Detailed Answer

a) First, JMB is a right triangle. We can use the Pythagorean theorem to find \( BJ \).

\[BJ = \sqrt{1^2 + 1^2} = 1.4\text{cm}\]

We can also use Pythagoras on triangle \( BIJ \):

\[BI = \sqrt{1.5^2 + \sqrt{2}} = 2.0\text{cm}\]

b) We have eight identical triangular faces with height \(IM = \sqrt{BI^2 - BM^2}\)

The surface of one triangle is \(\frac{1}{2} \times BA \times IM = \frac{1}{2} \times 2 \times \sqrt{2.0^2 - 1^2} = 1\text{cm}^2\)

So, the total surface of the dice is \(8\text{cm}^2\).

c) The volume of the dice is the volume of two identical pyramids, such as:

\[V_{\text{pyramid}} = \frac{1}{3} \times 2 \times 2 \times 1.5 = 2\text{cm}^3\]

\[V_{\text{dice}} = 2 \times 2 = 4\text{cm}^3\]

So, the resin needed will be \(1.2 \times 4 = 4.8\text{g}\).

Detailed Answer

a)

\[\frac{AI}{DI} = \tan(63^\circ)\]

Distance \( DI \) is 1cm, so \(AI = 1 \times \tan(63^\circ) = 2.0\text{cm}\)

b) Using Pythagoras:

\[AD = \sqrt{1^2 + 2^2} = \sqrt{5} = 2.2\text{cm}\]

Note: We could have also used \(\cos(63^\circ)\) or \(\sin(63^\circ)\)

c) The area of our trapezium is given by:

\[A = \frac{1}{2} \times (DC + AB) \times AI = \frac{1}{2} \times (4+2) \times 2 = 6\text{cm}^2\]

d) By applying the formula from the question:

\[V = 6 \times 6 = 36\text{cm}^3\]

e) There are 6 faces to cover in foil: 2 trapeziums, 1 rectangle on top, one on the bottom, and two rectangles on the side.

Area of the 2 trapeziums: \(2 \times 6 = 12\text{cm}^2\)

Top rectangle: \(2 \times 6 = 12\text{cm}^2\)

Bottom rectangle: \(4 \times 6 = 24\text{cm}^2\)

One side rectangle: \(AD \times 6 = 2.2 \times 6 = 13\text{cm}^2\). It has to be multiplied by 2 later on, since we have two of them.

Total surface area of the chocolate: \(12 + 12 + 24 + 13 \times 2 = 74\text{cm}^2\)

Detailed Answer

a) \(tan(α) = \frac{40}{10} = 76°\)

b) \(cos(α) = \frac{10}{BC} \implies BC = \frac{10}{cos(76°)} = 41cm\)

Note: we could have also used Pythagoras.

c) Area of our cone:

\[\pi * AC^2 + \pi * AC * BC\]

However, it is a hat, so the base has no fabric.

So, the fabric is only: \(\pi * AC * BC = \pi * 10 * 41 = 1300cm^2\)

d) Volume of our cone: \(\frac{1}{3} * \pi * AC^2 * AB = \frac{1}{3} * \pi * 10^2 * 40 = 4200cm^3\)

e)

\[\frac{4200}{1000} = 4.2\]

\[4.2 * 2 = 8.4\]

The wizard can pull eight magic rabbits out of the hat.

Detailed Answer

a) \( KA \) is half of the diagonal of the square. The diagonal \( AC \) has for distance \( \sqrt{4^2 + 4^2} = \sqrt{32} \). So, \( KA = \frac{\sqrt{32}}{2} = 2.82 \approx 3 \, \text{cm} \).

b) \( AKI \) is a right triangle. \[ \tan(α) = \frac{KI}{KA} \implies KI = KA * tan(α) = 2.82... * \tan 55^\circ = 4.0...cm \approx 4cm \].

c) We will repeat the two previous steps for the top part of the partial pyramid. Length of half the diagonal: \( \frac{\sqrt{2^2 + 2^2}}{2} = \frac{\sqrt{8}}{2} \approx 1.41 \, \text{cm} \).

\( LI \) is \( \sqrt{2} \tan 56^\circ = 2.0... \approx 2 \, \text{cm} \).

d) Volume of the upside-down pyramid:

\[ \frac{1}{3} * Surface_{base} * \text{height} = \frac{1}{3} * 4^2 * KJ = \frac{1}{3} * 4^2 * 3 = 16 \, \text{cm}^3 \].

The top part is one pytamid subtracted from a larger pyramid: \( \frac{1}{3} * 4^2 * KI\)

Large pyramid volume: \( \frac{1}{3} * 4^2 * 4.0... = 21.5...cm^3\)

Small pyramid volume: \( \frac{1}{3} * 2^2 * * LI = \frac{1}{3} * 2^2 * * 2.0... = 2.6...cm^3\)

Volume of the top part: \( 21.5... - 2.6... = 18.8...cm^3\)

Volume of the diamond: \( 16 + 18.8... = 34.8... = 35cm^3\)

Detailed Answer

a) (i) & (ii) Volume of the sphere: \(\frac{4}{3} * \pi * AO^3 = 113.0... \approx 110cm^3\)

Volume of the cone: \(\frac{1}{3} * \pi * AM^2 * MP = \frac{1}{3} * \pi * 3^2 = 51.8... \approx 52cm^3\)

b) (i) & (ii) Surface of the sphere: \(4 * \pi * AO^2 = 113.0... \approx 110cm^3\)

Surface of the cone: \(\pi * AM * MP\)

\[AP = \sqrt{3^2 + 5.5^2} = 6.26...\]

Surface of the cone: \(\pi * 3 * 6.26... = 59.0... \approx 60cm^2\)

c) Price of option A: \(110 * 5 \ (cents) = $5.5\)

Cost of paint A: \(110 * 1 \ (cent) = $1.1\)

So, \(\frac{5.5}{1.1} = 5\) drinks would pay for the paint.

Price of option B: \(52 * 5 \ (cents) = $2.6\)

Cost of paint A: \(60 * 1 \ (cent) = $0.6\)

Cost of paint A: \(110 * 1 \ (cent) = $1.1\)

So, \(\frac{2.6}{0.6} = 4.3... \approx 4\) drinks would pay for the paint.

Detailed Answer

a) (i) & (ii) \[V_{Normal} = \pi * 3.4^2 * 15.7 = 570.1... \approx 570cm^3\]

\[V_{Mini} = \pi * 3.31^2 * 9.6 = 330.4... \approx 330cm^3\]

b) Volume in a set of 6 Normal: \(6 * 570 = 3420cm^3\)

Mini cans needed \(\frac{3420}{330.4...} = 10.3...\)

We need a set of 11 mini cans to have at least as much volume. However, the number must be even. Our choices are:

\[10 * 330.4... \approx 3300cm^3\]

And:

\[12 * 330.4... \approx 3900cm^3\]

We get the closest volume for a set of 10 Mini cans.

c) Surface of a Normal can: \(2 * \pi * 3.4 * 15.7 = 335.3...cm^2\)

Surface in a set of Normal cans: \(335.3... * 6 = 2012.3... \approx 2000cm^2\)

Surface of a Mini can: \(2 * \pi * 3.31 * 9.6 = 199.6......cm^2\)

Surface in a set of Mini cans: \(199.6... * 10 = 1996.5... \approx 2000cm^2\)

The two sets use about the same amount of aluminium.

Detailed Answer

a) (i) \(sin(63°) = \frac{AI}{AC} \implies AC = \frac{AI}{sin(63°)} = 2.2cm\)

a) (ii) \(Volume = Area \ of \ the \ triangle * AD =\frac{1}{2} * BC * AI * AD = \frac{1}{2}*2*2*3 = 6m^3\)

b) (i) We consider the right triangle with hypothenuse BC:

\[BC = \sqrt{MB^2 + MC^2}\]

\[MC = \frac{(2-1)}{2} = 0.5\]

\[BC = \sqrt{0.5^2 + 1^2} = 1.1m\]

b) (ii) Trapezium area: \(\frac{1}{2} * (DC + AB) * IJ = \frac{1}{2}(2+1)*2 = 1.5m^2\)

Volume: \(1.5 * 3 = 4.5m^3\)

b) (iii) We add the volume of the lower part, which is: \(2 * 3 = 6m^3\)

Total volume: \(6 + 4.5 = 10.5m^3\)

c) The volume of tent 1 can be expressed as: \(\frac{1}{2} * BC * AI * CF\)

We solve the equation: \(\frac{1}{2} * BC * AI * CF = 10.5m^3\) for BC.

\[BC = \frac{2*10.5}{AI*CF} = \frac{2*10.5}{2*3} = 3.5m\]

d) Area of the tent:

Two identical triangles of surface: \(\frac{1}{2} * 3.5 * 2 = 3.5m^2\)

A rectangular base: \(BC * CF = 3.5 * 3 = 10.5m^2\)

Two side triangles: \(AC * AD = 2.65... = 7.97... = 8m^2\)

(We recalculate AC = \( \sqrt{IC^2 + AI^2} = \sqrt{(\frac{3.5}{2})^2 + 2^2} = 2.65...m\))

Total area: \(2 *3.5+10.5+2*7.97... = 33.4... = 33m^2\)

Detailed Answer

a) There are many ways to solve it. We will use that fact that this is an isoceles triangle. The two other angles have values of \(\frac{180-119}{2} = 30.5°\)

Then, we can use the sine rule:

\[\frac{120}{sin(30.5°)} = \frac{AC}{sin(119°)}\]

\[AC = \frac{sin(119°)}{sin(35°)} * 120 = 210cm\]

b) Surface of the side triangles: \(\frac{1}{2} * 120 * 120 * sin(119°) = 6300cm^2\)

Surface of the rectangles: \(50 * 120 = 6000cm^2\)

Total surface: \(12300cm^2\)

c) We will consider the triangle ADB. The angle \(\hat{B}\) is \(\frac{α}{2}\)

\[AD = \frac{230}{2} = 115cm\]

\[sin(\hat{B}) = \frac{AD}{AB} \implies \hat{B} = sin^{-1}(\frac{AD}{AB}) \implies α = 2sin^{-1}(\frac{115}{120}) = 150°\]

d) This is still the same chair, so the surface is also the same.

Detailed Answer

a) Volume of the main part: \(4 * 4 * 6 = 96dm^3\)

Volume of the roof: \(Area \ of \ the \ triangle * length\)

Area of the triangle:\(\frac{1}{2} * BC * IJ\)

\[IJ = \sqrt{IC^2 - JC^2} = \sqrt{3^2 - 2^2} = \sqrt{5}\]

So, the area of the triangle is: \(\frac{1}{2} * 4 * \sqrt{5} = 4.4...dm^3\)

The volume of the roof: \(4.4... * 6\)

Total doghouse volume: \(96 + 4.4... * 6 = 120dm^3\)

b) (i) \[FG * FE + \frac{1}{2} * \pi * (\frac{FG}{2})^2 = 2 * 2 + \frac{1}{2} * \pi * 1^2 = 5.5... = 5.6dm^2\]

b) (ii) Surface of the parallelepiped without its top (as the inside of the doghouse is hollow):

\[(4*4) * 2 + (4*3)*2 + (4*3) = 68dm^2\]

Surface with door removed: \(68 - 5.6... = 62.4... = 62dm^2\)

b) (iii) Surface of the roof: \(2 * \sqrt{5} + 2 * (6*3) = 40.4... = 40dm^2\)