Volume of dam

$V = A_bL$

$V = [ \, \frac{1}{2}(4 + 12)(15) + 4(4) \, ](50)$

$V = 136(50)$

$V = 6800 ~ \text{ft.}^3$

$W = \gamma V$

$W = 150(6800)$

$W = 1\,020\,000 ~ \text{lb}$

$W = 1\,020\,000 ~ \text{lb} \times \dfrac{1 ~ \text{short ton}}{2000 ~ \text{lb}}$

$W = 510 ~ \text{short tons}$ *answer*