Question 971976
In the American definition, a trapezoid has two parallel sides {{{drawing(300,200,-1.2,1.2,-0.3,1.3,
red(arc(0,0,2,2,180,360)),
line(-1,0,-0.5,0.866),line(1,0,0.5,0.866),
line(-1,0,1,0),line(-0.5,0.866,0.5,0.866),
arrow(-0.5,0.866,0,0.866),arrow(-0.5,0,0,0)
)}}} .
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 {{{60^o}}},
and could be thought of as made of three equilateral triangles: {{{drawing(300,200,-1.2,1.2,-0.3,1.3,
red(arc(0,0,2,2,180,360)),
triangle(-1,0,-0.5,0.866,0,0),triangle(1,0,0.5,0.866,0,0),
triangle(-0.5,0.866,0.5,0.866,0,0)
)}}} .
In a semicircle of radius {{{r=d/2}}} ,the area of that trapezoid is
{{{area=3*(1/2)r*r*sin(60^o)=3*(1/2)*r^2*(sqrt(3)/2)=3*r^2sqrt(3)/4=3*(r/2)^2*sqrt(3)=3sqrt(3)(d/4)^2=3sqrt(3)d^2/16}}} .
 
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.
 
{{{drawing(300,200,-1.2,1.2,-0.4,1.2,
red(arc(0,0,2,2,180,360)),locate(0.45,0.45,r),
triangle(-1,0,-0.707,0.707,0,0),green(triangle(1,0,0.707,0.707,0.707,0)),
green(rectangle(0.707,0,0.66,0.05)),triangle(1,0,0.707,0.707,0,0),
triangle(-0.707,0.707,0.707,0.707,0,0),locate(0.15,0.15,red(alpha)),
locate(-0.25,0.15,red(alpha)),red(arc(0,0,0.6,0.6,180,225)),
red(arc(0,0,0.6,0.6,-45,0)),red(arc(0,0,0.5,0.5,-135,-45)),
locate(-0.22,0.4,red(180^o-2alpha)),locate(0.72,0.37,green(h)),
green(arrow(-0.3,-0.1,-1,-0.1)),green(arrow(-0.7,-0.1,0,-0.1)),
locate(-0.7,-0.1,green(r=d/2)),locate(0.3,-0.1,green(x)),
green(arrow(0,-0.1,0.707,-0.1)),green(arrow(0.707,-0.1,0,-0.1))
)}}} .
 
Using {{{green(x)}}} :
{{{h=sqrt(r^2-x^2)}}}
{{{area=A=(2r+2x)sqrt(r^2-x^2)/2=(r+x)sqrt(r^2-x^2)}}}
{{{dA/dx=1*sqrt(r^2-x^2)+(r+x)*(1/2sqrt(r^2-x^2))*(-2x)}}}
{{{dA/dx=sqrt(r^2-x^2)+(r+x)*x/sqrt(r^2-x^2))}}}
{{{dA/dx=(r^2-x^2)/sqrt(r^2-x^2)-(rx+x^2)/sqrt(r^2-x^2))}}}
{{{dA/dx=(r^2-x^2-rx-x^2)/sqrt(r^2-x^2))}}}
{{{dA/dx=(-2x^2-rx+r^2)/sqrt(r^2-x^2))}}}
{{{dA/dx>0}}} between both roots of {{{-2x^2-rx+r^2=0}}}<--->{{{2x^2+rx-r^2=0}}}
Solving that quadratic equation we realize that the solutions are
{{{x=-r}}} (which, making {{{x<0}}} does not make sense, and
{{{x=r/2=d/4}}} , which makes sense and makes {{{dA/dx=0}}} ,
marking the maximum for {{{A=(r+x)sqrt(r^2-x^2)}}} .
So, that maximum is
{{{area[max]=(r+r/2)sqrt(r^2-(r/2)^2)}}}
{{{area[max]=(3r/2)sqrt(r^2-r^2/4)}}}
{{{area[max]=(3r/2)sqrt(3r^2/4)}}}
{{{area[max]=(3r/2)sqrt(3)(r/2))}}}
{{{area[max]=3sqrt(3)(r/2)^2)}}}
{{{area[max]=3sqrt(3)(r/4)^2)}}}
{{{area[max]=highlight(3sqrt(3)d^2/16)}}}
 
Using {{{red(alpha)}}} :
The trapezoid is made of three isosceles triangles.
They have legs of length {{{r}}} ,
and vertex angles measuring {{{red(alpha)}}} , {{{red(180^o-2alpha)}}} , and {{{red(alpha)}}} .
The area of the trapezoid is the sum of the triangles' areas, so it is
{{{area=A=r^2sin(alpha)/2+r^2sin(180^o-2alpha)/2+r^2sin(alpha)/2}}}
{{{A=r^2sin(alpha)+(r^2/2)sin(180^o-2alpha)}}}
{{{dA/d(alpha)=r^2(-cos(alpha))+(r^2/2)(-cos(180^o-2alpha))(-1)}}}
{{{dA/d(alpha)=-r^2cos(alpha)+(r^2/2)cos(180^o-2alpha)}}}
{{{dA/d(alpha)=-r^2cos(alpha)+(r^2/2)cos(180^o)cos(2alpha)+sin(180^o)sin(2alpha)}}}
{{{dA/d(alpha)=-r^2cos(alpha)+(r^2/2)(-1)cos(2alpha)+0*sin(2alpha)}}}
{{{dA/d(alpha)=-r^2cos(alpha)-(r^2/2)cos(2alpha)}}}
{{{dA/d(alpha)=-r^2cos(alpha)-(r^2/2)(2cos^2(alpha)-1)}}}
{{{dA/d(alpha)=-(r^2/2)(2cos(alpha)+2cos^2(alpha)-1)}}}
{{{dA/d(alpha)=0}}} for {{{2cos(alpha)+2cos^2(alpha)-1=0}}}
and {{{dA/d(alpha)>=0}}} between both roots,
so {{{A[max]}}} happens for the greatest root of {{{2cos(alpha)+2cos^2(alpha)-1=0}}} .
Making {{{cos(alpha)=u}}} for short, the equation is
{{{2u^2+2u-1=0}}} and the roots are
{{{u = (-1 +- sqrt(1^2-4*2*(-1)))/(2*2) }}}
{{{u =(-1 +- sqrt(1+8))/4=(-1 +- sqrt(9))/4=(-1 +- 3)/4}}}
The roots are {{{u[1]=(-1-3)/4=(-4)/4=-1<(-1+3)/4=2/4=1/2}}} ,
so {{{area}}} increases with {{{cos(alpha)}}} ,
between {{{cos(alpha)=0}}}<-->{{{alpha=90^o}}} , and
{{{cos(alpha)=1/2}}}<-->{{{alpha=60^o}}} ,
and the maximum is at {{{cos(alpha)=1/2}}}<-->{{{alpha=60^o}}} , where
{{{area[max]=r^2sin(alpha)+r^2sin(180^o-2alpha)/2}}}
{{{area[max]=r^2sin(60^o)+r^2sin(180^o-2*60^o)/2}}}
{{{area[max]=r^2sin(60^o)+r^2sin(60^o)/2}}}
{{{area[max]=(3r^2/2)sin(60^o)}}}
{{{area[max]=(3r^2/2)(sqrt(3)/2)}}}
{{{area[max]=(3/4)r^2sqrt(3)}} , and since {{{r=d/2}}} ,
{{{area[max]=(3/4)(d/2)^2sqrt(3)=(3/4)(d^2/4)sqrt(3)=highlight(3sqrt(3)d^2/16)}}}