Question

Find the length of the height of the right trapezoid shown below, if it has the greatest possible area and its perimeter is equal to 6 units.

Answer 1

Answer:

The height of the right trapezoid is   $\frac{6}{5+\sqrt{3}}\ units$

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

$P=AB+BC+CD+DH+HA$

we have

$P=6\ units$

$AB=BC=CH=HA=x$ ---> because is a square

substitute

$6=x+x+CD+DH+x$

$6=3x+CD+DH$ -----> equation A

In the right triangle CDH

$sin(30\°)=\frac{CH}{CD}$

$sin(30\°)=\frac{1}{2}$

so

Remember  that $CH=x$

$\frac{1}{2}=\frac{x}{CD}$

$CD=2x$

$tan(30\°)=\frac{CH}{DH}$

$tan(30\°)=\frac{\sqrt{3}}{3}$

so

$\frac{\sqrt{3}}{3}=\frac{x}{DH}$

$DH=x\sqrt{3}$

substitute the values in the equation A

$6=3x+CD+DH$ -----> equation A

$CD=2x$

$DH=x\sqrt{3}$

$6=3x+2x+x\sqrt{3}$

$6=5x+x\sqrt{3}$

$6=x[5+\sqrt{3}]$

$x=\frac{6}{5+\sqrt{3}}\ units$