IB Maths AI SL 3 Geometry and Trigonometry

Practise IB Maths AI SL geometry and trigonometry through diagrams, bearings, radians, sine and cosine rules, vectors and technology-supported solutions.

Covered syllabus points

3.1 Geometry and trigonometry - SL content

[Maximum number: 7]

A hollow chocolate box is manufactured in the form of a right prism with a regular hexagonal base. The height of the prism is $h \mathrm{~cm}$ , and the top and base of the prism have sides of length $x \mathrm{~cm}$ .

Figure for Question 3 Geometry and trigonometry question 1 — IB Maths AI SL

(a)

Given that $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$ , show that the area of the base of the box is equal to $\frac{3 \sqrt{3} x^{2}}{2}$ .

[ 2 ]

evidence of splitting diagram into equilateral triangles M1
area $=6\left(\frac{1}{2} x^{2} \sin 60^{\circ}\right) \quad$ A1
$=\frac{3 \sqrt{3} x^{2}}{2}$

Note:The AG line must be seen for the final A1 to be awarded.

(b)

Given that the total external surface area of the box is $1200 \mathrm{~cm}^{2}$ , show that the volume of the box may be expressed as $V=300 \sqrt{3} x-\frac{9}{4} x^{3}$ .

[ 5 ]

total surface area of prism $1200=2\left(3 x^{2} \frac{\sqrt{3}}{2}\right)+6 x h$ M1A1

Note: Award M1 for expressing total surface areas as a sum of areas of rectangles and hexagons, and A1 for a correctly substituted formula, equated to 1200.

\begin{aligned} & h=\frac{400-\sqrt{3} x^{2}}{2 x} \\ & \text { volume of prism }=\frac{3 \sqrt{3}}{2} x^{2} \times h \\ & =\frac{3 \sqrt{3}}{2} x^{2}\left(\frac{400-\sqrt{3} x^{2}}{2 x}\right) \\ & =300 \sqrt{3} x-\frac{9}{4} x^{3} \end{aligned}

Note:The AG line must be seen for the final A1 to be awarded.

[Maximum number: 18]

The Voronoi diagram below shows four supermarkets represented by points with coordinates A(0,0), B(6,0), C(0,6) and D(2,2). The vertices X, Y, Z are also shown. All distances are measured in kilometres.

Figure for Question 3 Geometry and trigonometry question 2 — IB Maths AI SL

(a)

Find the midpoint of $[\mathrm{BD}]$ .

[ 2 ]

$\left(\frac{2+6}{2}, \frac{2+0}{2}\right)$
(4,1)
(M1)
A1

Note: Award AO if parentheses are omitted in the final answer.

(b)

Find the equation of (XZ).

The equation of ( XY ) is y=2-x and the equation of (YZ) is y=0.5 x+3.5.

[ 4 ]

attempt to substitute values into gradient formula
$\left(\frac{0-2}{6-2}=\right)-\frac{1}{2}$
therefore the gradient of perpendicular bisector is 2

\text { so } y-1=2(x-4) \quad(y=2 x-7)

(c)

Find the coordinates of X .

The coordinates of Y are (-1,3) and the coordinates of Z are (7, 7).

[ 3 ]

(d)

Determine the exact length of [YZ].

[ 2 ]

attempt to use distance formula

\begin{aligned} & \mathrm{YZ}=\sqrt{(7-(-1))^{2}+(7-3)^{2}} \\ & =\sqrt{80}(4 \sqrt{5}) \end{aligned}

(e)

Given that the exact length of [X Y] is $\sqrt{32}$ , find the size of $X \hat{Y} Z$ in degrees.

[ 4 ]

METHOD 1 (cosine rule)
length of XZ is $\sqrt{80}(4 \sqrt{5}, 8.94427 \ldots)$

Note: Accept 8.94 and 8.9. attempt to substitute into cosine rule

\cos X \hat{Y} Z=\frac{80+32-80}{2 \times \sqrt{80} \sqrt{32}} \quad(=0.316227 \ldots)

Note: Award A1 for correct substitution of XZ, YZ, $\sqrt{32}$ values in the cos rule. Exact values do not need to be used in the substitution.

(\mathrm{X} \hat{\mathrm{Y}} \mathrm{Z}=) 71.6^{\circ}\left(71.5650 \ldots{ }^{\circ}\right)

Note: Last A1 mark may be lost if prematurely rounded values of XZ, YZ and/or XY are used.METHOD 2 (splitting isosceles triangle in half)
length of XZ is $\sqrt{80}(4 \sqrt{5}, 8.94427 \ldots)$
Note: Accept 8.94 and 8.9.
required angle is $\cos ^{-1}\left(\frac{\sqrt{32}}{2 \sqrt{80}}\right)$
Note: Award A1 for correct substitution of XZ (or YZ), $\frac{\sqrt{32}}{2}$ values in the cos rule. Exact values do not need to be used in the substitution.
$(\mathrm{X} \hat{\mathrm{Y}} \mathrm{Z}=) 71.6^{\circ}\left(71.5650^{\circ}\right)$
Note: Last A1 mark may be lost if prematurely rounded values of XZ, YZ and/or XY are used.

(f)

Hence find the area of triangle XYZ.

A town planner believes that the larger the area of the Voronoi cell XYZ, the more people will shop at supermarket D.

[ 2 ]

$\quad($ area $=) \frac{1}{2} \sqrt{80} \sqrt{32} \sin 71.5650 \ldots \quad$ OR $\quad($ area $=) \frac{1}{2} \sqrt{32} \sqrt{72}$ $=24 \mathrm{~km}^{2}$

(g)

State one criticism of this interpretation.

[ 1 ]

[Maximum number: 4]

Zheng models the height of a section of a rollercoaster.
She uses the function

f(x)=0.001 x^{3}-0.12 x^{2}+4.05 x+3, \text { for } 0 \leq x \leq 60

where x is the horizontal distance from the left-hand end of the section and f(x) is the vertical height. Distances are measured in metres.

Zheng's model is shown on the following graph of y=f(x).

Figure for Question 3 Geometry and trigonometry question 3 — IB Maths AI SL

Point O has coordinates (0,0).

Find the angle of elevation from O to M .

Zheng wants to place advertising boards in front of this section of the rollercoaster. At four positions along the rollercoaster track, she records the horizontal distance from O and the vertical height of the track.

Her values are shown in the table and represented on the graph.

Table for Question 3 Geometry and trigonometry question 3 — IB Maths AI SL

attempt to use inverse tangent
$\tan ^{-1}\left(\frac{y}{x}\right)$ $61.7^{\circ}\left(61.6907 \ldots{ }^{\circ}\right)$

[Maximum number: 5]

The Scheveningen Ferris wheel's lowest point is 8 m above sea level, and its highest point is 45 m above sea level.

Figure for Question 3 Geometry and trigonometry question 4 — IB Maths AI SL

(a)

Show that the radius of the Ferris wheel is 18.5 m .

[ 1 ]

$\frac{45-8}{2} \quad A1$ $18.5(\mathrm{~m})$ AG
Note: Use of 18.5 in the calculation is reverse engineering, and falls short of the demand of the command term "Show that'. In such cases, award A0.

(b)

Calculate the circumference of the Ferris wheel.

There are pods, equally spaced around the wheel, that carry passengers.

[ 1 ]

$\quad(2 \pi \times 18.5=) 116(116.2389 \ldots, 37 \pi)(\mathrm{m})$ .

(c)

When the wheel rotates $10^{\circ}$ , find the distance that a pod travels along the circumference.

The height in metres, above sea level, of a particular pod is modelled by the function:

h(t)=a \sin (b t)+d, \text { for } a, b>0

where t is the time, measured in minutes.
The wheel takes 15 minutes to complete 1 revolution.

[ 3 ]

recognition the distance is the arc length

\begin{aligned} & \frac{10^{\circ}}{360^{\circ}} \times 2 \pi \times 18.5 \text { OR } \frac{116.2389 \ldots}{36} \\ & =3.23\left(3.228859 \ldots, \frac{37 \pi}{36}\right)(\mathrm{m}) \end{aligned}

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Covered syllabus points

3 Geometry and trigonometry question 1

Question (a)

Question (b)

3 Geometry and trigonometry question 2

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question (f)

Question (g)

3 Geometry and trigonometry question 3

3 Geometry and trigonometry question 4

Question (a)

Question (b)

Question (c)