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Without using a calculator, find the hypotenuse of a right triangle that has legs with lengths 2.64 and 4.95.

 Nov 13, 2015

Best Answer 

 #1
avatar+104855 
+16

Let's pretend that the legs are  264 and 495 units in length

 

264   factors as  11 * 24  ....so  2.64 =  11 * .24

 

And

 

495  factors as  11 * 45........so   4.95  = 11 * .45

 

And a Pythagorean Triple triangle of  8 - 15 - 17 ...... exists....so.....   a triangle  of   .08  - .15 - .17 also exists

 

And notice that   3*.08  = .24     and 3*.15 = .45.......so  3* .17 =  .51   ....which implies that....

 

[ 3* .08]^2 + [3 * .15]^2  = [ 3 * .17]*2..........which further implies that......

 

[ .24]^2 + [ .45]^2   = [.51]^2    .........which further implies that....

 

[11* .24]^2  + [ 11 * .45] ^2   = [ 11 * .51]^2

 

So....the hypotenuse must be   11 * .51  = [ 10 + 1] [.51]   = 5.1  + .51   =  5.61 units in length

 

 

cool cool cool

 Nov 13, 2015
 #1
avatar+104855 
+16
Best Answer

Let's pretend that the legs are  264 and 495 units in length

 

264   factors as  11 * 24  ....so  2.64 =  11 * .24

 

And

 

495  factors as  11 * 45........so   4.95  = 11 * .45

 

And a Pythagorean Triple triangle of  8 - 15 - 17 ...... exists....so.....   a triangle  of   .08  - .15 - .17 also exists

 

And notice that   3*.08  = .24     and 3*.15 = .45.......so  3* .17 =  .51   ....which implies that....

 

[ 3* .08]^2 + [3 * .15]^2  = [ 3 * .17]*2..........which further implies that......

 

[ .24]^2 + [ .45]^2   = [.51]^2    .........which further implies that....

 

[11* .24]^2  + [ 11 * .45] ^2   = [ 11 * .51]^2

 

So....the hypotenuse must be   11 * .51  = [ 10 + 1] [.51]   = 5.1  + .51   =  5.61 units in length

 

 

cool cool cool

CPhill Nov 13, 2015

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