First \(BC=\sqrt{20^2-13^2}=\sqrt{231}\) cm. (a) The two triangular ends together have area \(13\sqrt{231}\). Three rectangles have area \(24(13+20+\sqrt{231})\). Total \(1354.3513\ldots=1350\) cm² to 3 significant figures. (b) \(V=\tfrac12(13)\sqrt{231}(24)=2370\) cm³ to 3 significant figures. (c) B is the perpendicular projection of A onto the base. \(BF=\sqrt{231+24^2}=\sqrt{807}\). \(\tan\theta=13/\sqrt{807}\), so \(\theta=24.6^\circ\).