+0  
 
0
47
1
avatar+378 

Find all values of t such that t - 1, t + 1, and 4 could be the lengths of the sides of a right triangle.

waffles  Oct 22, 2017
Sort: 

1+0 Answers

 #1
avatar+78762 
+2

If "t" has to be an integer...it is impossible to have integer sides with "4" as a hypotenuse

 

So.......I will assme that either

 

4^2 + ( t - 1)^2  = ( t + 1)^2     or    4^2 + (t + 1)^2  = (t - 1)^2

 

But.......for the second to be true, t  = -4, which would give us negative side lengths

 

So.....we have

 

4^2 + ( t - 1)^2  = ( t + 1)^2

 

16 -2t  =  2t

 

16  = 4t   →   t  = 4

 

So....the other two sides are 3 and 5

 

If there are no restrictions on t, we have that

 

( t - 1)^2  +  ( t + 1)^2  = 4^2

 

2t^2  + 2  =  16

 

t^2 =  7   →   t =  √7

 

So.....the sides are   √7 - 1 , √7 + 1 ,  4

 

 

cool cool cool

CPhill  Oct 22, 2017
edited by CPhill  Oct 22, 2017

5 Online Users

avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details