IB Maths AA SL 3.1 Three Dimensional Geometry Questions

Practise finding 3D distances, midpoints, lengths, angles, volumes and surface areas from coordinates or solid diagrams using geometry and correct units.

Exam points

Use three-dimensional coordinates to calculate distances and midpoints between points, showing the coordinate differences and resulting exact or numerical value.
Find lengths, angles, volumes or surface areas in three-dimensional solids by identifying right-angled triangles and applying the appropriate geometric formula.

[Maximum number: 5]

All lengths in this question are in centimetres.
A solid metal ornament is in the shape of a right pyramid, with vertex V and square base ABCD . The centre of the base is X . Point V has coordinates (1,5,0) and point A has coordinates (-1,1,6).

Figure for Question IB Maths AA SL 3.1 Three Dimensional Geometry Questions question 1 — IB Maths AA SL

(a)

Find AV .

[ 2 ]

attempt to use the distance formula to find AV

\begin{aligned} & \sqrt{(1-(-1))^{2}+(5-1)^{2}+(0-6)^{2}} \\ & =7.48331 \ldots \\ & =7.48(\mathrm{~cm})(=\sqrt{56} \text { or } 2 \sqrt{14}) \end{aligned}

(b)

Find the height of the pyramid, VX.

A second ornament is in the shape of a cuboid with a rectangular base of length $2 x \mathrm{~cm}$ , width $x \mathrm{~cm}$ and height $y \mathrm{~cm}$ . The cuboid has the same volume as the pyramid.

[ 3 ]

METHOD 1 equating volume of pyramid formula to 57.2
$\frac{1}{3} \times 5.11 \ldots{ }^{2} \times h=57.2$
(A1)
$h=6.54886 \ldots$ $h=6.55(\mathrm{~cm})$
METHOD 2
Let M be the midpoint of AB ]
$A V^{2}=A M^{2}+M X^{2}+X V^{2}$ $\Rightarrow X V=\sqrt{7.48 \ldots^{2}-\left(\frac{5.11 \ldots}{2}\right)^{2}-\left(\frac{5.11 \ldots}{2}\right)^{2}}$ $h=6.54886 \ldots$ $h=6.55(\mathrm{~cm})$

All question bank results loaded

IB Maths AA SL 3.1 Three Dimensional Geometry Questions question 1

Question (a)

Question (b)