IB Maths AI HL 3.1 Geometry and Trigonometry Sl Content Questions

Practise IB Mathematics AI HL 3.1 by combining vectors, trigonometric identities, loci and technology geometry in extended models.

Exam points

Solve 3D coordinate and solid-geometry problems using distances, midpoints, lengths, angles, volumes and surface areas.
Apply triangle trigonometry, bearings, elevation/depression, arc/sector formulae and radian measure to geometric contexts.
Use perpendicular bisectors and Voronoi structures to determine loci, nearest regions, boundaries and applied locations.

Covered syllabus points

SL 3.1—Three-dimensional geometry SL 3.2—Triangle trigonometry SL 3.3—Applications of trigonometry SL 3.4—Arc length and sector area SL 3.5—Perpendicular bisectors SL 3.6—Voronoi diagrams

[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.

[ 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)

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.

Figure for Question (c) — IB Maths AI HL

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

[ 2 ]

(i)

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

[ 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

[Maximum number: 2]

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

[Maximum number: 7]

This question considers how the assessment of the Air Quality Index (AQI) for a school depends on the method chosen by the person doing the assessing.
Air quality for a district is measured at three monitoring stations. The positions of these stations on a coordinate system with units in kilometres are A(0,5), B(8,9) and C(8,1).
A Voronoi diagram is constructed with the three stations as sites.

(a)

Find the equation of the perpendicular bisector of [BC].

[ 2 ]

(b)

Find the mid-point of [SB].

[ 2 ]

attempt to find a midpoint

\left(\frac{5+8}{2}, \frac{6+9}{2}\right)

(6.5,7.5)

(c)

Show that the equation of the perpendicular bisector of $[\mathrm{SB}]$ is y=14-x.

[ 3 ]

gradient is $\frac{9-6}{8-5}=1$
gradient of perpendicular bisector is -1
substituting their gradient and their mid-point into the equation of a line

y-7.5=-1(x-6.5) \text { OR } 7.5=-1 \times 6.5+c

y=14-x

Note: If A1A1M1 have been awarded but there is then incorrect work after the final M1 such that the AG does not follow, do not award the final M1.

(d)

Given that the equation of the perpendicular bisector of $[\mathrm{AB}]$ is y=15-2 x, find the coordinates of vertex V.

A school, S, is situated in the district at the point with coordinates (5,6).

[ 2 ]

equating the correct lines
5=15-2 x OR sketch graph

x=5 \quad y=5 \quad \text { OR }(5,5)

(e)

State which station is closest to the school.

The principal of the school is concerned about the air quality in the area. Air quality is measured by the Air Quality Index (AQI). In this district, values less than 50 are taken to indicate good air quality.

The principal contacts the local environmental agency requesting an AQI value for her school. They tell her the mean AQI reading from the closest station to the school.

[ 1 ]

(f)

Write down the type of interpolation being used by the environmental agency.

The principal obtains the mean AQI value from each of the three stations.

[ 1 ]

(g)

Explain why the principal might not accept that the air quality around the school can be classed as "good".

The principal decides to obtain an expected value for the AQI at the school that uses all the available information. To do this, she uses an alternative method: the natural neighbour algorithm. This algorithm has two stages.

The first stage is to create a new Voronoi diagram with the school as an extra site. This is shown in the following diagram with the edges of the previous diagram shown by dashed lines.

Figure for Question (g) — IB Maths AI HL

The second stage is to estimate the AQI value at the school, W, by using the formula

W=\frac{w_{A} a_{A}+w_{B} a_{B}+w_{C} a_{C}}{T}

In the formula, $a_{A}$ is the area within the new cell that has been taken from the cell surrounding site A , shown as region P on the diagram, and $w_{A}$ is the mean AQI value from site A . This is given as 132 in the table of mean AQI values above. Similarly for sites B and C . T is the total area of the new cell around S .

[ 1 ]

(h)

Hence find $a_{B}$ (the area of region Q on the diagram).

The areas of regions P and R are $a_{A}=13.7 \mathrm{~km}^{2}$ and $a_{C}=6.9 \mathrm{~km}^{2}$ respectively.

[ 5 ]

attempt to find where this line meets the other two lines
(9,5)
14-x=15-2 x
(1,13)

Note: These two marks are un-implied due to the "hence" in the question. At least one needs to be seen for the (M1) to be awarded.

\begin{aligned} \text { Area } & =\frac{1}{2} \times 4 \times 8 \\ & =16\left(\mathrm{~km}^{2}\right) \end{aligned}

Note: Award (M1) for any valid method which is completed (for example cosine rule, followed by area of triangle formula or using one of the other sides as base and correctly finding the perpendicular height) which leads to an answer, even if the answer is not correct due to a calculation error. (the side lengths of the triangle are 4, 11.3, 8.94 and the angles $45^{\circ}, 18.4^{\circ}, 116.6^{\circ}$ )

Note: Award (M0)A0 for an unsupported answer of 16.

(i)

Use the natural neighbour algorithm to show that an estimate for the expected AQI value at the school, W, is 94.4.

The principal is still concerned that this method is underestimating the AQI value at the school, as the school is situated close to a busy traffic intersection. She decides to take her own readings ( x ) over a period of 60 days. Her results are summarized as

\bar{x}=97.8, s_{n-1}=17.2, n=60 .

The principal assumes that the daily AQI values at the school can be modelled by a normal distribution and that each value is independent of any other value.

[ 2 ]

attempt to substitute the values into given formula M1

\begin{aligned} & W=\frac{132 \times 13.7+49 \times 16+125 \times 6.9}{16+13.7+6.9} \\ & W=94.4(94.3961 \ldots) \end{aligned}

All question bank results loaded

Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question 2

Question (a)

Question (b)

Question (c)

Question 3

Question (a)

Question (b)

Question (c)

Question (i)

Question (d)

Question 4

Question 5

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question (f)

Question (g)

Question (h)

Question (i)