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Study 3.1 Geometry and trigonometry - SL content conceptsOpen the corresponding Concept lesson for explanations, examples and exam connections.

[Maximum number: 7]

Kailash manufactures drink containers in the shape of a cuboid. The container has a square top and a square base of length, $l\,\mathrm{cm}$ . Its height, $d\,\mathrm{cm}$ , is three times the length of the base.

diagram not to scale

(a)

Calculate the total external surface area of the container.

[ 3 ]

Finding the area of a square and the area of a non-square rectangle: $5\times5$ and $5\times15$ .
$(5\times5+5\times15+5\times15)\times2$ $=350\,\mathrm{cm^2}$
Note: Units are required.

(b)

To reduce environmental impact, Kailash is trying to minimize the amount of material needed for the production of the $375\,\mathrm{cm^3}$ container.
He is willing to change the shape to a cylinder with radius $r\,\mathrm{cm}$ , and height $h\,\mathrm{cm}$ , as shown below.

Figure for Question (b) — IB Maths AI HL

The cylindrical container of drink must also hold $375\,\mathrm{cm^3}$ .
Find an expression for the height, h, of the container in terms of r.

[ 2 ]

$\pi r^2h=375$ $h=\frac{375}{\pi r^2}$
Note: Award at most A0A1 for a final answer $h=\frac{V}{\pi r^2}$ .

(c)

Let the total external surface area be $A\,\mathrm{cm^2}$ .
Show that $A=2\pi r^2+\frac{750}{r}$ .

[ 2 ]

Attempt to substitute their h into the curved surface area $2\pi rh$ .
$A=2\pi r^2+2\pi r\left(\frac{375}{\pi r^2}\right)$ $A=2\pi r^2+\frac{750}{r}$

[Maximum number: 6]

This question compares possible designs for a new computer network between multiple school buildings, and whether they meet specific requirements.
A school's administration team decides to install new fibre-optic internet cables underground. The school has eight buildings that need to be connected by these cables. A map of the school is shown below, with the internet access point of each building labelled A-H.

Jonas is planning where to install the underground cables. He begins by determining the distances, in metres, between the underground access points in each of the buildings.

He finds $\mathrm{AD}=89.2 \mathrm{~m}, \mathrm{DF}=104.9 \mathrm{~m}$ and $\mathrm{A} \hat{\mathrm{DF}}=83^{\circ}$ .

(a)

Find AF .

[ 3 ]

$\quad \mathrm{AF}^{2}=89.2^{2}+104.9^{2}-2(89.2)(104.9) \cos 83$
(M1)(A1)
Note: Award (M1) for substitution into the cosine rule and (A1) for correct substitution.

\mathrm{AF}=129 \mathrm{~m} \quad(129.150 \ldots)

(b)

The cost for installing the cable directly between A and F is $\$ 21310$ .

Find the cost per metre of installing this cable.

[ 2 ]

$21310 \div 129.150 \ldots$
$ 165

(c)

Jonas estimates that it will cost $\$ 110$ per metre to install the cables between all the other buildings.

State why the cost for installing the cable between A and F would be higher than between the other buildings.

[ 1 ]

[Maximum number: 9]

The following question compares the distance and direction between cities on a flat surface to the distance and direction between cities on a sphere.
Consider a model where the cities of Bogotá, Moscow, and Nairobi lie on a flat surface. In this model, Nairobi is 6000 km due south of Moscow and Bogotá is 12500 km due west of Nairobi, as shown in the following diagram.

(a)

Find the distance from Bogotá to Moscow.

[ 2 ]

attempt to use Pythagoras' theorem

\begin{aligned} & \sqrt{12500^{2}+6000^{2}} \\ & =13900(\mathrm{~km}) \end{aligned}

Note: Answers to more than 3 sf are not appropriate and should be penalised. A1 will be given only for 13900 km ( 3 sf ) or 14000 km (nearest 500 km ).

(b)

Find the bearing of Moscow from Bogotá. Give your answer in degrees.

In reality, these three cities lie on the curved surface of the Earth which will change the distances and directions found in part (a).

Now consider a curved model using a coordinate system (x, y, z) with its origin, O , at the centre of the Earth. The units of this system are thousands of kilometres and the Earth is modelled as a sphere with radius 6000 km . The North Pole, P, lies on the z-axis, and Nairobi, N, is modelled as being on the equator and lying on the y-axis.

P has position vector $\overrightarrow{\mathrm{OP}}=\boldsymbol{p}=\left(\begin{array}{l}0 \\ 0 \\ 6\end{array}\right)$ and N has position vector $\overrightarrow{\mathrm{ON}}=\boldsymbol{n}=\left(\begin{array}{l}0 \\ 6 \\ 0\end{array}\right)$ .

[ 3 ]

attempt to use an appropriate inverse trig ratio
e.g. $\arctan \left(\frac{12500}{6000}\right)$ or $\arctan \left(\frac{6000}{12500}\right)$
identifying the correct angle
$=(0) 64.4^{\circ}$ OR $064^{\circ}(64.3589 \ldots)$

Note: The two method marks can be awarded independently. The first can come from any geometry leading to an inverse trigonometric ratio. The second requires recognizing which angle is the bearing, possibly on a diagram. 64.4 appearing in intermediate working is not sufficient.

(c)

Show that the distance between P and N along the arc from P to N is $3000 \pi \mathrm{~km}$ .

Point A, which is also on the equator, has position vector $\boldsymbol{a}=\left(\begin{array}{l}6 \\ 0 \\ 0\end{array}\right)$ as shown in the
following diagram.

Figure for Question (c) — IB Maths AI HL

P, N and A , and the arcs connecting them, form a spherical triangle.
The angle at vertex A is defined as the angle between the vectors $\boldsymbol{a} \times \boldsymbol{p}$ and $\boldsymbol{a} \times \boldsymbol{n}$ .

[ 2 ]

attempt to use arc length formula $s=r \theta$ OR $\frac{90^{\circ}}{360}$ of circumference
$=6000\left(\frac{\pi}{2}\right)$ OR $\frac{1}{4} \times 2 \pi \times 6000$ $=3 \pi$ thousand $\mathrm{km} \quad$ OR $\quad 3000 \pi \mathrm{~km}$

(d)

Show that $\theta=57.3^{\circ}$ , correct to three significant figures.

Bogotá, B , is west of Nairobi and has position vector $\overrightarrow{\mathrm{OB}}=\boldsymbol{b}=\left(\begin{array}{c}6 \sin 120^{\circ} \\ 6 \cos 120^{\circ} \\ 0\end{array}\right)$ .

[ 2 ]

attempt to use $\frac{\theta}{360} \times 2 \pi r=6$ OR $s=r \theta$

\frac{\theta}{360} \times 2 \pi \times 6=6 \quad \text { OR } \quad 6=6 \theta \Rightarrow \theta=1 \text { radian }

$\theta=57.2957 \ldots{ }^{\circ}$

Note: The fourth significant figure must be seen somewhere to award the A1.

$\theta=57.3^{\circ}$ to 3 significant figures

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IB Maths AI HL 3.1 Geometry and Trigonometry Sl Content Topic Practice

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