Mini-Lesson

Geometry & Trigonometry (AI SL)

4 missions. Guess, learn, then check. Your place is saved.

Mission 1What's the longest pencil that fits in a 20 × 12 × 9 cm box?
Mission 2How can you find a mountain's height without climbing it?
Mission 330 cm pizza cut in 8, 24 cm cut in 6: which slice has more crust?
Mission 4Four phone masts: where in town is the signal weakest?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

What's the longest pencil that fits in a 20 × 12 × 9 cm box?

Make a guess: What's the longest pencil that fits in a 20 × 12 × 9 cm box?

Geometry · 3-D shapes

Distance and midpoint in 3-D

Points in space have three coordinates $(x, y, z)$ (x, y, z).
Distance between $(x_{1}, y_{1}, z_{1})$ (x₁, y₁, z₁) and $(x_{2}, y_{2}, z_{2})$ (x₂, y₂, z₂):

d = (x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} + (z_{1} - z_{2})^{2}

Midpoint: $(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2}, \frac{z _{1} + z _{2}}{2})$ ((x₁+x₂)/2, (y₁+y₂)/2, (z₁+z₂)/2)
In a cuboid $a \times b \times c$ a × b × c, Pythagoras on the base gives $a^{2} + b^{2}$ √(a²+b²); Pythagoras again with the height gives the space diagonal $a^{2} + b^{2} + c^{2}$ √(a²+b²+c²).

Worked example Geometry · 3-D shapes

Space diagonal and its angle with the base

Question

A cuboid measures 8 cm by 6 cm by 5 cm. Find the length of a space diagonal and the angle it makes with the base.

1
Base diagonal by Pythagoras: $8^{2} + 6^{2} = 10$ √(8²+6²)=10 cm
2
Space diagonal in the vertical right-angled triangle: $1 0^{2} + 5^{2} = 125 = 11.2$ √(10²+5²)=√125=11.2 cm (3 s.f.)
3
The angle with the base is between the diagonal and its projection, the base diagonal: $tan θ = \frac{5}{10}$ tan θ=5/10

Answer

11.2

cm and

θ = 26. 6^{\circ}

(3 s.f.)

⚠️ Watch out

the angle with the base uses the base diagonal (10 cm), not an edge;

tan^{- 1} \frac{5}{8}

gives the wrong

32. 0^{\circ}

.

Calculate

?

A

is

(2, - 1, 5)

and

B

is

(- 4, 1, 1)

.

M

is the midpoint of

A B

. Find the distance from

M

to

C (1, 6, 0)

. Give an exact answer.

Geometry · 3-D shapes

Angles in 3-D

The angle between a line and a plane is the angle between the line and its projection on the plane (the "shadow" straight below it).
Drop a perpendicular from a point on the line to the plane: this makes a right-angled triangle containing the angle.
The angle between two intersecting lines is found in the triangle that both lines lie in.
At SL, 3-D angle questions use right-angled trigonometry: sketch the triangle flat, label it, then use SOH CAH TOA.

Quick check

?

V A BC D

is a right pyramid on a square base

A BC D

.

M

is the centre of the base, directly below

V

, and

N

is the midpoint of

A B

. Which angle is the angle between the edge

V A

and the base?

Geometry · 3-D shapes

Volume and surface area

Solid	Volume	Surface area
Right pyramid	$\frac{1}{3} \times$ 1/3× base area $\times$ × height	base + triangular faces
Right cone	$\frac{1}{3} π r^{2} h$ 1/3π r²h	curved: $π r l$ π rl ( $l$ l = slant height)
Sphere	$\frac{4}{3} π r^{3}$ 4/3π r³	$4 π r^{2}$ 4π r²
Hemisphere	$\frac{2}{3} π r^{3}$ 2/3π r³	curved: $2 π r^{2}$ 2π r²

These formulas are in the IB formula booklet. For a combined solid, add the volumes, but count only the surfaces on the outside.
Slant height of a cone: $l = r^{2} + h^{2}$ l=√(r²+h²).

Worked example Geometry · 3-D shapes

A cone on a hemisphere

Question

A toy is a cone of radius 3 cm and height 4 cm fixed to a hemisphere of radius 3 cm. Find its volume and its total surface area.

1
Cone volume: $\frac{1}{3} π (3^{2}) (4) = 12 π$ 1/3π(3²)(4)=12π
2
Hemisphere volume: $\frac{2}{3} π (3^{3}) = 18 π$ 2/3π(3³)=18π, so total volume $= 30 π = 94.2$ =30π=94.2 cm³ (3 s.f.)
3
Slant height: $l = 3^{2} + 4^{2} = 5$ l=√(3²+4²)=5, so the curved cone surface is $π (3) (5) = 15 π$ π(3)(5)=15π
4
Curved hemisphere surface: $2 π (3^{2}) = 18 π$ 2π(3²)=18π; the flat circles are joined inside, so they don't count

Answer

volume

94.2

cm³, surface area

33 π = 104

cm² (3 s.f.)

⚠️ Watch out

using the whole sphere's

4 π r^{2}

gives

51 π

; only half the sphere is there.

Calculate

?A right pyramid has a square base of side 10 cm and vertical height 12 cm. Find the angle between a sloping edge and the base. Give your answer in degrees to 3 significant figures.

Calculate

?A grain silo is a cylinder of radius 4 m and height 10 m with a hemisphere of radius 4 m on top. Find the volume of the silo to 3 significant figures.

m³

Mission check

?What's the longest pencil that fits in a 20 × 12 × 9 cm box?

Mission 2

How can you find a mountain's height without climbing it?

Make a guess: How can you find a mountain's height without climbing it?

Geometry · trigonometry

Right-angled trigonometry, elevation and bearings

In a right-angled triangle: $sin θ = \frac{opp}{hyp}$ sin θ=opp/hyp, $cos θ = \frac{adj}{hyp}$ cos θ=adj/hyp, $tan θ = \frac{opp}{adj}$ tan θ=opp/adj. Use $sin^{- 1}$ sin⁻¹, $cos^{- 1}$ cos⁻¹, $tan^{- 1}$ tan⁻¹ to find angles.
Angle of elevation: measured up from the horizontal. Angle of depression: measured down from the horizontal.
Bearings are measured clockwise from north, written with three figures (e.g. 040°).
Always sketch and label a diagram from the words first.

Quick check

?From the top of a vertical cliff 40 m high, the angle of depression of a boat is 25°. How far is the boat from the foot of the cliff, to 3 significant figures?

Geometry · trigonometry

Sine rule, cosine rule and area

Label the side opposite angle $A$ A as $a$ a, and so on.
Sine rule: $\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$ a/(sin A)=b/(sin B)=c/(sin C). Use it with a side and its opposite angle known.
Cosine rule: $c^{2} = a^{2} + b^{2} - 2 ab cos C$ c²=a²+b²−2abcos C, or $cos C = \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab}$ cos C=(a²+b²−c²)/2ab. Use it with two sides and the angle between them, or with all three sides.
Area $= \frac{1}{2} ab sin C$ =1/2absin C ( $C$ C is the angle between sides $a$ a and $b$ b).

Worked example Geometry · trigonometry

Cosine rule, then area

Question

In triangle

PQR

,

PQ = 8

cm,

PR = 11

cm and angle

QPR = 4 0^{\circ}

. Find

QR

and the area of the triangle.

1
Two sides and the included angle, so use the cosine rule: $Q R^{2} = 8^{2} + 1 1^{2} - 2 (8) (11) cos 4 0^{\circ}$ QR²=8²+11²−2(8)(11)cos 40°
2
Evaluate: $Q R^{2} = 50.18 \dots$ QR²=50.18…, so $QR = 7.08$ QR=7.08 cm (3 s.f.)
3
Area with the same two sides and included angle: $\frac{1}{2} (8) (11) sin 4 0^{\circ}$ 1/2(8)(11)sin 40°

Answer

QR = 7.08

cm, area

= 28.3

cm² (3 s.f.)

⚠️ Watch out

with your GDC in radian mode,

cos 40

gives

QR = 17.4

cm; set degree mode for degree questions.

Sort it

?Which rule do you use first?

Tap a statement, then tap the group it belongs in. Then press Check.

Worked example Geometry · trigonometry

Height from two angles of elevation

Question

From

A

, the angle of elevation of the top

T

of a tower is 20°. From

B

, 500 m closer on level ground, it is 32°. Find the height of the tower.

1
In triangle $A BT$ ABT: angle $A BT = 18 0^{\circ} - 3 2^{\circ} = 14 8^{\circ}$ ABT=180°−32°=148°, so angle $A TB = 18 0^{\circ} - 2 0^{\circ} - 14 8^{\circ} = 1 2^{\circ}$ ATB=180°−20°−148°=12°
2
Sine rule: $\frac{BT}{sin 2 0 ^{\circ}} = \frac{500}{sin 1 2 ^{\circ}}$ BT/(sin 20°)=500/(sin 12°), so $BT = 822.5$ BT=822.5 m
3
Right-angled triangle below $T$ T: height $= BT sin 3 2^{\circ}$ =BTsin 32°

Answer

height

= 436

m (3 s.f.)

⚠️ Watch out

the angle at

T

is the difference of the two elevations,

3 2^{\circ} - 2 0^{\circ}

, not their sum.

Calculate

?A ship sails 15 km from port

P

on a bearing of 040° to

Q

, then 9 km on a bearing of 110° to

R

. Find the distance

PR

. Give your answer to 3 significant figures.

km

Calculate

?A triangular field has sides 50 m, 70 m and 90 m. Find its area. Give your answer to the nearest square metre.

m²

Mission check

?How can you find a mountain's height without climbing it?

Mission 3

30 cm pizza cut in 8, 24 cm cut in 6: which slice has more crust?

Make a guess: 30 cm pizza cut in 8, 24 cm cut in 6: which slice has more crust?

Geometry · circles

Arc length and sector area

A sector is a "pizza slice" of a circle; its curved edge is an arc.
A sector with angle $θ^{\circ}$ θ° is the fraction $\frac{θ}{360}$ θ/360 of the whole circle, so

arc length = \frac{θ}{360} \times 2 π r sector area = \frac{θ}{360} \times π r^{2}

Perimeter of a sector $=$ = arc $+ 2 r$ + 2r (two radii).

Worked example Geometry · circles

A sector in degrees

Question

A sector has radius 9 cm and angle 140°. Find its arc length, area and perimeter.

1
Fraction of the circle: $\frac{140}{360} = \frac{7}{18}$ 140/360=7/18
2
Arc length: $\frac{7}{18} \times 2 π (9) = 7 π = 22.0$ 7/18× 2π(9)=7π=22.0 cm (3 s.f.)
3
Area: $\frac{7}{18} \times π (9^{2}) = 31.5 π = 99.0$ 7/18× π(9²)=31.5π=99.0 cm² (3 s.f.)
4
Perimeter: $7 π + 2 (9) = 40.0$ 7π+2(9)=40.0 cm (3 s.f.)

Answer

arc

22.0

cm, area

99.0

cm², perimeter

40.0

cm

⚠️ Watch out

the perimeter includes both straight radii, not just the arc.

Calculate

?A sector of a circle of radius 6 cm has area 30 cm². Find the angle of the sector in degrees, to 3 significant figures.

Geometry · circles

Segments and compound shapes

A segment is the region between a chord and its arc.
Segment area $=$ = sector area $-$ − triangle area, with the triangle area $\frac{1}{2} r^{2} sin θ$ 1/2r²sin θ (two radii and the angle between them).
For a shape made of sectors, rectangles and triangles: find each piece, then add or subtract.
Work with unrounded values and round only the final answer to 3 significant figures.

Calculate

?A circle has centre

O

and radius 10 cm. Points

A

and

B

lie on the circle with angle

A OB = 8 0^{\circ}

. Find the area of the segment between chord

A B

and the minor arc

A B

. Give your answer to 3 significant figures.

cm²

Calculate

?A fan-shaped badge is a sector of radius 12 cm. Its perimeter is 40 cm. Find its area.

cm²

Mission check

?30 cm pizza cut in 8, 24 cm cut in 6: which slice has more crust?

Mission 4

Four phone masts: where in town is the signal weakest?

Make a guess: Four phone masts: where in town is the signal weakest?

Geometry · Voronoi diagrams

Perpendicular bisectors

The perpendicular bisector of $A B$ AB is the set of points equally distant from $A$ A and $B$ B.
It passes through the midpoint $(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$ ((x₁ + x₂)/2, (y₁ + y₂)/2).
Its gradient is the negative reciprocal of the gradient of $A B$ AB: if $A B$ AB has gradient $m$ m, the bisector has gradient $- \frac{1}{m}$ −1/m.
Then use $y - y_{1} = m (x - x_{1})$ y − y₁ = m(x − x₁) through the midpoint. If $A B$ AB is horizontal, the bisector is vertical ( $x =$ x = constant), and vice versa.

Worked example Geometry · Voronoi diagrams

The boundary between two masts

Question

Masts are at

A (1, 1)

and

D (8, 8)

, in km. Find the equation of the perpendicular bisector of

A D

.

1
Midpoint: $(\frac{1 + 8}{2}, \frac{1 + 8}{2}) = (4.5, 4.5)$ ((1 + 8)/2, (1 + 8)/2) = (4.5, 4.5)
2
Gradient of $A D$ AD: $\frac{8 - 1}{8 - 1} = 1$ (8 − 1)/(8 − 1) = 1
3
Perpendicular gradient: $- \frac{1}{1} = - 1$ −1/1 = −1
4
Line through the midpoint: $y - 4.5 = - 1 (x - 4.5)$ y − 4.5 = −1(x − 4.5)

Answer

y = - x + 9

⚠️ Watch out

the bisector goes through the midpoint, not through

A

or

D

. Using

(1, 1)

gives

y = - x + 2

, which is wrong.

Calculate

?A line segment lies on the line

y = 3 x - 4

and has midpoint

(2, 2)

. Its perpendicular bisector meets the

x

-axis at

(k, 0)

. Find

k

.

Geometry · Voronoi diagrams

Voronoi diagrams

A Voronoi diagram splits a region into cells, one around each site. Every point in a cell is nearer its own site than any other.
Edges are parts of perpendicular bisectors between pairs of sites. A vertex is where three edges meet: it is equidistant from three sites.
Nearest neighbour interpolation: every point in a cell is given the same value as its site (for example, the rainfall measured there).
Exams give the site coordinates; you will not construct bisectors.

Quick check

?Rain gauges at the four sites

A (1, 1)

,

B (11, 1)

,

C (1, 7)

and

D (8, 8)

record 12 mm, 20 mm, 16 mm and 9 mm. Use nearest neighbour interpolation to estimate the rainfall at

H (4, 6)

.

Geometry · Voronoi diagrams

Adding a site; the toxic waste dump

Adding a site: draw the bisectors between the new site and its neighbours; the new cell is carved from theirs.
Toxic waste dump problem: find the point in the region as far as possible from every site (a dump, or the weakest signal).
Within the region, the farthest point from every site is at a Voronoi vertex or on the region's boundary; exams put it at a vertex. Find each vertex's distance to one of its sites: the largest wins.

Worked example Geometry · Voronoi diagrams

Where is the weakest signal?

Question

In the town

0 \leq x \leq 12

,

0 \leq y \leq 10

(km), masts

A (1, 1)

,

B (11, 1)

,

C (1, 7)

and

D (8, 8)

give a Voronoi diagram with vertices

P (5, 4)

and

Q

. The edge between cells

A

and

B

is

x = 6

, and between

A

and

D

it is

y = - x + 9

. Find

Q

, and the point of the town farthest from every mast.

1
Solve $x = 6$ x = 6 with $y = - x + 9$ y = −x + 9: $Q = (6, 3)$ Q = (6, 3), equidistant from $A$ A, $B$ B and $D$ D
2
Distance from $Q$ Q to $A$ A: $5^{2} + 2^{2} = 29 = 5.39$ √(5² + 2²) = √29 = 5.39 km
3
Distance from $P$ P to $A$ A (one of $A$ A, $C$ C, $D$ D): $4^{2} + 3^{2} = 5$ √(4² + 3²) = 5 km
4
Compare: $5.39 > 5$ 5.39 > 5

Answer

Q (6, 3)

, 5.39 km from its three nearest masts.

⚠️ Watch out

measure from the vertex to one of its own three sites.

Q

to

C

is 6.40 km, but

A

,

B

and

D

are nearer.

Calculate

?For the town and masts above, the cell of mast

A

has vertices

(0, 0)

,

(6, 0)

,

(6, 3)

,

(5, 4)

and

(0, 4)

. Find the area of the cell of

A

.

km²

Calculate

?A fifth mast

E

is built at

(11, 6)

. The boundary between the new cells of

D (8, 8)

and

E

has equation

y = m x + c

. Find

c

as a decimal.

Mission check

?Four phone masts: where in town is the signal weakest?

Recap

Big ideas

3-D distance: $(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} + (z_{1} - z_{2})^{2}$ √((x₁−x₂)²+(y₁−y₂)²+(z₁−z₂)²); a cuboid's space diagonal is $a^{2} + b^{2} + c^{2}$ √(a²+b²+c²).
The angle between a line and a plane is the angle with its projection: find the right-angled triangle and use trigonometry.
Combined solids: add volumes; count only the outside surfaces.
Right-angled triangles: SOH CAH TOA. Elevation is up from the horizontal; depression is down; bearings are clockwise from north.
Sine rule needs a side with its opposite angle; cosine rule needs two sides and the included angle, or three sides. Area $= \frac{1}{2} ab sin C$ =1/2absin C.
Break a real problem into triangles and sketch a labelled diagram first.
Arc $= \frac{θ}{360} \times 2 π r$ =θ/360× 2π r and sector $= \frac{θ}{360} \times π r^{2}$ =θ/360×π r²; perimeter adds two radii.
Segment $=$ = sector $-$ − triangle, with the triangle $\frac{1}{2} r^{2} sin θ$ 1/2r²sin θ.
The perpendicular bisector of $A B$ AB passes through the midpoint with gradient $- \frac{1}{m}$ −1/m; every point on it is equidistant from $A$ A and $B$ B.
Voronoi cells hold the points nearest each site; edges lie on bisectors; vertices are equidistant from three sites. Nearest neighbour interpolation gives a cell its site's value.
Toxic waste dump: test the vertices and choose the one farthest from its three sites.

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Geometry & Trigonometry (AI SL) for IB Diploma Mathematics: Applications & Interpretation SL. 🎉

Mission 1What's the longest pencil that fits in a 20 × 12 × 9 cm box?
Mission 2How can you find a mountain's height without climbing it?
Mission 330 cm pizza cut in 8, 24 cm cut in 6: which slice has more crust?
Mission 4Four phone masts: where in town is the signal weakest?

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Specification points and sources for this lesson are listed at the bottom of the page.