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March 8th, 2017, 03:25 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 rightangled 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 07:41 AM. 
March 8th, 2017, 04:27 AM  #2 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 1,881 Thanks: 608 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, 06:09 AM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 8,425 Thanks: 570 
You'd think the ratio should be given as 3:4, not 3:5, so based on 345 right triangle... Havva looksee here: How to Solve Problems with the Altitude0nHypotenuse Theorem  dummies 
March 8th, 2017, 07:30 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 16,579 Thanks: 1198 
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, 11:27 AM  #5 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 8,425 Thanks: 570 
As a point of interest: smallest primitive Pythagorean triangle where BM, CM and AM are integers: 65156169 ; BM=25, CM=144, AM=60. Smallest nonprimitive: 152025 ; 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 306090 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: 16,579 Thanks: 1198 
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 Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 8,425 Thanks: 570  
March 13th, 2017, 10:59 PM  #10 
Global Moderator Joined: Dec 2006 Posts: 16,579 Thanks: 1198 
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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hypotenuse, mystery, triangle, trigonometry 
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