calculusgeometrysupremum-and-infimum
Three sides of a trapezium are each equal to $k$ cm.Prove that the greatest possible area of the trapezium is $\frac{3\sqrt3 k^2}{4}$ sq cm.
I let two non-parallel sides and one of the parallel sides as $k$(shorter one).I know that area of the trapezium is $\frac{1}{2}\times $sum of parallel sides $\times$ height.But in this question,neither height is given nor longest side is given.How should i formulate the equation of the area?
The midline divides the trapezoid into two equally high trapezoids. Its length is also the mean of the length of the two parallel sides. (You can see that by extending the nonparallel sides until they intersect in a point. The two parallel sides and the midline then forms the base line in three similar triangles).
Call the long parallel side $AB$, and the short one $CD$ with $AD$ and $BC$ being the two other sides of the trapezoid. Further let $r$ be the radius of the circle. The trapezoid then has height $2r$. Further, let $E, F, G, H$ be the places the circle touches the trapezoid at $AB$, $BC$, $CD$ and $AD$ respectively.
The hypothesis about the midline dividing the area (note that the two parts are also trapezoids) then dictates that $$ 3\frac{AB + \frac{AB + CD}{2}}{2}r = 5\frac{CD + \frac{AB + CD}{2}}{2}r \\ {3}AB + 3\frac{AB + CD}{2} -5CD -5\frac{AB+CD}{2} = 0\\ 2AB - 6CD = 0\\ AB = 3CD $$ so the midline has length $2CD$.
Since the sides touch the circle we must have the following: $$ AE = AH\\ BE = BF\\ CF = CG\\ DG = DH $$ Since $DH + AH + BF + CF = 16$, then we must have $AE+BE + CG + DG = 16$. But this also equals $4CD$, so $CD = 4$. The long parallel side $AB$ must therefore be $12$.
Quadrilaterals are not rigid: you can fix a side and move one of the other vertices freely while respecting all side lengths. The area changes when you move the vertex.
(image from http://www.mathsisfun.com/definitions/rigid.html)
Best Answer
You can parametrize the trapezium via its height $h$. Then the longer side is $$ 2 \sqrt{k^2-h^2} + k $$ and its area is $$ h (\sqrt{k^2-h^2} + k)\ .$$ The derivative for $h$ is $$k - \frac{h^2}{\sqrt{-h^2 + k^2}} + \sqrt{-h^2 + k^2}\ ,$$ which is zero for $h=0$ or $h=\pm \frac{\sqrt{3}}{2}k$. $h=\frac{\sqrt{3}}{2}k$ gives the proposed optimum, which you can be sure about after also checking the border case $h=k$ (which gives an area of $k^2$).