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

Dabae
Nov 13, 2015

#1**+15 **

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

CPhill
Nov 13, 2015

#1**+15 **

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

CPhill
Nov 13, 2015