1. Identify the solids and find surface area and volume: 1. Cone: height $h=30$, radius $r=13$. - Surface area formula: $A=\pi r (r + l)$ where $l=\sqrt{r^2 + h^2}$ is the slant height. - Volume formula: $V=\frac{1}{3} \pi r^2 h$. Calculate $l$: $$l=\sqrt{13^2 + 30^2} = \sqrt{169 + 900} = \sqrt{1069} \approx 32.7$$ Surface area: $$A=\pi \times 13 \times (13 + 32.7) = \pi \times 13 \times 45.7 \approx 1865.36$$ units². Volume: $$V=\frac{1}{3} \pi \times 13^2 \times 30 = \frac{1}{3} \pi \times 169 \times 30 = 5310\pi \approx 16678.41$$ units³. 2. Rectangular prism: length $l=10$, width $w=8$, height $h=6$. - Surface area: $$A=2(lw + lh + wh)=2(10 \times 8 + 10 \times 6 + 8 \times 6)=2(80 + 60 + 48)=2 \times 188=376$$ km². - Volume: $$V = lwh = 10 \times 8 \times 6 = 480$$ km³. 3. Pentagonal pyramid: slant height $l=12.2$, apothem $a=5.2$, base edge $e=6$. - Area of base (pentagon): $$A_b = \frac{5}{2} \times e \times a = \frac{5}{2} \times 6 \times 5.2 = 78$$ in². - Lateral surface area: $$A_l = \frac{1}{2} \times \text{perimeter} \times l = \frac{1}{2} \times 5 \times 6 \times 12.2 = 183$$ in². - Total surface area: $$A = A_b + A_l = 78 + 183 = 261$$ in². - Height computed by Pythagoras: $$h=\sqrt{l^2 - a^2} = \sqrt{12.2^2 - 5.2^2} = \sqrt{148.84 - 27.04} = \sqrt{121.8} \approx 11.04$$ in. - Volume: $$V = \frac{1}{3} A_b h = \frac{1}{3} \times 78 \times 11.04 \approx 287.04$$ in³. 4. Cylinder: height $h=8$, diameter $d=14$, radius $r=7$. - Surface area: $$A=2\pi r^2 + 2\pi rh = 2\pi \times 7^2 + 2\pi \times 7 \times 8 = 98\pi + 112\pi = 210\pi \approx 659.73$$ km². - Volume: $$V = \pi r^2 h = \pi \times 7^2 \times 8 = 392\pi \approx 1231.5$$ km³. 5. Sphere: radius $r=8$ in. - Surface area: $$A=4\pi r^2 = 4\pi \times 8^2 = 256\pi \approx 804.25$$ in². - Volume: $$V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi 512 = \frac{2048}{3}\pi \approx 2144.66$$ in³. 6. Pentagonal prism: sides 12 yd, 8 yd, height 8.3 yd. - Area of base pentagon: treat as composed of triangles, or use apothem if known. Since info limited, assume base area $A_b=\frac{1}{2} \times \text{perimeter} \times \text{apothem}$. - Perimeter $P = 5 \times 12 = 60$ yd assuming all sides 12 yd. Surface area and volume would require more data, thus no accurate calculation here. B. Solve problems: 1. Cube edge $2$ cm, rectangular prism $6\times4\times4$ cm. - Cube volume: $2^3=8$ cm³. - Prism volume: $6 \times 4 \times 4 =96$ cm³. - Number of cubes: $$\frac{96}{8} =12$$ cubes. 2. Aquarium $2 m \times 3 m \times 4 m$. - Volume in m³: $2 \times 3 \times 4=24$ m³. - 1 m³ = 1000 liters. - Liters needed: $$24 \times 1000=24000$$ liters. 3. Balloon diameter $18$ cm, radius $r=9$ cm. - Volume of sphere: $$V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi 9^3=\frac{4}{3}\pi 729=972\pi \approx 3053.63$$ cm³. 4. Cone and cylinder have same height and diameter. - Volume of cone: $V_c=36.5$ cm³. - Volume of cylinder $V_{cy}$ relates as: $$V_c = \frac{1}{3} V_{cy} \Rightarrow V_{cy} = 3 \times 36.5=109.5$$ cm³. 5. Sphere radius $4$ cm, cylinder height $8$ cm (same as sphere diameter). - Sphere volume: $$V_s=\frac{4}{3}\pi 4^3=\frac{4}{3}\pi 64=\frac{256}{3}\pi \approx 268.08$$ cm³. - Cylinder radius equals sphere radius $r=4$ cm, - Cylinder volume: $$V_{cy}=\pi r^2 h=\pi 4^2 \times 8=128\pi \approx 402.12$$ cm³. - Volume inside cylinder but outside sphere: $$402.12 - 268.08=134.04$$ cm³. 6. Cylindrical can diameter $10$ cm, radius $r=5$ cm, height $20$ cm, sand filled up to $18$ cm. - Volume sand: $$V_s=\pi r^2 h_s=\pi 5^2 \times 18=450\pi \approx 1413.72$$ cm³. - Rectangular container $10 \times 30 \times 20=6000$ cm³. - Empty space volume: $$6000 - 1413.72=4586.28$$ cm³. Final answers summarized: 1. Cone SA $\approx 1865.36$, V $\approx 16678.41$. 2. Rectangular prism SA $=376$, V $=480$. 3. Pentagonal pyramid SA $=261$, V $\approx 287.04$. 4. Cylinder SA $\approx 659.73$, V $\approx 1231.5$. 5. Sphere SA $\approx 804.25$, V $\approx 2144.66$. B1. 12 cubes. B2. 24000 liters. B3. Balloon volume $\approx 3053.63$. B4. Cylinder volume $=109.5$. B5. Volume inside cylinder but outside sphere $\approx 134.04$. B6. Volume empty space $\approx 4586.28$.