1. ## 45-45-90 Triangle

Hi

Principle 5: The length of a leg opposite a 45 degree angle equals one-half the length of the hypotenuse times the square root of 2.

$a=\tfrac{1}{2}c\sqrt{2}$

why is this so? when I solve for 'a' in $c^{2}=a^{2}+a^{2}$ I get $\frac{c}{\sqrt{2}}=a$

Thanks.

2. Originally Posted by norifurippu
Hi

Principle 5: The length of a leg opposite a 45 degree angle equals one-half the length of the hypotenuse times the square root of 2.

$a=\tfrac{1}{2}c\sqrt{2}$

why is this so? when I solve for 'a' in $c^{2}=a^{2}+a^{2}$ I get $\frac{c}{\sqrt{2}}=a$

Thanks.
When your final answer has a radical in the denominator, we must 'rationalize' the denominator. By that, I mean, remove the radical from the denominator. How do we do that? We simply mutiply both numerator and denominator by the radical. Same as multiplying by 1. Value doesn't change, just how the fraction looks. See:

$\frac{c}{\sqrt2}\cdot\frac{\sqrt2}{\sqrt2}=\frac{c \sqrt2}{2}=\frac{1}{2}c\sqrt2$

3. Thanks!