Prisms, cylinders and surface area

(a) $108\text{ cm}^3$

1. Volume of a prism $=$ area of cross-section $\times$ length $= 12 \times 9 = 108$.

• M1 Writing $12 \times 9$.
• A1 $108\text{ cm}^3$.

(b) $60\text{ cm}^3$

1. $5 \times 4 \times 3 = 60$.

• B1 $60\text{ cm}^3$.

(a) $132\text{ cm}^2$

1. Two triangular ends.
   $$2 \times \tfrac{1}{2} \times 3 \times 4 = 12$$
2. Three rectangles, one for each side of the triangle.
   $$(3 + 4 + 5) \times 10 = 120$$
3. Add.
   $$12 + 120 = 132$$

• P1 Area of both triangular ends, 12.
• P1 Areas of the three rectangles, 30, 40 and 50.
• A1 $132\text{ cm}^2$.

(a) $63\pi$

1. $$\pi \times 3^{2}\times 14/2$$
2. A cylinder has volume $\pi r^{2}h$, so the full volume is $\pi$ $\times$ $3^{2}$ $\times$ 14 = $126\pi$ cm³.
3. Half of this volume is $63\pi$ cm³.

• P1 Establishing $\pi \times 3^{2}\times 14/2$ or an equivalent valid method.
• A1 Correct answer: $63\pi$

(a) $120$ cm³

1. Find the triangular cross-section area.
   $$6\times 8/2$$
2. Multiply by the length of the prism.
   $$24\times 5$$
3. Therefore $120$ cm³.

• P1 Find the triangular cross-section area.
• P1 Multiply by the length of the prism.
• A1 Correct answer: $120$ cm³

(b) $168$ cm²

1. Add the two triangular ends and rectangles along all three triangle sides.
   $$2\times 24+5\times (6+8+10)$$
2. Therefore $168$ cm².

• P1 Add the two triangular ends and rectangles along all three triangle sides.
• A1 Correct answer: $168$ cm²

(a) $72\pi$

1. Multiply the circular base area by the height.
   $$\pi \times 3^{2}\times 8$$
2. Therefore $72\pi$.

• M1 Multiply the circular base area by the height.
• A1 Correct answer: $72\pi$

(b) $66\pi$

1. Add two circular ends and the curved area.
   $$2\pi \times 3^{2}+2\pi \times 3\times 8$$
2. Therefore $66\pi$.

• M1 Add two circular ends and the curved area.
• A1 Correct answer: $66\pi$

(a) $6$ cm

1. Divide volume by base area.
   $$324/(6\times 9)$$
2. Therefore $6$ cm.

• M1 Divide volume by base area.
• A1 Correct answer: $6$ cm

(b) $234$ cm²

1. Include the base once and the two pairs of side faces.
   $$6\times 9+2\times 6\times (6+9)$$
2. Therefore $234$ cm².

• P1 Include the base once and the two pairs of side faces.
• A1 Correct answer: $234$ cm²

(a) 22

1. Volume of the tank.
   $$\pi \times 30^2 \times 80 = 72\,000\pi = 226\,194.6\ldots\text{ cm}^3$$
2. Volume of one container.
   $$20 \times 20 \times 25 = 10\,000\text{ cm}^3$$
3. Divide.
   $$226\,194.6 \div 10\,000 = 22.6\ldots$$
4. Only 22 containers can be filled completely.

• P1 Volume of the cylinder, $\pi \times 30^2 \times 80$.
• P1 Volume of a container, 10 000.
• P1 Dividing the tank's volume by a container's volume.
• A1 Correct answer: 22 containers.

(a) $557\text{ cm}^2$

1. Find the height from the volume.
   $$\pi \times 5^2 \times h = 1000 \;\Rightarrow\; h = \frac{1000}{25\pi} = 12.732\ldots$$
2. Curved surface.
   $$2\pi \times 5 \times 12.732\ldots = 400$$
3. Two ends.
   $$2 \times \pi \times 5^2 = 50\pi = 157.08\ldots$$
4. Total.
   $$400 + 157.08\ldots = 557.08\ldots$$

• P1 Rearranging $\pi r^2 h = 1000$ to find $h$.
• P1 The curved surface area $2\pi r h$ with their $h$.
• P1 Adding both circular ends, $2\pi r^2$.
• A1 $557\text{ cm}^2$ (awrt 557).