Question 971976: find the area of the largest trapezoid that can be inscribed a semicircle of diameter "d" in terms of "d"?. what is the length of the bottom base "x" of the same trapezoid ??
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! In the American definition, a trapezoid has two parallel sides  .
I am assuming that the drawing above shows what was meant: a trapezoid inscribed in a semicircle.
Intuitively, we could assume that the largest polygon that can be inscribed in a circle is the regular polygon of its kind,
so that the largest inscribed triangle would be an equilateral triangle,
the largest quadrilateral would be a square,
the largest pentagon a regular pentagon,
the largest hexagon a regular hexagon, and so on.
If that is so, the largest trapezoid inscribed in a semicircle may be half of a hexagon.
That trapezoid would have base angles measuring  ,
and could be thought of as made of three equilateral triangles:  .
In a semicircle of radius  ,the area of that trapezoid is
 .
There may be a very simple proof supporting that maximum area value,
but I was taught a lot of math past the fifth grade, and that spoiled me forever,
so I can only offer a complicated proof or two.
I just cannot think of a proof that does not involve calculus.
 .
Using  :
 between both roots of  <--->
Solving that quadratic equation we realize that the solutions are
 (which, making  does not make sense, and
 , which makes sense and makes  ,
marking the maximum for  .
So, that maximum is
Using  :
The trapezoid is made of three isosceles triangles.
They have legs of length  ,
and vertex angles measuring ,  , and .
The area of the trapezoid is the sum of the triangles' areas, so it is
 for 
and  between both roots,
so  happens for the greatest root of .
Making  for short, the equation is
 and the roots are
The roots are  ,
so  increases with  ,
between  <--> , and
 <--> ,
and the maximum is at <--> , where
 ,
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