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
$\displaystyle x^2 + y^2 = z^2$
$\displaystyle x^2 + x^2 = z^2$ <-- Since y = x
$\displaystyle 2x^2 = z^2$
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
$\displaystyle z = \sqrt{2} \cdot x = \sqrt{2} \cdot 5 = 5 \sqrt{2}$.
Example: You are given z = 20. Thus $\displaystyle x = \frac{z}{\sqrt{2}} = \frac{20}{\sqrt{2}}$
(Which simplifies to $\displaystyle 10 \sqrt{2}$.)
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 $\displaystyle x^2 + y^2 = z^2$ and you can work from there.
-Dan
Hello, aikenfan!
You did quite good . . . the ones you skipped are rather tricky.
For 1 - 10:Code:* * * *45°* z * * * * y * * * 45° * * * * * * * * * x
$\displaystyle \begin{array}{ccccc} & x & y & z \\ \\
3) & {\color{blue}10} & 10 & {\color{blue}10\sqrt{2}} & \text{Correct!} \\ \\
6) & {\color{blue}6\sqrt{2}} & {\color{blue}6\sqrt{2}} & 12 & \text{Right!} \\ \\
9) & {\color{blue}-} & {\color{blue}-} & 5\sqrt{10} & \end{array}$
For the 45-45-90 triangle, think of the sides as: .$\displaystyle \begin{array}{ccc}x & = &a \\ y & = & a \\ z & = & a\sqrt{2}\end{array}$
So, if we know one side, we can find the others.
In #3, we are given: $\displaystyle y \,= \,10$
Then: .$\displaystyle x \,= \,10,\;z \,= \,10\sqrt{2}$
In #6, we are given: $\displaystyle z \,=\,a\sqrt{2} \,=\,12$
Divide both sides by $\displaystyle \sqrt{2}\!:\;\;\frac{a\sqrt{2}}{\sqrt{2}}\:=\:\fra c{12}{\sqrt{2}}\quad\Rightarrow\quad a \:=\:\frac{12}{\sqrt{2}}$
Rationalize: .$\displaystyle a \:=\:\frac{12}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt {2}} \:=\:\frac{12\sqrt{2}}{2} \:=\:6\sqrt{2}$
Therefore: .$\displaystyle x \,=\,y\,=\,6\sqrt{2}$
In #9, we are given: .$\displaystyle z \,=\,a\sqrt{2}\,=\,5\sqrt{10}$
Divide both sides by $\displaystyle \sqrt{2}\!:\;\;\frac{a\sqrt{2}}{\sqrt{2}}\:=\:\fra c{5\sqrt{10}}{\sqrt{2}} \quad\Rightarrow\quad a \:=\:5\sqrt{5}$
. . $\displaystyle \begin{array}{cccc}9)\; & \;{\color{blue}5\sqrt{5}}\; & \;{\color{blue}5\sqrt{5}}\; & \;5\sqrt{10} \end{array}$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Try the same technique on the other problems.
For the 30-60-90 triangle, think of them as: .$\displaystyle \begin{array}{ccc}x & = & a\sqrt{3} \\ y & = & a \\ z & = & 2a\end{array}$