IB Maths AI SL 3.1 Geometry and Trigonometry Sl Content Questions

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

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]

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 $x \mathrm{~cm}$ .

(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.

(a)

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

[ 2 ]

$\left(\frac{2+6}{2}, \frac{2+0}{2}\right)$
(4,1)
Note: Award AO if parentheses are omitted in the final answer.

(b)

Find the equation of (XZ).

[ 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)

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

Find the coordinates of X .

[ 3 ]

identifying the correct equations to use:

$y=2-x\quad\text{and}\quad y=2x-7$

evidence of solving their correct equations or of finding intersection point graphically

(3,-1)

Note:Accept an answer expressed as " x=3, y=-1 ".

(d)

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

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.

[ 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)

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

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).

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.

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

All question bank results loaded

Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question (f)

Question (g)

Question 3