Help with Special Triangle Problem

**mathguy20** · May 22nd 2010, 05:21 AM

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

**skeeter** · May 22nd 2010, 05:48 AM

given $AB = 2$

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

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

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

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

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

**mathguy20** · May 22nd 2010, 12:10 PM

Originally Posted by skeeter

given $AB = 2$

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

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

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

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

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

Thanks!

So if you can just confirm for me:

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

Is this correct and in simplest form?

**mathguy20** · May 27th 2010, 04:03 AM

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

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