IB Mathematics AI SL 3.1 Geometry and Trigonometry Question Bank

Practise IB Mathematics AI SL 3.1 by solving triangles, radians, coordinate geometry, trigonometric models and technology-supported diagrams.

Exam points

Apply sine, cosine and area rules to triangle and navigation contexts with correct units.
Use radians, arc length, sector area, bearings and coordinates to model geometric situations.
Solve and interpret trigonometric equations and graphs with technology while checking branches.

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]

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 3.1 Geometry and trigonometry - SL content 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 3.1 Geometry and trigonometry - SL content 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 3.1 Geometry and trigonometry - SL content 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 3.1 Geometry and trigonometry - SL content question 3 — IB Maths AI SL

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

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Covered syllabus points

3.1 Geometry and trigonometry - SL content question 1

Question (a)

Question (b)

3.1 Geometry and trigonometry - SL content question 2

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question (f)

Question (g)

3.1 Geometry and trigonometry - SL content question 3