1. **Problem 1: Composite solid with cylinder and hemispheres** We have a right circular cylinder with radius $r=4$ in and height $h=12$ in, capped by two hemispheres each of radius $r=4$ in. **Find:** Surface area and volume of the solid. **Formulas:** - Surface area of cylinder (without top and bottom): $A_{cyl} = 2\pi r h$ - Surface area of sphere: $A_{sphere} = 4\pi r^2$ - Surface area of hemisphere: $A_{hemi} = 2\pi r^2$ - Volume of cylinder: $V_{cyl} = \pi r^2 h$ - Volume of sphere: $V_{sphere} = \frac{4}{3}\pi r^3$ - Volume of hemisphere: $V_{hemi} = \frac{2}{3}\pi r^3$ **Step 1:** Calculate surface area. - The cylinder's top and bottom are covered by hemispheres, so only lateral area of cylinder counts: $A_{cyl} = 2\pi (4)(12) = 96\pi$ - Two hemispheres make a full sphere, so total hemisphere surface area is $A_{sphere} = 4\pi (4)^2 = 64\pi$ - Total surface area $= A_{cyl} + A_{sphere} = 96\pi + 64\pi = 160\pi$ **Step 2:** Calculate volume. - Volume of cylinder: $V_{cyl} = \pi (4)^2 (12) = 192\pi$ - Volume of two hemispheres = volume of one sphere: $V_{sphere} = \frac{4}{3}\pi (4)^3 = \frac{4}{3}\pi 64 = \frac{256}{3}\pi$ - Total volume $= V_{cyl} + V_{sphere} = 192\pi + \frac{256}{3}\pi = \frac{576}{3}\pi + \frac{256}{3}\pi = \frac{832}{3}\pi$ **Final answers:** - Surface area $= 160\pi \approx 502.65$ square inches - Volume $= \frac{832}{3}\pi \approx 871.57$ cubic inches 2. **Problem 2: Sphere inscribed in cylinder** A sphere of radius $r=8$ in is inscribed in a right circular cylinder. **Find:** Lateral area of the cylinder. **Step 1:** The sphere fits exactly inside the cylinder, so the cylinder's radius $R$ equals the sphere's radius $8$ in. **Step 2:** The height $h$ of the cylinder equals the diameter of the sphere: $h = 2r = 16$ in. **Step 3:** Lateral surface area of cylinder is $A = 2\pi R h = 2\pi (8)(16) = 256\pi$ **Final answer:** Lateral area $= 256\pi \approx 804.25$ square inches 3. **Problem 3: Figure with trapezoidal and rectangular faces** The figure has four equilateral trapezoidal faces and two rectangular faces. Given: - Trapezoid bases: $b_1=4$ ft, $b_2=2$ ft - Trapezoid side (leg): $3$ ft - Height of trapezoid: $4$ ft - Length of rectangular base: $6$ ft **Step 1:** Calculate area of one trapezoidal face. - Area $= \frac{1}{2}(b_1 + b_2)h = \frac{1}{2}(4+2)(4) = \frac{1}{2}(6)(4) = 12$ sq ft **Step 2:** Total area of four trapezoidal faces = $4 \times 12 = 48$ sq ft **Step 3:** Calculate area of one rectangular face. - Dimensions: height $= 3$ ft (side of trapezoid), length $= 6$ ft - Area $= 3 \times 6 = 18$ sq ft **Step 4:** Total area of two rectangular faces = $2 \times 18 = 36$ sq ft **Step 5:** Total surface area = trapezoidal faces area + rectangular faces area = $48 + 36 = 84$ sq ft **Final answers:** 1. Surface area $= 160\pi$, Volume $= \frac{832}{3}\pi$ 2. Lateral area $= 256\pi$ 3. Total surface area $= 84$ sq ft