Solution
= Solution
The two radii enclose the <angle> $2\pi/3$, so the triangle has <area> $T=\tfrac12r^2\sin(2\pi/3)=\sqrt3r^2/4$. From the sector calculation, $r^2=3B/\pi$. Consequently
$$
\boxed{T=\frac{3\sqrt3}{4\pi}\,B.}
$$
= Solution
The two radii enclose the <angle> $2\pi/3$, so the triangle has <area> $T=\tfrac12r^2\sin(2\pi/3)=\sqrt3r^2/4$. From the sector calculation, $r^2=3B/\pi$. Consequently
$$
\boxed{T=\frac{3\sqrt3}{4\pi}\,B.}
$$