Year 11 General Shape And Measurement

Problem-Solving And Modelling

5 practice questions 2 video lessons Theory + worked examples

Theory

This page is about combining measurement techniques — perimeter, area, volume, surface area, Pythagoras, similarity and scale — and applying them to real-world modelling problems. The challenge is choosing the right tools and using them in the right order.

Problem-solving and modelling doesn't introduce new formulas. Instead, it tests your ability to combine the measurement techniques from this chapter (perimeter, area, volume, surface area, Pythagoras, similarity, scale) and apply them to real-world situations.

The challenge is choosing the right tools and using them in the right order. Many problems also require unit conversion (cm to m, m $^{3}$ to L) and a final round-up step when the answer counts whole items needed.

A composite shape is one that doesn't match a standard formula. There are two strategies for handling them: split-and-add (divide into recognisable pieces and add their measurements) or enclose-and-subtract (find the bounding shape's measurement and subtract the cut-out pieces). Choose whichever gives fewer, simpler pieces.

The four-step workflow for any modelling problem.

Two strategies for composite shapes: split-and-add or enclose-and-subtract.

Common problem types and which quantity to compute:

Problem type	Quantity to find
Fencing, edging	Perimeter $\times$ cost per m
Painting, paving, turf	Area $\times$ cost per m $^{2}$
Concrete, soil, water fill	Volume $\times$ cost per m $^{3}$
Filling/draining tanks	Capacity $\div$ flow rate
Sloping or roof problems	Pythagoras to find slant length
Scaling tank/statue costs	$k^{2}$ (area) or $k^{3}$ (volume)

Unit conversions (always match units before substituting):

Quantity	Conversions
Length	$1$ m $= 100$ cm $= 1000$ mm; $1$ km $= 1000$ m
Area	$1$ m $^{2} = 10 000$ cm $^{2}$ ; $1$ ha $= 10 000$ m $^{2}$
Volume / capacity	$1$ m $^{3} = 1000$ L $= 1$ kL; $1$ L $= 1000$ cm $^{3}$

Pythagoras for slants. For a right triangle with legs $a$ and $b$ and hypotenuse $c$ : $c^{2} = a^{2} + b^{2}$ . Use this whenever a roof slope, ramp, or slant face is needed but only the horizontal and vertical pieces are given.

The four-step workflow

Read carefully. Sketch the situation. Identify the shape (rectangle, prism, cylinder, composite\ldots), then list the given measurements and what's being asked.
Set up. Choose the formulas you need. For composite shapes, decide whether to split-and-add or enclose-and-subtract.
Solve. Compute step by step. Show working clearly so units and intermediate values are visible.
Check the answer. Is the unit correct? Is the size sensible? Have you rounded as the question asks (decimal places, or up to the next whole item)?

Sense check. A typical pool holds tens of thousands of litres; a typical room area is tens of square metres; a typical garden plot perimeter is tens of metres. If your answer is off by a factor of $100$ or $1000$ , check your units.

Example 1 — Concrete Cost

A rectangular slab is

8

m long,

5

m wide, and

0.15

m thick. Concrete costs

$ 220

per m

^{3}

. Find the total cost.

Solution

Find the volume of the slab, then multiply by the cost rate.

$V$	$=$	$8 \times 5 \times 0.15$
$V$	$=$	$6 m^{3}$
cost	$=$	$6 \times 220$
cost	$=$	$$ 1 320$

cost = 1320

Example 2 — Filling a Tank

A cylindrical tank has radius

1.5

m and height

2

m. Water flows in at

60

L per minute. Find the time to fill the tank, in minutes (to

2

dp).

Solution

Find the volume in m $^{3}$ , convert to litres, then divide by the flow rate.

$V$	$=$	$π r^{2} h$
$V$	$=$	$π \times {1.5}^{2} \times 2$
$V$	$=$	$4.5 π$
$V$	$\approx$	$14.137 m^{3}$
capacity	$\approx$	$14 137 L$
time	$=$	$\frac{14 137}{60}$
time	$\approx$	$235.62 min$

time \approx 235.62

Example 3 — Roof with Pythagoras

A roof has cross-section a triangle with base

8

m and peak

3

m above the centre. The house is

12

m long. Find the total area of the two sloping sides.

Solution

Find the slant of one half-roof by Pythagoras, then compute each sloping side.

$s$	$=$	$\sqrt{4^{2} + 3^{2}}$
$s$	$=$	$\sqrt{25} = 5 m$
one side	$=$	$5 \times 12 = 60 m^{2}$
total	$=$	$2 \times 60$
total	$=$	$120 m^{2}$

total = 120

Example 4 — Scaling Up a Tank

A small fuel can holds

8

L. A larger similar can has a linear scale ratio of

2 : 5

compared to the small one. Find the capacity of the larger.

Solution

Capacity scales with volume — cube the linear ratio.

volume ratio	$=$	$2^{3} : 5^{3} = 8 : 125$
$\frac{8}{V_{large}}$	$=$	$\frac{8}{125}$
$V_{large}$	$=$	$125 L$

V_{large} = 125 L

Common pitfalls

Not rounding up for whole items. If the question asks "how many bags" or "how many litres of paint", round up to ensure enough material. Even

7.1

bags means buying

8

bags.

Mixed units in volume conversions. Always convert to the same unit system before substituting.

1

^{3} = 1000

L, not

100

L. Getting this wrong gives an answer off by a factor of

10

Wrong scaling power. Costs that scale with surface area use

k^{2}

; costs that scale with volume use

k^{3}

. Picking the wrong power gives an answer off by a whole order of magnitude.

Forgetting Pythagoras for slants. Roof areas, ramp lengths, and cone surface areas often need a slant length found from the horizontal and vertical pieces. If only base and height are given, suspect Pythagoras.

Frequently asked questions

What is the four-step problem-solving workflow?

Read carefully and sketch the situation. Set up by choosing the right formulas and deciding whether to split and add or enclose and subtract. Solve step by step with clear working. Check the answer: units correct, size sensible, rounded as requested.

When do I use perimeter, area, volume or surface area?

Use perimeter for fencing and edging. Use area for painting, paving and turfing. Use volume for concrete, soil, water and other fills. Use surface area for wrapping, plating and coatings. Capacity (litres) and volume (cubic metres) are the same quantity in different units.

How do I know when to round up versus round to a decimal place?

If the answer is a count of whole items needed (bags of soil, litres of paint, tiles), always round up because you need enough material. If the answer is a measurement (length, area, time), round to the decimal place requested in the question.

What is the difference between split-and-add and enclose-and-subtract for composite shapes?

Split-and-add divides the shape into recognisable pieces (rectangles, triangles, semicircles) and adds their areas or volumes. Enclose-and-subtract draws a bounding rectangle around the shape and subtracts the cut-out pieces. Pick whichever splits the problem into the fewest, easiest pieces.

How do I convert between cubic metres and litres?

1 cubic metre equals 1000 litres equals 1 kilolitre. Also, 1 litre equals 1000 cubic centimetres. So a tank with a volume of 0.5 m cubed holds 500 litres.

How do I work out how long a tank takes to fill?

Find the tank's capacity in the same unit as the flow rate (usually litres). Then divide capacity by flow rate. For example, a 600 L tank filling at 50 L per minute takes 600 divided by 50 which is 12 minutes.