Three-dimensional Pythagoras and trigonometry

(a) 13 cm

1. Diagonal of the 3 by 4 base.
   $$\sqrt{3^2 + 4^2} = 5$$
2. Then with the height 12.
   $$\sqrt{5^2 + 12^2} = \sqrt{169} = 13$$

• M1 Base diagonal 5 (or $3^2 + 4^2$).
• M1 $\sqrt{3^2 + 4^2 + 12^2}$ or $\sqrt{5^2 + 12^2}$.
• A1 Correct answer: 13 cm.

(a) $13$ cm

1. $$3^{2}+4^{2}$$
2. $$5^{2}+12^{2}$$
3. Pythagoras gives the diagonal of the 3 cm by 4 cm face as $\sqrt{}$($3^{2}$ + $4^{2}$) = 5 cm.
4. The space diagonal is the hypotenuse of a triangle with perpendicular sides 5 cm and 12 cm.
5. Its length is $\sqrt{}$($5^{2}$ + $12^{2}$) = $\sqrt{169}$ = 13 cm.

• P1 Establishing $3^{2}+4^{2}$ or an equivalent valid method.
• P1 Establishing $5^{2}+12^{2}$ or an equivalent valid method.
• A1 Correct answer: $13$ cm

(a) $17.1^\circ$

1. The projection of $AF$ on the base is $AC$.
   $$AC = \sqrt{12^2 + 5^2} = 13$$
2. $$\tan\theta = \frac{4}{13} \;\Rightarrow\; \theta = 17.10^\circ$$

• P1 Finding $AC = 13$.
• P1 $\tan\theta = \frac{4}{13}$.
• A1 Correct answer: $17.1^\circ$.

(a) $42.0$°

1. $$\sqrt{6^{2}+8^{2}}$$
2. The required angle is between the rod and this projection. In the right triangle, tan θ = 9/10.
3. The projection of the rod onto the base is the base diagonal, $\sqrt{}$($6^{2}$ + $8^{2}$) = 10 cm.
4. The required angle is between the rod and this projection. In the right triangle, tan θ = 9/10.
5. θ = $\tan^{-1}$(9/10) = $41.987212...^\circ$, so the angle is $42.0^\circ$ to 1 decimal place.

• P1 Establishing $\sqrt{6^{2}+8^{2}}$ or an equivalent valid method.
• P1 The required angle is between the rod and this projection. In the right triangle, tan θ = 9/10.
• A1 Correct answer: $42.0$°

(a) $26$ cm

1. Find the base diagonal using the perpendicular base sides.
   $$\sqrt{6^{2}+8^{2}}$$
2. Use the base diagonal with the vertical height.
   $$\sqrt{10^{2}+24^{2}}$$
3. Therefore $26$ cm.

• P1 Find the base diagonal using the perpendicular base sides.
• P1 Use the base diagonal with the vertical height.
• A1 Correct answer: $26$ cm

(b) $67.4$°

1. In the vertical cross-section, opposite is height and adjacent is base diagonal.
2. Therefore $67.4$°.

• P1 In the vertical cross-section, opposite is height and adjacent is base diagonal.
• A1 Correct answer: $67.4$°

(a) $7.35$ cm

1. The centre-to-vertex distance is half the base diagonal.
   $$\sqrt{3^{2}+3^{2}}$$
2. Combine that horizontal distance with the vertical height.
   $$\sqrt{18+6^{2}}$$
3. Therefore $7.35$ cm.

• P1 The centre-to-vertex distance is half the base diagonal.
• P1 Combine that horizontal distance with the vertical height.
• A1 Correct answer: $7.35$ cm

(b) $80.5$ cm²

1. Use the face slant height from half a base side and vertical height.
   $$4\times 0.5\times 6\sqrt{3^{2}+6^{2}}$$
2. Therefore $80.5$ cm².

• P1 Use the face slant height from half a base side and vertical height.
• A1 Correct answer: $80.5$ cm²

(a) $61.9^\circ$

1. $AM$ is half the diagonal of the base.
   $$AC = \sqrt{6^2 + 6^2} = \sqrt{72}, \quad AM = \tfrac{1}{2}\sqrt{72} = \sqrt{18}$$
2. In right-angled triangle $VMA$:
   $$\cos\theta = \frac{\sqrt{18}}{9} = 0.4714\ldots$$
3. $$\theta = 61.87^\circ$$

• P1 $AM = \sqrt{18}$ (4.24...).
• P1 $\cos\theta = \frac{AM}{9}$ (or another correct ratio with the height).
• A1 Correct answer: $61.9^\circ$.

(b) $95.2\text{ cm}^3$

1. Height.
   $$VM = \sqrt{9^2 - 18} = \sqrt{63} = 7.937\ldots$$
2. $$\tfrac{1}{3} \times 36 \times \sqrt{63} = 12\sqrt{63} = 95.24\ldots$$

• P1 Height $\sqrt{63}$.
• P1 $\frac{1}{3} \times 36 \times$ their height.
• A1 $95.2\text{ cm}^3$.

(a) $512$ cm²

1. $$640\times 3/160$$
2. $$\sqrt{12^{2}+8^{2}}$$
3. $$\sqrt{12^{2}+5^{2}}$$
4. The base area is 160 cm². From 640 = (1/3) $\times$ 160 $\times$ h, the vertical height is h = 12 cm.
5. For a face with base 16 cm, the face height is $\sqrt{}$($12^{2}$ + $5^{2}$) = 13 cm. The two such faces total 2 $\times$ $\frac{1}{2}$ $\times$ 16 $\times$ 13 = 208 cm².
6. For a face with base 10 cm, the face height is $\sqrt{}$($12^{2}$ + $8^{2}$) = $4\sqrt{13}$ cm. The two such faces total $40\sqrt{13}$ cm².
7. Total surface area = 160 + 208 + $40\sqrt{13}$ = 512.222051... cm², giving 512 cm² to 3 significant figures.

• P1 Establishing $640\times 3/160$ or an equivalent valid method.
• P1 Establishing $\sqrt{12^{2}+8^{2}}$ or an equivalent valid method.
• P1 Establishing $\sqrt{12^{2}+5^{2}}$ or an equivalent valid method.
• A1 Correct answer: $512$ cm²