# Thread: Special Right Triangles

1. ## Special Right Triangles

Could someone please check my work? I was assigned multiples of 3...so there arent very many..I also skipped a few that I did not understand...

2. Originally Posted by aikenfan
Could someone please check my work? I was assigned multiples of 3...so there arent very many..I also skipped a few that I did not understand...
In all of your problems you are dealing with a specific kind of right triangle: one where the two legs are equal. So
$x^2 + y^2 = z^2$

$x^2 + x^2 = z^2$ <-- Since y = x

$2x^2 = z^2$

If you are given one of the legs, you know the other one is equal to the given leg. If you are looking then for z then solve the above equation for z.

If you are given z then you know the two legs are equal and equal to x in the above equation. So solve it for x.

Example: You are given y = 5. Thus x = y = 5. And
$z = \sqrt{2} \cdot x = \sqrt{2} \cdot 5 = 5 \sqrt{2}$.

Example: You are given z = 20. Thus $x = \frac{z}{\sqrt{2}} = \frac{20}{\sqrt{2}}$
(Which simplifies to $10 \sqrt{2}$.)

Doing the other kinds of triangles involves a similar process, though, of course, x and y may not be equal. Either way we still have $x^2 + y^2 = z^2$ and you can work from there.

-Dan

3. Hello, aikenfan!

You did quite good . . . the ones you skipped are rather tricky.

For 1 - 10:
Code:
*
* *
*45°*
z  *     *
*       * y
*         *
* 45°       *
* * * * * * * *
x

$\begin{array}{ccccc} & x & y & z \\ \\
3) & {\color{blue}10} & 10 & {\color{blue}10\sqrt{2}} & \text{Correct!} \\ \\

6) & {\color{blue}6\sqrt{2}} & {\color{blue}6\sqrt{2}} & 12 & \text{Right!} \\ \\

9) & {\color{blue}-} & {\color{blue}-} & 5\sqrt{10} & \end{array}$

For the 45-45-90 triangle, think of the sides as: . $\begin{array}{ccc}x & = &a \\ y & = & a \\ z & = & a\sqrt{2}\end{array}$
So, if we know one side, we can find the others.

In #3, we are given: $y \,= \,10$

Then: . $x \,= \,10,\;z \,= \,10\sqrt{2}$

In #6, we are given: $z \,=\,a\sqrt{2} \,=\,12$

Divide both sides by $\sqrt{2}\!:\;\;\frac{a\sqrt{2}}{\sqrt{2}}\:=\:\fra c{12}{\sqrt{2}}\quad\Rightarrow\quad a \:=\:\frac{12}{\sqrt{2}}$

Rationalize: . $a \:=\:\frac{12}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt {2}} \:=\:\frac{12\sqrt{2}}{2} \:=\:6\sqrt{2}$

Therefore: . $x \,=\,y\,=\,6\sqrt{2}$

In #9, we are given: . $z \,=\,a\sqrt{2}\,=\,5\sqrt{10}$

Divide both sides by $\sqrt{2}\!:\;\;\frac{a\sqrt{2}}{\sqrt{2}}\:=\:\fra c{5\sqrt{10}}{\sqrt{2}} \quad\Rightarrow\quad a \:=\:5\sqrt{5}$

. . $\begin{array}{cccc}9)\; & \;{\color{blue}5\sqrt{5}}\; & \;{\color{blue}5\sqrt{5}}\; & \;5\sqrt{10} \end{array}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Try the same technique on the other problems.

For the 30-60-90 triangle, think of them as: . $\begin{array}{ccc}x & = & a\sqrt{3} \\ y & = & a \\ z & = & 2a\end{array}$

4. I think that I have got it now, I am going to repost my work that I have changed to make sure that I am on the right track....