1. Problem: Find the area of the lower base of a frustum of a regular pyramid given volume $V=93.3333$ m³, upper base dimensions $2.5 \times 4$ m, and altitude $h=4$ m. Step 1: Calculate the area of the upper base $A_1$: $$A_1 = 2.5 \times 4 = 10 \text{ m}^2$$ Step 2: Use the volume formula for a frustum of a pyramid: $$V = \frac{h}{3}(A_1 + A_2 + \sqrt{A_1 A_2})$$ where $A_2$ is the area of the lower base. Step 3: Substitute known values: $$93.3333 = \frac{4}{3}(10 + A_2 + \sqrt{10 A_2})$$ Step 4: Multiply both sides by $\frac{3}{4}$: $$70 = 10 + A_2 + \sqrt{10 A_2}$$ Step 5: Rearrange: $$A_2 + \sqrt{10 A_2} = 60$$ Step 6: Let $x = \sqrt{A_2}$, then $A_2 = x^2$ and equation becomes: $$x^2 + \sqrt{10} x = 60$$ Step 7: Rewrite as quadratic in $x$: $$x^2 + 3.1623 x - 60 = 0$$ Step 8: Solve using quadratic formula: $$x = \frac{-3.1623 \pm \sqrt{3.1623^2 + 4 \times 60}}{2} = \frac{-3.1623 \pm 16.4317}{2}$$ Step 9: Positive root: $$x = \frac{13.2694}{2} = 6.6347$$ Step 10: Calculate $A_2$: $$A_2 = x^2 = 6.6347^2 = 44.0045 \text{ m}^2$$ \boxed{44.0045}\text{ m}^2 2. Problem: Find the lower base diameter of a frustum of a right circular cone with altitude $h=7$ m, volume $V=51.31$ m³, and upper base diameter $d_1=2$ m. Step 1: Calculate upper radius $r_1$: $$r_1 = \frac{d_1}{2} = 1 \text{ m}$$ Step 2: Let lower radius be $r_2$. Volume formula for frustum of cone: $$V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)$$ Step 3: Substitute known values: $$51.31 = \frac{1}{3} \pi \times 7 (1^2 + r_2^2 + 1 \times r_2)$$ Step 4: Simplify: $$51.31 = \frac{7 \pi}{3} (1 + r_2^2 + r_2)$$ Step 5: Multiply both sides by $\frac{3}{7 \pi}$: $$\frac{51.31 \times 3}{7 \pi} = 1 + r_2^2 + r_2$$ Step 6: Calculate left side: $$\approx \frac{153.93}{21.991} = 7.0$$ Step 7: Rearrange: $$r_2^2 + r_2 + 1 = 7$$ $$r_2^2 + r_2 - 6 = 0$$ Step 8: Solve quadratic: $$r_2 = \frac{-1 \pm \sqrt{1 + 24}}{2} = \frac{-1 \pm 5}{2}$$ Step 9: Positive root: $$r_2 = 2$$ Step 10: Lower diameter: $$d_2 = 2 r_2 = 4 \text{ m}$$ \boxed{4.0000}\text{ m} 3. Problem: Compute volume of a cylindrical garbage can with height $h=5$ feet and radius $r=14$ inches. Step 1: Convert radius to feet: $$r = \frac{14}{12} = 1.1667 \text{ ft}$$ Step 2: Volume formula for cylinder: $$V = \pi r^2 h$$ Step 3: Calculate volume: $$V = \pi (1.1667)^2 \times 5 = \pi \times 1.3611 \times 5 = 21.3628 \text{ ft}^3$$ \boxed{21.3628}\text{ ft}^3 4. Problem: Find height $h$ of a cylindrical tank with open top, surface area $S=78$ m², and diameter $d=\frac{2}{3} h$. Step 1: Express radius $r$: $$r = \frac{d}{2} = \frac{1}{3} h$$ Step 2: Surface area of open top cylinder: $$S = \pi r^2 + 2 \pi r h$$ Step 3: Substitute $r$: $$78 = \pi \left(\frac{h}{3}\right)^2 + 2 \pi \left(\frac{h}{3}\right) h = \pi \frac{h^2}{9} + \frac{2 \pi h^2}{3}$$ Step 4: Simplify: $$78 = \pi h^2 \left(\frac{1}{9} + \frac{2}{3}\right) = \pi h^2 \left(\frac{1}{9} + \frac{6}{9}\right) = \pi h^2 \frac{7}{9}$$ Step 5: Solve for $h^2$: $$h^2 = \frac{78 \times 9}{7 \pi} = \frac{702}{7 \pi} = \frac{100.2857}{\pi} = 31.9363$$ Step 6: Calculate $h$: $$h = \sqrt{31.9363} = 5.6503 \text{ m}$$ \boxed{5.6503}\text{ m} 5. Problem: Find distance between bases (height) of a right prism with volume $V=600$ cm³ and hexagonal bases with side length $s=8$ cm. Step 1: Area of regular hexagon: $$A = \frac{3 \sqrt{3}}{2} s^2 = \frac{3 \sqrt{3}}{2} \times 64 = 166.2760 \text{ cm}^2$$ Step 2: Volume formula: $$V = A \times h$$ Step 3: Solve for $h$: $$h = \frac{V}{A} = \frac{600}{166.2760} = 3.6081 \text{ cm}$$ \boxed{3.6081}\text{ cm} 6. Problem: Find ratio of volume to lateral area of a right prism with equilateral triangular bases, base edge $20$ cm, and height $60$ cm. Step 1: Area of equilateral triangle: $$A = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \times 400 = 173.2051 \text{ cm}^2$$ Step 2: Volume: $$V = A \times h = 173.2051 \times 60 = 10392.3051 \text{ cm}^3$$ Step 3: Lateral area (perimeter times height): $$P = 3 \times 20 = 60 \text{ cm}$$ $$Lateral\ Area = P \times h = 60 \times 60 = 3600 \text{ cm}^2$$ Step 4: Ratio volume to lateral area: $$\frac{V}{Lateral\ Area} = \frac{10392.3051}{3600} = 2.8868$$ \boxed{2.8868}