Pyramids, cones, spheres and composite solids

(a) $905\text{ cm}^3$

1. Substitute $r = 6$.
   $$\tfrac{4}{3} \times \pi \times 6^3 = 288\pi = 904.77\ldots$$
2. Correct to 3 significant figures: $905\text{ cm}^3$.

• M1 $\frac{4}{3} \times \pi \times 6^3$.
• A1 $905\text{ cm}^3$ (awrt 905).

(a) $120\text{ cm}^3$

1. Base area.
   $$6 \times 6 = 36$$
2. $$\tfrac{1}{3} \times 36 \times 10 = 120$$

• M1 Base area 36 used with the height 10.
• A1 $120\text{ cm}^3$.

(a) $64$

1. Divide the large sphere volume by the volume of one small sphere.
   $$(4\pi \times 4^{3}/3)/(\frac{4\pi}{3})$$
2. Therefore $64$.

• P1 Divide the large sphere volume by the volume of one small sphere.
• A1 Correct answer: $64$

(b) $256\pi$

1. Multiply one small sphere’s area by the number of spheres.
   $$64\times 4\pi$$
2. Therefore $256\pi$.

• M1 Multiply one small sphere’s area by the number of spheres.
• A1 Correct answer: $256\pi$

(a) $9$ cm

1. Multiply volume by 3 and divide by base area.
   $$3\times 105/(5\times 7)$$
2. Therefore $9$ cm.

• M1 Multiply volume by 3 and divide by base area.
• A1 Correct answer: $9$ cm

(b) $3$ cm

1. A prism’s volume is base area times height.
   $$105/(5\times 7)$$
2. Therefore $3$ cm.

• M1 A prism’s volume is base area times height.
• A1 Correct answer: $3$ cm

(a) $314\text{ cm}^3$

1. $$\tfrac{1}{3} \times \pi \times 5^2 \times 12 = 100\pi = 314.15\ldots$$

• M1 $\frac{1}{3} \times \pi \times 5^2 \times 12$.
• A1 $314\text{ cm}^3$ (awrt 314).

(b) $283\text{ cm}^2$

1. Curved surface.
   $$\pi \times 5 \times 13 = 65\pi$$
2. Circular base.
   $$\pi \times 5^2 = 25\pi$$
3. Total.
   $$90\pi = 282.74\ldots$$

• P1 The curved surface area $\pi \times 5 \times 13$.
• P1 Adding the base, $\pi \times 5^2$.
• A1 $283\text{ cm}^2$ (awrt 283).

(a) $9$ cm

1. $$972\times 3/4$$
2. $$9^{3}$$
3. Substitute the volume: (4/3)$\pi r^{3}$ = $972\pi .$
4. Divide by $\pi$ and multiply by 3/4: $r^{3}$ = 729.
5. The positive cube root of 729 is 9, so the radius is 9 cm.

• P1 Establishing $972\times 3/4$ or an equivalent valid method.
• P1 Establishing $9^{3}$ or an equivalent valid method.
• A1 Correct answer: $9$ cm

(a) $\frac{416\pi}{3}$

1. $$8/(1-2/6)$$
2. $$\pi \times 6^{2}\times 12/3-\pi \times 2^{2}\times 4/3$$
3. The removed cone and original cone are similar with radius ratio 2:6 = 1:3, so their heights are also in ratio 1:3.
4. The frustum height is two equal height parts: each part is 8/2 = 4 cm. The removed and original cone heights are 4 cm and 12 cm.
5. Subtract cone volumes: V = (1/3)$\pi$ $\times$ $6^{2}$ $\times$ 12 $-$ (1/3)$\pi$ $\times$ $2^{2}$ $\times$ 4.
6. V = $144\pi$ $-$ $16\pi$/3 = $416\pi$/3 cm³.

• P1 Establishing $8/(1-2/6)$ or an equivalent valid method.
• P1 Establishing $\pi \times 6^{2}\times 12/3-\pi \times 2^{2}\times 4/3$ or an equivalent valid method.
• A1 Correct answer: $\frac{416\pi}{3}$

(a) $324\pi$

1. $$216/360\times 2\pi \times 15$$
2. $$18\pi /(2\pi )$$
3. $$\sqrt{15^{2}-9^{2}}$$
4. The sector arc becomes the base circumference: (216/360) $\times$ $2\pi$ $\times$ 15 = $18\pi$ cm.
5. Thus $2\pi r$ = $18\pi$ and the base radius is r = 9 cm. The sector radius becomes the slant height, 15 cm.
6. The vertical height follows from Pythagoras: h = $\sqrt{}$($15^{2}$ $-$ $9^{2}$) = 12 cm.
7. Volume = (1/3)$\pi$ $\times$ $9^{2}$ $\times$ 12 = $324\pi$ cm³.

• P1 Establishing $216/360\times 2\pi \times 15$ or an equivalent valid method.
• P1 Establishing $18\pi /(2\pi )$ or an equivalent valid method.
• P1 Establishing $\sqrt{15^{2}-9^{2}}$ or an equivalent valid method.
• A1 Correct answer: $324\pi$