Isosceles Right Triangle - Finding Legs based on Hypotenuse

[Masonan]

So I've been working on this one question for a long time now, so I decided to post it here.

Question: How does one find the length of the legs on an isosceles right triangle that has a hypotenuse of 14?

I've tried a number of things to solve this.

14^2=x^2+x^2
196=x^2+x^2
(sqr)196 (sqr)x^2
14=x^2+x
(sqr)14 (sqr)x^2
3.75=x+x
3.75=2x
1.87=x

or

14^2=x^2+x^2
196=x^2+x^2
196-x^2=x^2

I have no idea what to do. Help would greatly be appreciated.

 

[pka]

\(\displaystyle \begin{gathered} 2x^2 = \left( {14} \right)^2 \hfill \\ x^2 = \left( {\frac{{14}}{{\sqrt 2 }}} \right)^2 \hfill \\ x = \frac{14}{{\sqrt 2 }} \hfill \\ \end{gathered}\)

 

[Masonan]

Thank you very much... I finally understand this.

 

[Loren]

2x[sup:16umasdz]2[/sup:16umasdz] = 14[sup:16umasdz]2[/sup:16umasdz]

x[sup:16umasdz]2[/sup:16umasdz] = 14[sup:16umasdz]2[/sup:16umasdz]/2

\(\displaystyle x=\frac{14}{\sqrt2}\)???

 

[mmm4444bot]

... \(\displaystyle x=\frac{14}{\sqrt2}\)???

How about rationalizing the denominator? (This is a rhetorical question.)

\(\displaystyle \mbox{x = } 7 \cdot \sqrt{2}\)

~ Mark

 

[Mrspi]

Or, when you have

196 = x^2 + x^2, or
196 = 2x^2

why not divide both sides by 2?

98 = x^2

sqrt(98) = x

And simplify sqrt(98)...to 7*sqrt(2).

7 * sqrt(2) = x

You can kill a fly with a flyswatter...why use an atom bomb?

 

[Denis]

Well Mrspi, to each his/her own: I prefer pka's :wink: