Module 21 of 47Lesson 1 of 6 · AI 3.1 — 3D geometry: distance, midpoint, volume, surface area, angles

Course map · Topic 3: Geometry and Trigonometry

3.1 3D geometry: distance, midpoint, volume, surface area, angles

SL & HL11 worked questions

Most real objects are 3D: rooms, tanks, tents, silos. This subtopic is about measuring them — how far apart two points are in space, how much a solid holds (volume), how much material covers it (surface area), and the angles hidden inside it.

✏️ Start with the questions. Try each one on paper, then open the solution and compare.

Stuck? Open the notes ↓

✏️ Practice questions

Try each question on paper first. Then open the solution and compare every step.

Question 1●○○ Warm-up✍️ + 🖩 · Paper 1[4 marks]

The points

A(2,\,-1,\,4)

and

B(-3,\,5,\,1)

are given.

(a) Find the distance $AB$ .[2]
(b) Find the coordinates of $M$ , the midpoint of $[AB]$ .[2]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkDistance = Pythagoras in 3D

Find how far apart the points are in each direction: the change in

x

, the change in

y

and the change in

z

. Square each one, add them, then take the square root. Squaring makes every change positive, so the order of subtracting does not matter.

Step 2✍️ By handSubstitute into the formula

AB=\sqrt{(2-(-3))^2+(-1-5)^2+(4-1)^2}

AB=\sqrt{5^2+(-6)^2+3^2}=\sqrt{25+36+9}

AB=\sqrt{70}=8.37

(3 s.f.)A1

Step 3🖩 On the GDCCheck on the calculator

ctrl💻 Ctrl›x²💻 type ^2, then →›5^2+(-6)^2+3^2›ctrl+enter💻 Ctrl+Enter

Without ctrl+enter the calculator shows the exact answer

\sqrt{70}

. ctrl+enter gives the decimal.

√70 ≈ 8.3666

Part (b)

Step 1🧠 ThinkMidpoint = the average of each coordinate

The midpoint is exactly halfway. Halfway between two numbers is their average: add them and divide by 2. Do it for

x

, then

y

, then

z

Step 2✍️ By handAverage each coordinate

M=\left(\dfrac{2+(-3)}{2},\ \dfrac{-1+5}{2},\ \dfrac{4+1}{2}\right)

M=(-0.5,\ 2,\ 2.5)

Final answer (a)

AB=\sqrt{70}\approx 8.37

(b)

M(-0.5,\ 2,\ 2.5)

⚠️ Common mistake Writing

-1-5 = 4

. Be careful with negatives — put brackets round every coordinate before you subtract.

Question 2●○○ Warm-up✍️ + 🖩 · Paper 1[4 marks]

A football is a sphere with radius 11 cm.

(a) Find the volume of the football.[2]
(b) Find the surface area of the football.[2]

Show the full step-by-step solution

Step 1🧠 ThinkWhich formula?

Volume is the space inside (cm³). Surface area is the leather on the outside (cm²). Both formulas are in the formula booklet; we only need

r = 11

Part (a)

Step 1✍️ By handVolume of a sphere

V=\dfrac{4}{3}\pi\times11^3

V=5575.28\ldots \approx 5580\text{ cm}^3

Part (b)

Step 1✍️ By handSurface area of a sphere

A=4\pi\times11^2

A=1520.53\ldots \approx 1520\text{ cm}^2

Step 2🖩 On the GDCBoth on the calculator

(4/3)›×💻 *›π💻 type pi›×💻 *›11›^💻 Shift+6, then →›3›ctrl+enter💻 Ctrl+Enter

The π key is next to the letters. Press ctrl+enter, otherwise you get the exact answer

\dfrac{5324\pi}{3}

Final answer (a)

5580\text{ cm}^3

(3 s.f.) (b)

1520\text{ cm}^2

(3 s.f.)

⚠️ Common mistake Mixing up the two formulas. Volume has

r^3

(cubic units, cm³); area has

r^2

(square units, cm²).

Question 3●○○ Warm-up✍️ + 🖩 · Paper 1[6 marks]

A party hat is a right cone with base radius 4 cm and vertical height 9 cm.

(a) Find the slant height $l$ .[2]
(b) Find the volume of the cone.[2]
(c) Find the curved surface area of the cone (the card needed to make the hat).[2]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkFind the right-angled triangle

Cut the cone straight down the middle. You see a right-angled triangle: the height (9) going up, the radius (4) along the bottom, and the slant height

l

as the longest side (the hypotenuse).

Step 2✍️ By handPythagoras

l^2=4^2+9^2=97

l=\sqrt{97}=9.85\text{ cm}

(3 s.f.)A1

Part (b)

Step 1✍️ By handVolume uses the vertical height 9

V=\dfrac{1}{3}\pi\times4^2\times9

V=48\pi=151\text{ cm}^3

(3 s.f.)A1

Part (c)

Step 1✍️ By handCurved area uses the slant height

A=\pi\times4\times\sqrt{97}

A=124\text{ cm}^2

(3 s.f.)A1

Step 2🖩 On the GDCCalculator check

ctrl💻 Ctrl›x²💻 type ^2, then →›4^2+9^2›ctrl+enter💻 Ctrl+Enter

Use the full

\sqrt{97}

inside the area — do not type the rounded 9.85.

Final answer (a)

l\approx9.85\text{ cm}

(b)

151\text{ cm}^3

(c)

124\text{ cm}^2

⚠️ Common mistake Using

l

in the volume or

h

in the area. Volume → vertical height; curved area → slant height.

Question 4●●○ Standard✍️ + 🖩 · Paper 1[6 marks]

The diagram shows a cuboid

ABCDEFGH

with

AB = 8\text{ cm}

BC = 5\text{ cm}

and

CG = 4\text{ cm}

(a) Find the length $AC$ .[2]
(b) Find the length of the diagonal $AG$ .[2]
(c) Find the angle $\theta$ between $AG$ and the base $ABCD$ .[2]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkTwo right-angled triangles, one after the other

AC

lies flat on the floor of the box. Triangle

ABC

has a right angle at

B

, so Pythagoras gives

AC

Step 2✍️ By handTriangle ABC (floor)

AC^2=8^2+5^2=89

AC=\sqrt{89}=9.43\text{ cm}

Part (b)

Step 1🧠 ThinkNow stand up: triangle ACG

G

is straight above

C

, so the edge

CG

is vertical and the angle

ACG

90°

. Triangle

ACG

is right-angled at

C

Step 2✍️ By handTriangle ACG

AG^2=AC^2+CG^2=89+16=105

AG=\sqrt{105}=10.2\text{ cm}

(3 s.f.)A1

💡 Remember Shortcut for any cuboid:

\text{diagonal}=\sqrt{a^2+b^2+c^2}=\sqrt{64+25+16}

. It is the 3D distance formula again.

Part (c)

Step 1🧠 ThinkWhich angle is it?

The “shadow” of

AG

on the floor is

AC

. So

\theta

is angle

GAC

in triangle

ACG

. From

A

: the opposite side is

CG = 4

, the adjacent side is

AC=\sqrt{89}

. Opposite and adjacent → tan.

Step 2✍️ By handUse tan

\tan\theta=\dfrac{4}{\sqrt{89}}

\theta=23.0°

(3 s.f.)A1

Step 3🖩 On the GDCCalculator in DEG mode

trig💻 type sin( cos( tan(›tan⁻¹›4÷√(89)›enter💻 Enter

📍 Where to find it on the TI-Nspire

⌨️ square root: ctrl then x² on the keypad, or type sqrt(

Check that DEG is shown at the top of the screen. If it says RAD, the answer will be 0.401 — wrong.

DEG mode: θ = 22.98°

Final answer (a)

9.43\text{ cm}

(b)

10.2\text{ cm}

(c)

23.0°

⚠️ Common mistake Using triangle

ABG

or the angle at

G

. The angle with the base is always at the point where the line touches the base, between the line and its shadow.

Question 5●●○ Standard✍️ + 🖩 · Paper 2[8 marks]

A monument is a right pyramid

VABCD

with a square base of side 6 m. The vertex

V

is 8 m vertically above

O

, the centre of the base.

(a) Find the volume of the pyramid.[2]
(b) Find the length of the sloping edge $VC$ .[3]
(c) Find the angle $\alpha$ between $VC$ and the base.[2]
(d) Each triangular face has a height (from $V$ to the middle of a base edge) of $8.544\text{ m}$ . Find the total surface area of the pyramid.[1]

Show the full step-by-step solution

Part (a)

Step 1✍️ By handOne third × base area × height

V=\dfrac{1}{3}\times(6\times6)\times8

V=96\text{ m}^3

Part (b)

Step 1🧠 ThinkFind OC first

O

is the centre of the square, so

OC

is half of the diagonal

AC

. The diagonal of a 6 by 6 square is

\sqrt{6^2+6^2}

. Then triangle

VOC

is right-angled at

O

(the height is vertical).

Step 2✍️ By handTwo uses of Pythagoras

AC=\sqrt{6^2+6^2}=\sqrt{72}\Rightarrow OC=\dfrac{\sqrt{72}}{2}=4.243

VC^2=8^2+4.243^2=64+18=82

VC=\sqrt{82}=9.06\text{ m}

(3 s.f.)A1

Part (c)

Step 1🧠 ThinkAngle between the edge and its shadow

The shadow of

VC

on the base is

OC

. So

\alpha

is angle

VCO

. From

C

: opposite

= VO = 8

, adjacent

= OC

Step 2✍️ By handUse tan

\tan\alpha=\dfrac{8}{\sqrt{18}}

\alpha=62.1°

Step 3🖩 On the GDCCalculator (DEG)

Part (d)

Step 1✍️ By handSquare base + four equal triangles

A=6^2+4\times\dfrac{1}{2}\times6\times8.544=36+102.53=139\text{ m}^2

💡 Remember The 8.544 m is the slant height of a face:

\sqrt{8^2+3^2}

. It is not the edge

VC

— the face height goes to the middle of the edge, the edge goes to the corner.

Final answer (a)

96\text{ m}^3

(b)

9.06\text{ m}

(c)

62.1°

(d)

139\text{ m}^2

Question 6●●○ Standard✍️ + 🖩 · Paper 2[6 marks]

A grain silo on a farm near Francistown is a cylinder of radius 3 m and height 10 m, with a hemisphere on top.

(a) Find the volume of the silo.[3]
(b) The outside of the silo (not the base) is to be painted. Find the area to be painted.[3]

Show the full step-by-step solution

Step 1🧠 ThinkSplit the solid into shapes you know

The silo is a cylinder plus half a sphere. Work out each part separately, then add.

Part (a)

Step 1✍️ By handCylinder + hemisphere

Cylinder:

\pi\times3^2\times10=90\pi=282.74

Hemisphere:

\dfrac{1}{2}\times\dfrac{4}{3}\pi\times3^3=18\pi=56.55

Total:

108\pi=339\text{ m}^3

(3 s.f.)A1

Part (b)

Step 1🧠 ThinkOnly the outside skin that can be seen

Paint goes on the curved wall of the cylinder and the curved dome. No paint on the base, and no paint on the circle where the dome sits (it is inside).

Step 2✍️ By handCurved cylinder + curved hemisphere

Wall:

2\pi\times3\times10=60\pi

Dome:

\dfrac{1}{2}\times4\pi\times3^2=18\pi

Total:

78\pi=245\text{ m}^2

(3 s.f.)A1

Step 3🖩 On the GDCCalculator

108›×💻 *›π💻 type pi›ctrl+enter💻 Ctrl+Enter

Final answer (a)

339\text{ m}^3

(b)

245\text{ m}^2

⚠️ Common mistake Adding the area of the circle between the cylinder and the dome. It is hidden inside the silo, so it is not painted.

Question 7●●○ Standard🖩 GDC · Paper 1[5 marks]

Volume questions can also be asked backwards: you are given the volume and must find a length.

(a) A spherical water container holds exactly 2 litres. Find its radius in cm.[3]
(b) A conical cup has radius 6 cm and volume 500 cm³. Find its height.[2]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkChange litres to cm³ first

1 litre = 1000 cm³, so 2 litres = 2000 cm³. Now we know the volume and want

r

: we work the formula backwards.

Step 2✍️ By handRearrange the formula

\dfrac{4}{3}\pi r^3=2000

r^3=\dfrac{3\times2000}{4\pi}=477.46

r=\sqrt[3]{477.46}=7.82\text{ cm}

(3 s.f.)A1

Step 3🖩 On the GDCOr let the calculator solve it

menu💻 click menu on keypad›3: Algebra›1: Numerical Solve›(4/3)·π·r^3=2000,r›enter💻 Enter

nSolve finds the value of

r

that makes the equation true.

Part (b)

Step 1✍️ By handSame idea for the cone

\dfrac{1}{3}\pi\times6^2\times h=500

h=\dfrac{3\times500}{36\pi}=13.3\text{ cm}

(3 s.f.)A1

Step 2🖩 On the GDCCheck with nSolve

Final answer (a)

r\approx7.82\text{ cm}

(b)

h\approx13.3\text{ cm}

⚠️ Common mistake Putting 2 (litres) straight into the formula. Units must match: the answer is in cm, so the volume must be in cm³.

Question 8●●● Exam-level✍️ + 🖩 · Paper 2[7 marks]

A classroom is a cuboid 6 m long, 4 m wide and 3 m high. Take the corner

O

as the origin, with

A(6,0,0)

B(6,4,0)

and

C(0,4,0)

on the floor. A lamp

L

hangs at the centre of the ceiling. A straight wire runs from

L

to the floor corner

B

N

is the point on the floor directly below

L

(a) Write down the coordinates of $L$ .[2]
(b) Find the length of the wire $LB$ .[2]
(c) Find the angle $\theta$ that the wire makes with the floor.[3]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkCentre of the ceiling

The centre is halfway along the length (

6\div2=3

) and halfway across the width (

4\div2=2

). The ceiling is 3 m up, so

z=3

. (It is the midpoint of the ceiling diagonal from

(0,0,3)

(6,4,3)

Step 2✍️ By handWrite it down

L(3,\ 2,\ 3)

A1A1

Part (b)

Step 1✍️ By hand3D distance from L(3, 2, 3) to B(6, 4, 0)

LB=\sqrt{(6-3)^2+(4-2)^2+(0-3)^2}=\sqrt{9+4+9}

LB=\sqrt{22}=4.69\text{ m}

(3 s.f.)A1

Part (c)

Step 1🧠 ThinkUse the triangle LNB

N(3,2,0)

is straight below

L

, so

LN=3

m is vertical and angle

LNB=90°

NB

is the shadow of the wire on the floor, so

\theta

is angle

LBN

. We know the opposite side (

LN = 3

) and the hypotenuse (

LB=\sqrt{22}

) → sin.

Step 2✍️ By handUse sin

\sin\theta=\dfrac{3}{\sqrt{22}}

M1A1

\theta=39.8°

(3 s.f.)A1

Step 3🖩 On the GDCCalculator (DEG)

trig💻 type sin( cos( tan(›sin⁻¹›3÷√(22)›enter💻 Enter

📍 Where to find it on the TI-Nspire

⌨️ square root: ctrl then x² on the keypad, or type sqrt(

✅ Check it Another way:

NB=\sqrt{3^2+2^2}=\sqrt{13}

and

\tan\theta=\dfrac{3}{\sqrt{13}}

also gives

39.8°

. ✔

Final answer (a)

L(3,2,3)

(b)

\sqrt{22}\approx4.69\text{ m}

(c)

39.8°

Question 9●●● Exam-level✍️ + 🖩 · Paper 2[9 marks]

A traditional rondavel in Maun has a cylindrical wall of radius 3.5 m and height 2.4 m. The thatched roof is a cone with the same radius and vertical height 3.2 m. Ignore the overhang of the roof.

(a) Find the volume of air inside the rondavel.[3]
(b) Find the slant height of the roof.[2]
(c) Thatch costs P 85 per m². Find the cost of thatching the roof, to the nearest pula.[2]
(d) Find the angle the roof makes with the horizontal.[2]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkCylinder + cone

The air fills the cylinder (walls) and the cone (roof). Both have

r = 3.5

. Use the right height for each part: 2.4 for the cylinder, 3.2 for the cone.

Step 2✍️ By handAdd the two volumes

Cylinder:

\pi\times3.5^2\times2.4=92.36

Cone:

\dfrac{1}{3}\pi\times3.5^2\times3.2=41.05

Total

=133\text{ m}^3

(3 s.f.)A1

Part (b)

Step 1✍️ By handPythagoras in the roof

l=\sqrt{3.5^2+3.2^2}

l=4.74\text{ m}

(3 s.f.)A1

Part (c)

Step 1✍️ By handCurved area × price

A=\pi\times3.5\times4.7424=52.145\text{ m}^2

Cost

=52.145\times85=\text{P }4432

Step 2🖩 On the GDCCalculator — keep full accuracy

Store the slant height: type √(3.5^2+3.2^2) then ctrl var (sto→) and call it l. Then use l in the next line.

Part (d)

Step 1🧠 ThinkAngle at the edge of the roof

In the right-angled triangle of the roof: opposite

=3.2

(height), adjacent

=3.5

(radius). Opposite and adjacent → tan.

Step 2✍️ By handUse tan

\tan x=\dfrac{3.2}{3.5}

x=42.4°

Final answer (a)

133\text{ m}^3

(b)

4.74\text{ m}

(c)

\text{P }4432

(d)

42.4°

⚠️ Common mistake Using the total height 5.6 m for the cone. The cone is only the roof: its height is 3.2 m.

Question 10●●● Exam-level✍️ + 🖩 · Paper 2[9 marks]

Three drones hover at the points

P(1,\,2,\,0)

Q(7,\,4,\,3)

and

R(3,\,-2,\,5)

(units: metres).

(a) Show that $PQ = 7$ .[2]
(b) Find $PR$ and $QR$ .[2]
(c) Find the midpoint of $[QR]$ .[1]
(d) Find the angle $Q\hat{P}R$ .[3]
(e) Find the area of triangle $PQR$ .[1]

Show the full step-by-step solution

Part (a)

Step 1🧠 Think“Show that” means: show every line

We are given the answer, 7. We must show the working that leads to exactly 7.

Step 2✍️ By handDistance formula

PQ=\sqrt{(7-1)^2+(4-2)^2+(3-0)^2}

=\sqrt{36+4+9}=\sqrt{49}=7

A1 AG

Part (b)

Step 1✍️ By handThe other two sides

PR=\sqrt{2^2+(-4)^2+5^2}=\sqrt{45}=6.71

QR=\sqrt{(-4)^2+(-6)^2+2^2}=\sqrt{56}=7.48

Part (c)

Step 1✍️ By handAverage the coordinates

\left(\dfrac{7+3}{2},\dfrac{4+(-2)}{2},\dfrac{3+5}{2}\right)=(5,\ 1,\ 4)

Part (d)

Step 1🧠 ThinkThree sides known → cosine rule

The three drones make a flat triangle in space. We know all three sides, so the cosine rule (subtopic 3.2) gives any angle. The angle at

P

is opposite the side

QR

Step 2✍️ By handCosine rule for the angle at P

\cos Q\hat{P}R=\dfrac{PQ^2+PR^2-QR^2}{2\times PQ\times PR}=\dfrac{49+45-56}{2\times7\times\sqrt{45}}

M1A1

Q\hat{P}R=66.1°

(3 s.f.)A1

Step 3🖩 On the GDCCalculator (DEG)

Part (e)

Step 1✍️ By handArea = ½ab sin C

\dfrac{1}{2}\times7\times\sqrt{45}\times\sin66.13°=21.5\text{ m}^2

Final answer (a)

\sqrt{49}=7

(b)

PR\approx6.71

QR\approx7.48

(c)

(5,1,4)

(d)

66.1°

(e)

21.5\text{ m}^2

Question 11●●○ Standard✍️ + 🖩 · Paper 1[6 marks]

A hemispherical bowl has an inside diameter of 24 cm.

(a) Find the capacity of the bowl in litres.[3]
(b) Neo fills bowls of soup from a full 20-litre pot. Find how many bowls he can fill completely.[2]
(c) Find the curved surface area of the bowl.[1]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkRadius first, then cm³, then litres

Diameter 24 cm → radius 12 cm. A hemisphere is half a sphere. The answer comes out in cm³; divide by 1000 to get litres.

Step 2✍️ By handHalf of a sphere

V=\dfrac{1}{2}\times\dfrac{4}{3}\pi\times12^3=1152\pi

V=3619.1\text{ cm}^3

3619.1\div1000=3.62\text{ litres}

Part (b)

Step 1✍️ By handDivide and round DOWN

20\div3.6191=5.53

5 full bowlsA1

Part (c)

Step 1✍️ By handHalf of 4πr²

A=2\pi\times12^2=905\text{ cm}^2

(3 s.f.)A1

Step 2🖩 On the GDCCalculator

Final answer (a)

3.62

litres (b) 5 bowls (c)

905\text{ cm}^2

⚠️ Common mistake Rounding 5.53 up to 6. The 6th bowl would only be half full — the question asks for bowls filled completely.

🆘 Stuck on a question? Open the notes, GDC toolkit and common mistakes

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

🎯 What you must know

A point in 3D has three coordinates $(x, y, z)$ . The distance and midpoint formulas are the 2D ones with a $z$ part added.
Volume = space inside (units cm³, m³). Surface area = area of the outside skin (units cm², m²). 1 litre = 1000 cm³, 1 m³ = 1000 litres.
Pyramids and cones hold exactly one third of the prism or cylinder with the same base and height: $V=\dfrac{1}{3}\times\text{base area}\times h$ .
For a cone, $h$ is the vertical height, $l$ is the slant height. They are linked by Pythagoras: $l^2 = r^2 + h^2$ .
A hemisphere is half a sphere: volume $\dfrac{2}{3}\pi r^3$ , curved surface $2\pi r^2$ .
To find an angle inside a solid, find a right-angled triangle that contains it, work out the sides you need with Pythagoras, then use sin, cos or tan.
The angle between a line and a plane (e.g. the floor) is the angle between the line and its shadow on that plane (the line straight below it).
Calculator in DEGREE mode for every angle in this topic (the screen shows DEG at the top).

Distance between two points

d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}

📘 In the formula booklet

Midpoint

\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2},\ \dfrac{z_1+z_2}{2}\right)

📘 In the formula booklet

Volume of a right pyramid

V=\dfrac{1}{3}Ah

📘 In the formula booklet

Volume of a right cone

V=\dfrac{1}{3}\pi r^2h

📘 In the formula booklet

Curved surface area of a cone

A=\pi rl

📘 In the formula booklet

Volume of a sphere

V=\dfrac{4}{3}\pi r^3

📘 In the formula booklet

Surface area of a sphere

A=4\pi r^2

📘 In the formula booklet

Cylinder (prior learning)

V=\pi r^2h,\quad \text{curved } A=2\pi rh

📘 In the formula booklet

🖩 Your GDC toolkit for this subtopic

Follow the steps on your own TI-Nspire CX II. The screens show what you should see.

Set DEGREE mode (do this once, before the exam starts): doc → Settings → Document Settings → Angle: Degree

Press doc.
Choose Settings, then Document Settings.
Press tab until Angle is highlighted, press click and choose Degree.
Press tab to OK and press enter. The top of the screen now shows DEG.

Set Angle to Degree

Get a decimal instead of an exact answer like √70 or

\dfrac{5324\pi}{3}

ctrl+enter💻 Ctrl+Enter

Press ctrl then enter to get the answer as a decimal.

√70 ≈ 8.3666

Worked example: Question 1 — Check on the calculator

Without ctrl+enter the calculator shows the exact answer

\sqrt{70}

. ctrl+enter gives the decimal.

Square root and π

ctrl💻 Ctrl›x²💻 type ^2, then →›π💻 type pi

Press ctrl.
Press x².
Press π.

Inverse trig (sin⁻¹, cos⁻¹, tan⁻¹): open the trig palette and pick it

trig💻 type sin( cos( tan(›tan⁻¹

Press trig.
Choose tan⁻¹.

DEG mode: θ = 22.98°

Worked example: Question 4 — Calculator in DEG mode

Check that DEG is shown at the top of the screen. If it says RAD, the answer will be 0.401 — wrong.

Solve an equation for an unknown length

menu💻 click menu on keypad›3: Algebra›1: Numerical Solve

Start on a Calculator page (ctrl doc adds one).
Press menu.
Choose 3: Algebra.
Choose 1: Numerical Solve.

The menu path

Worked example: Question 7 — Or let the calculator solve it

nSolve finds the value of

r

that makes the equation true.

⚠️ Where students lose marks

Using the slant height in the cone volume formula. Volume needs the vertical height $h$ ; curved surface needs the slant height $l$ .
Forgetting the flat faces: the total surface area of a cone is $\pi rl + \pi r^2$ ; of a hemisphere is $2\pi r^2+\pi r^2$ . Read whether the question wants curved or total.
Using the diameter as the radius. Halve it first!
Calculator in radian mode: $\tan^{-1}(0.4)$ gives 0.381 instead of 21.8°. Always check for DEG at the top of the screen.
Rounding a length early, then using it for the angle. Keep the full value in the calculator (use ans or store it).

IB DP Mathematics: Applications and Interpretation (AI) — SL and HL AI 3.1 — 3D geometry: distance, midpoint, volume, surface area, angles

IB DP Mathematics: Applications and Interpretation (AI) — SL and HL

3.1 3D geometry: distance, midpoint, volume, surface area, angles

✏️ Practice questions

🎯 What you must know

🖩 Your GDC toolkit for this subtopic

⚠️ Where students lose marks