Lesson Guide
Explanation
OC runs perimeter backwards. OC style: 64 cm of ribbon bent into a rectangle with two sides of 12 cm — find one of the other sides. Forward thinking says add sides; reverse thinking subtracts: the two known sides consume 24 cm, leaving 40 cm for the other two — 20 cm each. The trap answer 40 (forgetting there are TWO of them) is in the options, of course. Rectangle sides come in PAIRS: use it forwards and backwards.
Your Turn
A square is made from 36 cm of wire. Side length? And a rectangle from 50 cm of wire with long sides 15 cm — short side? Check: 9 cm; and (50 − 30) ÷ 2 = 10 cm.
Watch Out
After subtracting the known pair, DIVIDE BY 2 — the remaining wire covers two sides, not one.
Explanation
Burn this in: perimeter and area are INDEPENDENT. A 1×6 rectangle and a 2×3 rectangle both hold 6 squares of area — but their perimeters are 14 and 10! Long thin shapes carry more fence per grass; compact shapes carry less. OC asks both directions: 'same area, biggest perimeter?' (the thinnest one) and 'same perimeter, biggest area?' (the squarest one). Intuition points you; a quick computation confirms.
Your Turn
Rectangles with area 12: list all (1×12, 2×6, 3×4) and their perimeters. Which is biggest? Check: 26, 16, 14 → the 1×12.
Watch Out
Never assume bigger area ⇒ bigger perimeter. The exam question EXISTS because that assumption is false.
Explanation
A hard perimeter question: attach a 2×1 rectangle to a 5×3 rectangle so the perimeter grows by exactly 4 — which positions work? The physics: attaching hides the CONTACT edges on both shapes and adds the piece's other edges. Contact of length 2 (long side flat): new exposed edges 1+1+2 = 4, hidden 2+2 = 4 → net +2... careful accounting matters! Rule of thumb: perimeter change = (perimeter of piece) − 2 × (contact length). A 2×1 piece has P=6: contact 2 → +2; contact 1 → +4. So the answer is the positions touching by a SHORT side.
Your Turn
Attach a 1×1 square (P=4) to a big rectangle by one full side (contact 1). Perimeter change? Check: 4 − 2×1 = +2.
Watch Out
The seam hides edges on BOTH shapes — that is why the formula subtracts TWICE the contact length.
Explanation
A favourite late-paper question: a ruler has only marks at 0, 3, 4 and 9 — which length can you NOT draw between two marks? Every drawable length is a DIFFERENCE of two marks. Robot-list the pairs: 3, 4, 9 (from 0), 1 (4−3), 6 (9−3), 5 (9−4). Possible: 1, 3, 4, 5, 6, 9 — so of the lengths offered, 7 is the one you cannot draw. Difference-set thinking plus systematic listing: two skills you already own, combined. That's what late-paper OC questions are — combinations.
Your Turn
Marks at 0, 1, 4, 9: list all drawable lengths. Any gaps below 9? Check: 1,3,4,5,8,9 — missing 2, 6, 7.
Watch Out
List pairs in a fixed order (each mark against every LATER mark) or you'll miss one — there are 6 pairs for 4 marks.
Explanation
Units are where OC hides easy marks and easy errors. The conversions you must own cold: ×10 (mm→cm), ×100 (cm→m), ×1000 (m→km, g→kg, mL→L). Questions lean on 2½ kg = 2500 g and on reading scales — where the first job is always 'what is ONE gap worth?' (the practice-test jug counted in 10 mL steps; the walking grid in 10 m squares). Golden rule: convert everything to one unit BEFORE any arithmetic. Quiz and exercises!
Your Turn
A path is 1 km 250 m. A snail has crawled 800 m. How far to go, in metres? Check: 1250 − 800 = 450 m.
Watch Out
2 m 5 cm is 205 cm (not 250!). The missing tens place is the sneakiest unit trap of all.
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