Mathematics • Year 8 • Unit 3 • Lesson 5
Area — Mixed Challenge
Mix rectangles, squares, triangles, working backwards, and composite shapes. Six mixed problems, one "find the mistake", and one open-ended design challenge that connects perimeter and area.
1. Mixed problems — choose the right formula
Each question uses a different idea from Lesson 5. Decide whether it's a rectangle, triangle or composite BEFORE writing. Show your working. 3 marks each
1.1 A rectangle is 13 cm × 11 cm. Find A.
1.2 A triangle has base 18 m and perpendicular height 7 m. Find A.
1.3 A rectangle has A = 96 m² and width 6 m. Find the length.
1.4 A triangle has A = 48 cm² and perpendicular height 8 cm. Find the base.
1.5 An L-shape is a 7 cm × 6 cm rectangle with a 2 cm × 3 cm corner removed. Find the area.
1.6 A composite shape is a 10 cm × 4 cm rectangle with a triangular roof on top: the triangle has base 10 cm and perpendicular height 3 cm. Sketch the "house" shape and find its total area.
2. Find the mistake
A student is asked to find the area of an L-shaped floor: total dimensions 7 m × 6 m with a 2 m × 3 m corner removed. Their working is below. Exactly one line contains a mistake — spot it, explain why, and re-do correctly. 3 marks
Student's working — find the area of the L-shape:
Line 1: Total rectangle area = 7 × 6 = 42 m²
Line 2: Cut-out area = 2 × 3 = 6 m²
Line 3: L-shape area = 42 + 6 = 48 m²
Line 4: So the area is 48 m².
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected final answer.
Stuck? Revisit lesson § Card 8 — when a piece is REMOVED, you subtract its area, not add it.3. Open-ended challenge — same perimeter, different areas
This question has more than one valid answer. 4 marks
3.1 A council has exactly 24 m of fencing to build a rectangular dog enclosure. They want to know which rectangle gives the BIGGEST floor area for the dogs.
Find three different rectangles with perimeter exactly 24 m (whole-number side lengths only). For each:
- State the length and width.
- Check the perimeter is 24 m.
- Calculate the area.
Then: rank your three rectangles from smallest area to largest, and write one sentence explaining what shape of rectangle gives the maximum area for a given perimeter.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Rectangle 13 × 11
A = 13 × 11 = 143 cm².
1.2 — Triangle b = 18, h = 7
A = ½ × 18 × 7 = ½ × 126 = 63 m².
1.3 — Rectangle A = 96, w = 6
l = A ÷ w = 96 ÷ 6 = 16 m.
1.4 — Triangle A = 48, h = 8
From A = ½bh: 48 = ½ × b × 8 = 4b, so b = 48 ÷ 4 = 12 cm.
1.5 — L-shape 7 × 6 with 2 × 3 cut
Total = 7 × 6 = 42 cm². Cut = 2 × 3 = 6 cm². L-shape A = 42 − 6 = 36 cm².
1.6 — House shape (10 × 4 rect + 10-base 3-high triangle roof)
Rectangle = 10 × 4 = 40 cm². Triangle roof = ½ × 10 × 3 = 15 cm². Total A = 40 + 15 = 55 cm².
2 — Find the mistake
(a) The mistake is on Line 3 (carried into Line 4).
(b) The student ADDED the cut-out area to the total instead of SUBTRACTING it. The L-shape is what's LEFT after the cut, so you must subtract the removed piece, not add it.
(c) Corrected working:
Total rectangle area = 7 × 6 = 42 m²
Cut-out area = 2 × 3 = 6 m²
L-shape area = 42 − 6 = 36 m². ✓
Sanity check: the L-shape MUST be smaller than the full 42 m² rectangle, since a piece is missing. 36 < 42 ✓; 48 > 42 ✗.
3 — Same perimeter, different areas (sample solution)
Half-perimeter = l + w = 12 m. Whole-number pairs:
Rectangle A: 11 m × 1 m. P-check: 2(11) + 2(1) = 24 ✓. A = 11 m².
Rectangle B: 8 m × 4 m. P-check: 2(8) + 2(4) = 24 ✓. A = 32 m².
Rectangle C: 6 m × 6 m (a square). P-check: 4(6) = 24 ✓. A = 36 m².
Ranked smallest to largest: A (11 m²) < B (32 m²) < C (36 m²).
The square gives the maximum area for a given perimeter. The closer a rectangle is to a square, the bigger its area. Long, thin rectangles waste the perimeter on extra length without much area gain.
Marking: 1 mark per valid rectangle with P-check AND area (up to 3 marks). 1 bonus mark for ranking correctly AND stating that the square (or "closest to a square") gives the maximum area.