IB Maths AI HL 3.1 Three-dimensional geometry and vector reasoning Question Bank

Practise IB Maths AI HL 3.1 by analysing three-dimensional coordinates, vectors and measurements in applied geometry.

[Maximum number: 7]

Kailash manufactures drink containers in the shape of a cuboid. The container has a square top and a square base of length, $l\,\mathrm{cm}$ . Its height, $d\,\mathrm{cm}$ , is three times the length of the base.

diagram not to scale

(a)

Calculate the total external surface area of the container.

[ 3 ]

Finding the area of a square and the area of a non-square rectangle: $5\times5$ and $5\times15$ .
$(5\times5+5\times15+5\times15)\times2$ $=350\,\mathrm{cm^2}$
Note: Units are required.

(b)

To reduce environmental impact, Kailash is trying to minimize the amount of material needed for the production of the $375\,\mathrm{cm^3}$ container.
He is willing to change the shape to a cylinder with radius $r\,\mathrm{cm}$ , and height $h\,\mathrm{cm}$ , as shown below.

Figure for Question (b) — IB Maths AI HL

The cylindrical container of drink must also hold $375\,\mathrm{cm^3}$ .
Find an expression for the height, h, of the container in terms of r.

[ 2 ]

$\pi r^2h=375$ $h=\frac{375}{\pi r^2}$
Note: Award at most A0A1 for a final answer $h=\frac{V}{\pi r^2}$ .

(c)

Let the total external surface area be $A\,\mathrm{cm^2}$ .
Show that $A=2\pi r^2+\frac{750}{r}$ .

[ 2 ]

Attempt to substitute their h into the curved surface area $2\pi rh$ .
$A=2\pi r^2+2\pi r\left(\frac{375}{\pi r^2}\right)$ $A=2\pi r^2+\frac{750}{r}$

All question bank results loaded

SL 3.1—Three-dimensional geometry question 1

Question (a)

Question (b)

Question (c)