.856406^2 -or- 16^2 = (sin30 16)^2 * (sin60 16)^2 Please jump in and correct, this is my best guess attempt at the problem.
 March 12th, 2017, 10:13 AM #7 Global Moderator   Joined: Dec 2006 Posts: 18,718 Thanks: 1533 Why assume 60° slope?
 March 13th, 2017, 06:42 PM #8 Newbie   Joined: Feb 2017 From: Omaha, Nebraska Posts: 8 Thanks: 0 Math Focus: Algebra and Trigonometry Without assumptions... let hypotenuse line segment MC = x and, let hypotenuse line segment BM = x - 8: x + x - 8 = 2x - 8 = x = 4 = MC x + 8 = BM = 12 hypotenuse line BC = BM + MC = 16 So I've calculated the hypotenuse to be 16, not the 17 you've provided. (The question becomes...Where did you get the 17 as a length of the hypotenuse from?) hypotenuse slope = rise/run = AC/AB = 5/3 = 1.6667 rise is 2/3 greater than run. Rise and Run is a linear relationship, if you increase or decrease the Run then the Rise must increased or decreased by a proportional amount. A given for this problem is the ratio 3:5, meaning for every 3 the Run is incremented the Rise increments by 5. Thus the slope relationship of 1.6667. (...and without assumptions) let x be the Run, then, 1.6667x is the Rise. Solve for length x by: x + 1.6667x = 16 ==> (16 is the hypotenuse length) x = 16/2.6667 = 6 y = 1.6667x = 10 6^2 = 36 10^2 = 100 Then, C^2 = A^2 + B^2 as 256 = 36 + 100
March 13th, 2017, 07:27 PM   #9
Math Team

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Quote:
 Originally Posted by y2kevin let hypotenuse line segment MC = x and, let hypotenuse line segment BM = x - 8: x + x - 8 = 2x - 8 = x = 4 = MC x + 8 = BM = 12 hypotenuse line BC = BM + MC = 16
Kevin...are you serious?

 March 13th, 2017, 10:59 PM #10 Global Moderator   Joined: Dec 2006 Posts: 18,718 Thanks: 1533 There were various mistakes by y2kevin. If MC = x and BM = x - 8, adding gives 2x - 8 = BM + MC = BC, which is incompatible with x = 4. Also, BM = x - 8 rules out x + 8 = BM. It turns out that AB = 3√(17/2) and AC = 5√(17/2), so BC² = AB² + AC² = 9(17/2) + 25(17/2)= 34(17/2) = 17². Using y2kevin's values would give BC² = 6² + 10² = 36 + 100 = 136, not 256.

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