018i - Problem Solving with Area

Lesson 18 — Problem Solving and Consolidation: Area

Strand: Measurement | Descriptor: AC9M7M01 | Duration: 45 minutes

Learning Intentions

• To solve practical problems involving the areas of triangles and parallelograms.
• To work with unit conversions and costing in area contexts.

Success Criteria

I can:

1. Convert between , and correctly.
2. Solve multi-step area problems set in real contexts.
3. Calculate costs and quantities from an area, rounding sensibly for the context.
4. Justify whether an answer should be rounded up or down.

Warmup

(6 minutes — unit conversion, explicit instruction plus practice)

The key idea, stated and shown: because area is two-dimensional, the conversion factor is squared.

Demonstrate with a drawn square divided into a grid of millimetre squares.

Practice:

(Answers: ; ; ; .)

Activities

Activity 1 — Costing and Quantity Problems (14 min)

Pairs. Every answer requires a sentence, correct units, and a rounding decision.

Problem 1 — Painting. A triangular gable wall has base m and height m. Paint covers per litre and is sold only in L tins. Two coats are required. How many tins must be bought?

Problem 2 — Turf. A parallelogram-shaped lawn has base m and perpendicular height m. Turf costs 14$ per square metre. Find the cost.

Problem 3 — Tiling. A triangular patio has base m and height m. Tiles are cm by cm. How many tiles are needed, ignoring cuts and waste?

Problem 4 — Comparison. A rectangular banner is m by m. A triangular banner has base m and height m. Which has the greater area, and by how much?

Socratic scaffolding for Problem 1:

Prompt	Purpose
Understand: what is the final unknown?	A number of tins — a whole number.
What must be found first?	The wall area.
Find it.	.
Is that all the area to be painted?	No — two coats, so of coverage is needed.
How much paint does that need?	L.
So how many tins?	tins exactly.
Looking back — what if it had been ?	L, so 4 tins — you cannot buy part of a tin. Always round up for materials.

Answers: 1. tins. 2. , cost 277210\ \text{m}^20.25\ \text{m}^210 \div 0.25 = 403.6\ \text{m}^23.6\ \text{m}^2$ — they are equal.

Activity 2 — Working backwards from Area (10 min)

Explicit instruction, then practice.

I do. A triangular sail must have an area of . The mast (the height) is m. What must the base measure?

You do:

1. A parallelogram has area and height cm. Find the base.
2. A triangle has area and base m. Find the height.
3. A triangular flag has area and base m. Find its height.
4. A parallelogram has area and base cm. Find its height in centimetres.

(Answers: cm; m; m; , so cm.)

Activity 3 — Investigation: Maximum Area (8 min)

Pairs, grid paper.

A farmer has m of fencing to make a triangular pen against a long straight wall. The wall forms the base of the triangle (no fencing needed there), and the m is used for the other two sides.

Investigate which triangle gives the greatest area. Try several and record base, height and area in a table.

Socratic scaffolding:

Prompt	Purpose
Understand: what is fixed, and what varies?	The two slanted sides total m; the base and height both vary.
Try a simple case first.	Two sides of m each, meeting the wall symmetrically.
How would you record your results?	A table: base, height, area — so patterns become visible.
Try extremes.	Very flat (large base, tiny height) and very tall (small base). Both give small areas.
What do you notice?	The area peaks somewhere in the middle.
Where is the maximum?	When the two sides are perpendicular to each other — a right angle at the apex.
Carry it out	Legs of m each at right angles give .
Looking back	Is this surprising? Compare with a – split, or an – split.

Discussion: This is a genuine optimisation problem. Year 7 students are not expected to prove the result, but exploring it builds strong intuition about the trade-off between base and height.

Checks for Understanding

(5 minutes — exit ticket, collected)

1. Convert: (a) to (b) to .
2. A triangular garden has base m and height m. Mulch costs 9$ per square metre. Find the cost.
3. A parallelogram has area and base cm. Find its height.
4. A wall of area needs painting. One tin covers . How many tins must be bought?
5. Reasoning. Explain why is and not .

Answers: 1. (a) (b) ; 2. 9 = $270430 \div 12 = 2.53100100100 \times 100 = 10,000$ centimetre squares — the conversion factor is squared because area is two-dimensional.

Common Misconceptions

Misconception	How to pre-empt it
(using the linear factor).	Draw the grid. Reinforce that the factor is squared.
Rounding down when buying materials.	Ask “can you buy of a tin?” for every materials problem.
Rounding up when the question asks how many complete items fit.	Contrast the two directly: buying materials rounds up; fitting whole objects into a space rounds down.
Forgetting to double for two coats of paint.	Underline the words “two coats” in the question before starting.
Mixing units, e.g. base in metres and height in centimetres.	Require a conversion line as the first step of any mixed-unit problem.
Assuming a longer perimeter gives a larger area.	Activity 3 confronts this directly — equal fencing, very different areas.

Enrichment — Competition-Style Problems

E1 (AMC Junior style). A rectangular room is m by m. Tiles are cm by cm. How many whole tiles are needed to cover the floor?

Answer

Along the m side: tiles. Along the m side: tiles. Total tiles. (Check by area: ✓)

E2 (Kangaroo style). Two triangles have the same area. The first has base cm and height cm. The second has height cm. What is its base?

Answer

E3 (Challenge). A square has area . A triangle drawn inside it has the full side of the square as its base, and its apex on the opposite side. What is the triangle’s area, and what fraction of the square is it?

Answer

The square’s side is cm, so the triangle has base cm and height cm:

That is exactly half the square.

E4 (Challenge). A parallelogram-shaped field has base m and height m. Grass seed is sold in bags covering each, at 32$ per bag. Find the total cost.

Answer

Homework

1. Convert: (a) to (b) to (c) to (d) to .
2. A triangular window has base m and height m. Glass costs 85$ per square metre. Find the cost.
3. A parallelogram-shaped deck has base m and height m. Decking oil covers per litre and is sold in L tins. How many tins are needed for one coat?
4. A triangle has area and base cm. Find its height.
5. A rectangular lawn m by m contains a triangular flowerbed with base m and height m. Find the grassed area. Turf costs 11$ per square metre — find the cost of turfing the grassed part.
6. A triangular sign must have area . Its height is m. Find its base.
7. Reasoning. A student converts to . Explain the error and give the correct answer.
8. Challenge. A parallelogram and a triangle have the same base of cm. The parallelogram’s height is cm. What height must the triangle have for the two areas to be equal?

Answers: Q1 — (a) (b) (c) (d) . Q2 — 85 = $81.6030\ \text{m}^2327330 - 27 = 303\ \text{m}^2$33331.65\ \text{m}^2 = 50,000\ \text{cm}^212$ cm (double the parallelogram’s height).