Thread: Help with Special Triangle Problem

1. Help with Special Triangle Problem

Would appreciate any help on the following problem, Thank You!

2. given $\displaystyle AB = 2$

BC is the hypotenuse of the 45-45-90 triangle ...

$\displaystyle BC = (AB)\sqrt{2} = 2\sqrt{2}$

BC is also the longer leg of a 30-60-90 triangle ... BC is the short leg.

$\displaystyle BD = \frac{2\sqrt{2}}{\sqrt{3}} = \frac{2\sqrt{6}}{3}$

from this point, you should be able to finish up.

3. Originally Posted by skeeter
given $\displaystyle AB = 2$

BC is the hypotenuse of the 45-45-90 triangle ...

$\displaystyle BC = (AB)\sqrt{2} = 2\sqrt{2}$

BC is also the longer leg of a 30-60-90 triangle ... BC is the short leg.

$\displaystyle BD = \frac{2\sqrt{2}}{\sqrt{3}} = \frac{2\sqrt{6}}{3}$

from this point, you should be able to finish up.
Thanks!

So if you can just confirm for me:

$\displaystyle CD = \frac{2\sqrt{42}}{3}$ (Used Pythagorean Theorem)
and
$\displaystyle x = \frac{2\sqrt{7}}{\sqrt{3}}$ (Used Pythagorean Theorem again)

Is this correct and in simplest form?

4. bump.... Anybody able to confirm if my previous answer was correct?
Thanks!