The triangle need not be acute for the result given. The only triangles for which it fails seem to be isosceles triangles which have a measure of one or more of its sides as 2. Triangles with a side that is 2 must be scalene in order for the given remainder to be true.

clearvars

ct=0;

prms= vpa(primes(300));

prms(1)=[];

for asub=1:length(prms)

  a=prms(asub)

  for bsub=asub:length(prms)

    b=prms(bsub);

    for csub=bsub:length(prms)

      c=prms(csub);

      if c>=a+b

        break

      end

      s=(a+b+c)/2;

      A=sqrt(s*(s-a)*(s-b)*(s-c));

      r=A^2-floor(A^2);

      if abs(r-3/16)>.00000001

        fprintf('%6d %6d %6d %15.13f\n',[a b c r])

        ct=ct+1;

        if ct>30

          return

        end

      end

    end

  end

end

verifies the 3/16 remainder value for all odd-prime-side-length triangles, even obtuse ones. The triangles need not be acute for this to hold.

Initially the break was made as

      if c^2>=a^2+b^2

        break

      end

to assure all triangles were acute. The removal of the squaring changed it to verify even for obtuse triangles.

In fact, even if we allow the even prime, 2, as a side length, the 3/16 remainder remains so long as the triangle is not isosceles (including not equilateral).