Solve for x and y. Show your work.

Answer:
Step-by-step explanation:
We can find the value of x and y of the triangles using the ratio of 45°-45°-90° triangle and 30°-60°-90° triangle.
(let the length of the leg = a)
[tex]hypotenuse=\sqrt{a^2+a^2}[/tex]
[tex]=\sqrt{2a^2}[/tex]
[tex]=a\sqrt{2}[/tex]
[tex]\boxed{\bf leg:hypotenuse=1:\sqrt{2} }[/tex]
(let the length of the short leg = a, then the hypotenuse = 2a)
[tex]hypotenuse=\sqrt{short\ leg^2+long\ leg^2}[/tex]
[tex]2a=\sqrt{a^2+long\ leg^2}[/tex]
[tex](2a)^2=a^2+long\ leg^2[/tex]
[tex]long\ leg =\sqrt{4a^2-a^2}[/tex]
[tex]long\ leg =\sqrt{3a^2}[/tex]
[tex]long\ leg =a\sqrt{3}[/tex]
[tex]\boxed{\bf short\ leg:long\ leg:hypotenuse=1:\sqrt{3} :2 }[/tex]
For the 30°-60°-90° Triangle, given hypotenuse = 6
[tex]short\ leg:hypotenuse=1:2[/tex]
[tex]x:6=1:2[/tex]
[tex]2x=6[/tex]
[tex]\bf x=3[/tex]
For the 45°-45°-90° Triangle, given hypotenuse = 6
[tex]leg:hypotenuse=1:\sqrt{2}[/tex]
[tex]y : 6 = 1:\sqrt{2}[/tex]
[tex]\sqrt{2} y=6[/tex]
[tex]\displaystyle y=\frac{6}{\sqrt{2} }[/tex]
[tex]\bf y=3\sqrt{2}[/tex]