Answer:
The height of the right trapezoid is [tex]\frac{6}{5+\sqrt{3}}\ units[/tex]
Step-by-step explanation:
Let
x ----> the height of the right trapezoid in units
we know that
The perimeter of the figure is equal to
[tex]P=AB+BC+CD+DH+HA[/tex]
we have
[tex]P=6\ units[/tex]
[tex]AB=BC=CH=HA=x[/tex] ---> because is a square
substitute
[tex]6=x+x+CD+DH+x[/tex]
[tex]6=3x+CD+DH[/tex] -----> equation A
In the right triangle CDH
[tex]sin(30\°)=\frac{CH}{CD}[/tex]
[tex]sin(30\°)=\frac{1}{2}[/tex]
so
Remember that [tex]CH=x[/tex]
[tex]\frac{1}{2}=\frac{x}{CD}[/tex]
[tex]CD=2x[/tex]
[tex]tan(30\°)=\frac{CH}{DH}[/tex]
[tex]tan(30\°)=\frac{\sqrt{3}}{3}[/tex]
so
[tex]\frac{\sqrt{3}}{3}=\frac{x}{DH}[/tex]
[tex]DH=x\sqrt{3}[/tex]
substitute the values in the equation A
[tex]6=3x+CD+DH[/tex] -----> equation A
[tex]CD=2x[/tex]
[tex]DH=x\sqrt{3}[/tex]
[tex]6=3x+2x+x\sqrt{3}[/tex]
[tex]6=5x+x\sqrt{3}[/tex]
[tex]6=x[5+\sqrt{3}][/tex]
[tex]x=\frac{6}{5+\sqrt{3}}\ units[/tex]
