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Mathematics 21 Online
OpenStudy (anonymous):

one leg is x and the other is h and the hypotenuse is 2x and i have to solve for h and i got just x did i do this right?

OpenStudy (anonymous):

This is a special right triangle, 30-60-90

OpenStudy (anonymous):

yes and how can we use that to solve this problem ?

OpenStudy (anonymous):

What do we know about 30-60-90 right triangles?

OpenStudy (anonymous):

the hypotenuse is 2x the shortest side and the other leg is x times the squareroot of 3

OpenStudy (anonymous):

Exactly

OpenStudy (anonymous):

So what is h?

OpenStudy (anonymous):

h= x squareroot of 3

OpenStudy (anonymous):

right

OpenStudy (anonymous):

but cant you use pythagorean theorem as well

OpenStudy (anonymous):

Yes.

OpenStudy (anonymous):

but how can we do that b/c it says use an algebraic expression

OpenStudy (anonymous):

\[h=\sqrt{(2x)^2-x^2}=\sqrt{4x^2-x^2}=\sqrt{3x^2}=x \sqrt{3}\]

OpenStudy (anonymous):

see?

OpenStudy (anonymous):

yes i get it thanks a lot

OpenStudy (anonymous):

you're welcome

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