My Math Forum  

Go Back   My Math Forum > Math Forums > Math

Math General Math Forum - For general math related discussion and news


Reply
 
LinkBack Thread Tools Display Modes
March 8th, 2017, 03:25 AM   #1
Bio
Newbie
 
Joined: Mar 2017
From: Australia

Posts: 2
Thanks: 0

Question 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 07:41 AM.
Bio is offline  
 
March 8th, 2017, 04:27 AM   #2
Senior Member
 
Joined: Apr 2014
From: Glasgow

Posts: 1,988
Thanks: 646

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.
Benit13 is offline  
March 8th, 2017, 06:09 AM   #3
Math Team
 
Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 9,402
Thanks: 638

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
Denis is offline  
March 8th, 2017, 07:30 AM   #4
Global Moderator
 
Joined: Dec 2006

Posts: 17,158
Thanks: 1284

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.
skipjack is offline  
March 10th, 2017, 11:27 AM   #5
Math Team
 
Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 9,402
Thanks: 638

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.
Denis is offline  
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.
y2kevin is offline  
March 12th, 2017, 10:13 AM   #7
Global Moderator
 
Joined: Dec 2006

Posts: 17,158
Thanks: 1284

Why assume 60° slope?
skipjack is offline  
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
y2kevin is offline  
March 13th, 2017, 07:27 PM   #9
Math Team
 
Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 9,402
Thanks: 638

Quote:
Originally Posted by y2kevin View Post
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?
Denis is offline  
March 13th, 2017, 10:59 PM   #10
Global Moderator
 
Joined: Dec 2006

Posts: 17,158
Thanks: 1284

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.
skipjack is offline  
Reply

  My Math Forum > Math Forums > Math

Tags
hypotenuse, mystery, triangle, trigonometry



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Crime mystery shunya Elementary Math 3 November 10th, 2015 01:48 PM
The mystery of cos 15/sr280 Mark Heaton Art 4 February 28th, 2014 12:59 PM
Rubik's Cube Mystery mathmaniac New Users 93 May 13th, 2013 03:20 AM
A Pythagorean mystery icemanfan Math Events 2 April 9th, 2012 05:50 AM
RUBIK'S CUBE MYSTERY mathmaniac Applied Math 0 January 1st, 1970 12:00 AM





Copyright © 2017 My Math Forum. All rights reserved.