Module 22 of 46Lesson 1 of 9 · AA 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

Points in space, solids and the angles hidden inside them. On Paper 1 you must give exact answers (surds and multiples of π); on Paper 2 the GDC does the arithmetic, but you still need the right triangle and the right formula.

✏️ 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✍️ By hand · Paper 1 · no calculator[6 marks]

The points

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

and

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

are given.

(a) Find the exact distance $AB$ .[2]
(b) Find the coordinates of $M$ , the midpoint of $[AB]$ .[2]
(c) The point $C(k,\,1,\,2)$ is such that $AC = 3$ . Find the possible values of $k$ .[2]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkDistance = Pythagoras in 3D

Find the change in

x

, in

y

and in

z

. Square each one, add, 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{25+36+9}

AB=\sqrt{70}

Part (b)

Step 1✍️ By handMidpoint = average of each coordinate

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

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

Part (c)

Step 1🧠 ThinkAn unknown inside the formula

Write the distance

AC

with

k

in it and set it equal to 3. Square both sides to remove the root. A square equal to a number gives two answers (+ and −).

Step 2✍️ By handSet up and solve

(k-2)^2+(1-(-1))^2+(2-4)^2=9

(k-2)^2+8=9 \Rightarrow (k-2)^2=1

k-2=\pm1

k=1

k=3

✅ Check it

k=3

AC=\sqrt{1+4+4}=\sqrt{9}=3

✔

Final answer (a)

\sqrt{70}

(b)

M\left(-\dfrac{1}{2},\,2,\,\dfrac{5}{2}\right)

(c)

k=1

k=3

⚠️ Common mistake Writing

-1-5=4

. Put brackets round every coordinate before you subtract.

Question 2●○○ Warm-up✍️ By hand · Paper 1 · no calculator[5 marks]

A sphere has radius

3\text{ cm}

(a) Show that its volume and its surface area have the same numerical value, and state this value in terms of $\pi$ .[3]
(b) A second sphere has volume $288\pi\text{ cm}^3$ . Find its exact surface area.[2]

Show the full step-by-step solution

Step 1🧠 ThinkExact means “keep π”

On Paper 1 there is no calculator, so we leave

\pi

as a letter.

3^3 = 27

and

3^2 = 9

are easy by hand.

Part (a)

Step 1✍️ By hand

V=\dfrac{4}{3}\pi\times3^3=\dfrac{4}{3}\pi\times27=36\pi

M1 A1

A=4\pi\times3^2=36\pi

— the same number,

36\pi

A1 AG

Part (b)

Step 1🧠 ThinkWork the volume formula backwards

Set

\dfrac{4}{3}\pi r^3=288\pi

. Divide both sides by

\pi

, then multiply by

\dfrac{3}{4}

. Then use the radius in

4\pi r^2

Step 2✍️ By hand

r^3=288\times\dfrac{3}{4}=216 \Rightarrow r=6

A=4\pi\times36=144\pi\text{ cm}^2

Final answer (a) both equal

36\pi

(b)

144\pi\text{ cm}^2

⚠️ Common mistake Writing

r^3=288

and forgetting the

\dfrac{4}{3}

. Undo every part of the formula.

Question 3●○○ Warm-up✍️ By hand · Paper 1 · no calculator[6 marks]

A party hat is a right cone with base radius 6 cm and vertical height 8 cm.

(a) Find the slant height $l$ .[2]
(b) Find the exact volume of the cone.[2]
(c) Find the exact curved surface area of the cone.[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 (8) going up, the radius (6) along the bottom, and the slant height

l

as the hypotenuse.

Step 2✍️ By hand

l^2=6^2+8^2=36+64=100

l=10\text{ cm}

Part (b)

Step 1✍️ By handVolume uses the vertical height 8

V=\dfrac{1}{3}\pi\times6^2\times8=\dfrac{288\pi}{3}

V=96\pi\text{ cm}^3

Part (c)

Step 1✍️ By handCurved area uses the slant height 10

A=\pi\times6\times10

A=60\pi\text{ cm}^2

💡 Remember 6-8-10 is the 3-4-5 triangle doubled. Paper 1 questions often use these “nice” triangles: 3-4-5, 5-12-13, 8-15-17.

Final answer (a)

10\text{ cm}

(b)

96\pi\text{ cm}^3

(c)

60\pi\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✍️ By hand · Paper 1 · no calculator[6 marks]

The diagram shows a cuboid

ABCDEFGH

with

AB = 6\text{ cm}

BC = 3\text{ cm}

and

CG = 2\text{ cm}

(a) Find $AC$ , giving your answer in the form $a\sqrt{5}$ .[2]
(b) Show that $AG = 7\text{ cm}$ .[2]
(c) Find the exact value of $\sin\theta$ , where $\theta$ is the angle 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

. To simplify a surd, look for a square number factor:

45 = 9\times5

Step 2✍️ By hand

AC^2=6^2+3^2=45

AC=\sqrt{45}=\sqrt{9}\sqrt{5}=3\sqrt{5}\text{ cm}

Part (b)

Step 1🧠 ThinkNow stand up: triangle ACG

G

is straight above

C

, so angle

ACG = 90°

. Use Pythagoras again with

AC^2 = 45

(no need to use the surd).

Step 2✍️ By hand

AG^2=AC^2+CG^2=45+4=49

AG=\sqrt{49}=7\text{ cm}

A1 AG

Part (c)

Step 1🧠 ThinkWhich angle is it?

The shadow of

AG

on the floor is

AC

, so

\theta

is angle

GAC

. From

A

: opposite

= CG = 2

, hypotenuse

= AG = 7

. Opposite and hypotenuse → sin.

Step 2✍️ By hand

\sin\theta=\dfrac{CG}{AG}

\sin\theta=\dfrac{2}{7}

✅ Check it At home you can check on the GDC (DEG):

\sin^{-1}\left(\dfrac{2}{7}\right)\approx16.6°

and

\tan^{-1}\left(\dfrac{2}{3\sqrt{5}}\right)\approx16.6°

— the same angle.

Final answer (a)

3\sqrt{5}\text{ cm}

(b)

\sqrt{49}=7

(c)

\sin\theta=\dfrac{2}{7}

⚠️ Common mistake Using the angle at

G

. The angle with the base is at the point where the line meets the base.

Question 5●●○ Standard✍️ + 🖩 · Paper 2[9 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) Find the total surface area of the pyramid.[2]

Show the full step-by-step solution

Part (a)

Step 1✍️ By hand

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 the diagonal

AC=\sqrt{6^2+6^2}

. Triangle

VOC

is right-angled at

O

Step 2✍️ By hand

OC=\dfrac{\sqrt{72}}{2}=\sqrt{18}

VC^2=8^2+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

: opposite

VO = 8

, adjacent

OC=\sqrt{18}

Step 2🖩 On the GDCCalculator (DEG mode)

trig💻 type sin( cos( tan(›tan⁻¹›8/√(18)›enter💻 Enter

📍 Where to find it on the TI-Nspire

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

Step 3✍️ By hand

\tan\alpha=\dfrac{8}{\sqrt{18}} \Rightarrow \alpha=62.1°

M1 A1

Part (d)

Step 1🧠 ThinkHeight of a triangular face

Each face is a triangle with base 6. Its height goes from

V

to the middle of a base edge: that point is 3 m from

O

, so the face height is

\sqrt{8^2+3^2}=\sqrt{73}

Step 2✍️ By hand

A=6^2+4\times\dfrac{1}{2}\times6\times\sqrt{73}

A=36+12\sqrt{73}=139\text{ m}^2

(3 s.f.)A1

Step 3🖩 On the GDCCheck

Final answer (a)

96\text{ m}^3

(b)

9.06\text{ m}

(c)

62.1°

(d)

139\text{ m}^2

⚠️ Common mistake Using the edge

VC

as the height of a face. The face height goes to the middle of the edge, not the corner.

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, then add.

Part (a)

Step 1✍️ By hand

Cylinder:

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

Hemisphere:

\dfrac{2}{3}\pi\times3^3=18\pi

Total:

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

(3 s.f.)A1

Part (b)

Step 1🧠 ThinkOnly the outside skin

Paint goes on the curved wall and the curved dome. No paint on the base, and none on the hidden circle where the dome sits.

Step 2✍️ By hand

Wall:

2\pi\times3\times10=60\pi

Dome:

2\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

Press ctrl+enter to get decimals.

Final answer (a)

108\pi\approx339\text{ m}^3

(b)

78\pi\approx245\text{ m}^2

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

Question 7●●○ Standard✍️ + 🖩 · Paper 2[5 marks]

Volume questions can also be asked backwards.

(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

Step 2✍️ By hand

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

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

r=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 hand

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

h=\dfrac{1500}{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. The answer is in cm, so the volume must be in cm³.

Question 8●●● Exam-level✍️ By hand · Paper 1 · no calculator[8 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 $L$ and of $N$ .[2]
(b) Find the exact length of the wire $LB$ .[2]
(c) Let $\theta$ be the angle between the wire and the floor. Find the exact values of $\sin\theta$ and $\cos\theta$ .[4]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkCentre of the ceiling

Halfway along the length (

6\div2=3

) and halfway across the width (

4\div2=2

). The ceiling is 3 m up, so

z=3

N

is straight below, on the floor, so

z=0

Step 2✍️ By hand

L(3,\ 2,\ 3)

N(3,\ 2,\ 0)

A1 A1

Part (b)

Step 1✍️ By hand

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

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

Part (c)

Step 1🧠 ThinkUse the triangle LNB

LN=3

is vertical, so angle

LNB=90°

NB

is the shadow of the wire, so

\theta

is angle

LBN

. We need

NB

too, for the cosine.

Step 2✍️ By hand

NB=\sqrt{(6-3)^2+(4-2)^2}=\sqrt{13}

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

M1 A1

\cos\theta=\dfrac{NB}{LB}=\dfrac{\sqrt{13}}{\sqrt{22}}

✅ Check it

\sin^2\theta+\cos^2\theta=\dfrac{9}{22}+\dfrac{13}{22}=1

✔. At home on the GDC:

\sin^{-1}\left(\dfrac{3}{\sqrt{22}}\right)\approx39.8°

Final answer (a)

L(3,2,3)

N(3,2,0)

(b)

\sqrt{22}\text{ m}

(c)

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

\cos\theta=\dfrac{\sqrt{13}}{\sqrt{22}}

⚠️ Common mistake Using

LO

or another line as the shadow. The shadow of

LB

on the floor runs from

N

(below

L

) to

B

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

Both parts have

r = 3.5

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

Step 2✍️ By hand

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 hand

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

l=4.74\text{ m}

(3 s.f.)A1

Part (c)

Step 1✍️ By hand

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

Cost

=52.145\times85=\text{P }4432

Step 2🖩 On the GDCCalculator — keep full accuracy

Store the slant height as l (type l:= before it), then use l in the next line.

Part (d)

Step 1✍️ By handOpposite 3.2, adjacent 3.5 → 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✍️ By hand

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 hand

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 hand

\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. All three sides are known, so the cosine rule (subtopic 3.2) gives the angle at

P

, which is opposite

QR

Step 2✍️ By hand

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

M1 A1

Q\hat{P}R=66.1°

(3 s.f.)A1

Step 3🖩 On the GDCCalculator (DEG)

Part (e)

Step 1✍️ By hand

\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●●● Exam-level✍️ By hand · Paper 1 · no calculator[8 marks]

A toy is made from a cone of radius

r

cm and vertical height

2r

cm, fixed on top of a hemisphere of radius

r

cm (flat faces together).

(a) Show that the volume of the toy is $\dfrac{4}{3}\pi r^3$ .[3]
(b) The volume of the toy is $288\pi\text{ cm}^3$ . Find $r$ .[2]
(c) Find the exact total surface area of the toy, in the form $k\pi(2+\sqrt{5})$ .[3]

Show the full step-by-step solution

Step 1🧠 ThinkAlgebra instead of numbers

The method is the same as with numbers — just keep

r

as a letter. Volume = cone + hemisphere.

Part (a)

Step 1✍️ By hand

Cone:

\dfrac{1}{3}\pi r^2(2r)=\dfrac{2}{3}\pi r^3

Hemisphere:

\dfrac{2}{3}\pi r^3

Total:

\dfrac{2}{3}\pi r^3+\dfrac{2}{3}\pi r^3=\dfrac{4}{3}\pi r^3

A1 AG

Part (b)

Step 1✍️ By hand

\dfrac{4}{3}\pi r^3=288\pi \Rightarrow r^3=216

r=6

Part (c)

Step 1🧠 ThinkOutside skin = curved cone + curved hemisphere

The flat circles are glued together, so they are hidden. The cone has

r=6

and

h=12

; its slant height comes from Pythagoras.

Step 2✍️ By hand

l=\sqrt{6^2+12^2}=\sqrt{180}=6\sqrt{5}

Cone:

\pi\times6\times6\sqrt{5}=36\sqrt{5}\,\pi

; hemisphere:

2\pi\times6^2=72\pi

Total:

72\pi+36\sqrt{5}\,\pi=36\pi(2+\sqrt{5})\text{ cm}^2

Final answer (a) shown (b)

r=6

(c)

36\pi(2+\sqrt{5})\text{ cm}^2

⚠️ Common mistake Including the two flat circles in the surface area. Where solids are joined, those faces are inside.

🆘 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 (cm³, m³). Surface area = area of the outside skin (cm², m²). 1 litre = 1000 cm³.
Pyramids and cones: $V=\dfrac{1}{3}\times\text{base area}\times h$ . Cone: $h$ = vertical height, $l$ = slant height, $l^2 = r^2 + h^2$ .
Hemisphere: volume $\dfrac{2}{3}\pi r^3$ , curved surface $2\pi r^2$ .
Exact answers on Paper 1: leave $\sqrt{70}$ , $48\pi$ , $\dfrac{2}{7}$ as they are. Simplify surds: $\sqrt{45}=3\sqrt{5}$ .
To find an angle inside a solid, find a right-angled triangle that contains it, then use sin, cos or tan.
The angle between a line and a plane is the angle between the line and its shadow (projection) on the plane.

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

Sphere

V=\dfrac{4}{3}\pi r^3,\quad 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: 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 108π

ctrl+enter💻 Ctrl+Enter

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

Worked example: Question 5 — Calculator (DEG mode)

Inverse trig (sin⁻¹, cos⁻¹, tan⁻¹) from the trig palette

trig💻 type sin( cos( tan(›tan⁻¹

Press trig.
Choose tan⁻¹.

Worked example: Question 5 — Calculator (DEG mode)

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 $h$ ; curved surface needs $l$ .
Forgetting flat faces: total surface of a cone is $\pi rl + \pi r^2$ . Read whether the question wants curved or total.
Paper 1: writing 8.37 instead of $\sqrt{70}$ when the question says “exact”. Decimals lose the A mark.
Calculator in radian mode on Paper 2: $\tan^{-1}(0.4)$ gives 0.381 instead of 21.8°. Check for DEG.
Rounding a length early, then using it for the angle. Keep full accuracy (store values).

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL AA 3.1 — 3D geometry: distance, midpoint, volume, surface area, angles

IB DP Mathematics: Analysis and Approaches (AA) — 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