\(The\ volume\ of\ a\ cone:V=\frac{1}{3}\pi r^2 H=\frac{1}{3}A_bH\\\\r-the\ radius\ \\H-the\ height\ of\ the\ cone\\A_b-area\ of\ the\ base\\----------------------\\\\V=35\ ft^3;\ H=7\ ft\\\\therefore\\\\\frac{1}{3}A_b\cdot7=35\\\\\frac{7}{3}A_b=35\ \ \ \ |divide\ both\ sides\ by\ \frac{7}{3}\to multiply\ both\ sides\ by\ \frac{3}{7}\\\\A_b=\frac{35}{1}\cdot\frac{3}{7}=\frac{5}{1}\cdot\frac{3}{1}=5\cdot3=\boxed{15\ (ft^2)}\)
