MHB Minimum Perimeter of a Trapezoid: Find R & P

[Albert1]

a trapezoid $ABCD,$ with $\overline {AD}// \overline {BC}, \overline {AB}=\overline {CD}$, and diagonal $\overline {AC}=15=\overline {BD}$
if R is its maximum area ,please find :
(1)R
(2)find its minimum perimeter P

 

[Greg]

Spoiler

$$(1)\quad R=\dfrac{225}{2}$$

$$(2)\quad P=30\sqrt2$$

 

[Albert1]

Spoiler

$$(1)\quad R=\dfrac{225}{2}$$

$$(2)\quad P=30\sqrt2$$

your answers are correct ,please show your solution

 

[Greg]

Spoiler

I calculated the area and perimeter of a square with diagonals of 15 units.

 

[johng1]

First, I give a more definitive proof for the maximum area. Next, I show that the answer for minimum perimeter is wrong.

Spoiler

For any convex quadrilateral, the area is 1/2 the magnitude of the cross product of the diagonals. In this case $${225\over 2}|\sin(\theta)|$$
Here $\theta$ is the angle between the diagonals. This is obviously maximized when $\theta=\pi/2$. Note there are many different isosceles trapezoids with equal diagonals of 15 that attain this maximum area.

Next, consider the rectangle with vertices $A=(x_0,y_0)=(7.5\cos(\theta),7.5\sin(\theta))$, $B=(x_0,-y_0)$, $C=(-x_0,-y_0)$ and $D=(-x_0,y_0)$. The perimeter is then $p=30\cos(\theta)+4\sin(\theta)$. Thus $p$ can be arbitrarily close to 30 by choosing $\theta$ to be sufficiently close to 0. I believe, but can not prove, that any isosceles trapezoid with diagonals of 15 has perimeter at least 30, but no such trapezoid attains the minimum of 30.
Edit:
I feel a little foolish. This was definitely a case of the forest hiding the trees. In triangle $ABD$, $|AB|+|DA|>|BD|=15$. Similarly for the other two sides of the trapezoid. So the perimeter is strictly greater than 30.

 

[Albert1]

a trapezoid $ABCD,$ with $\overline {AD}// \overline {BC}, \overline {AB}=\overline {CD}$, and diagonal $\overline {AC}=15=\overline {BD}$
if R is its maximum area ,please find :
(1)R
(2)find its minimum perimeter P

my solution :

Spoiler

View attachment 5080