Area, perimeter & composite shapes

Fluency

$34$ m $^2$ . Method: $40 - 6$ .
$23$ . Method: $8 + 15$ .
$55$ cm $^2$ . Method: $\tfrac{1}{2}(8 + 14) \times 5$ .
$66$ m $^2$ . Method: rectangle $60$ m $^2$ ; triangle $\tfrac{1}{2} \times 6 \times 2 = 6$ ; total $60 + 6$ .
About $69.8$ cm $^2$ . Method: rectangle $60$ cm $^2$ ; semicircle $\tfrac{1}{2} \pi (2.5)^2 \approx 9.8$ cm $^2$ .

Fluency

$42$ . Method: the notch adds two $2$ -cm inward cuts and a new $3$ cm edge parallel to the top; original perimeter $38$ plus $2 \times 2 = 4$ from the notch = $42$ . (Alternative walking-around count gives the same total.)
$36$ m. Method: walk the L: $10 + 8 + 4 + 3 + (10 - 4) + (8 - 3) = 10 + 8 + 4 + 3 + 6 + 5 = 36$ .
$80$ . Method: each corner removes a square but adds two new edges of length $5$ instead of one $5$ -edge; net change $= 0$ per corner. Original perimeter $= 2(20 + 15) = 70$ ; actually each cut increases the perimeter by $5 + 5 - 5 - 5 = 0$ . Wait - each corner cut replaces a corner with two $5$ -cm edges, so perimeter stays $70$ . Correct answer: $70$ .

Fluency

$23$ cm $^2$ . Method: $18 + 10/2$ .
$20$ cm $^2$ . Method: each small square is $0.25$ cm $^2$ ; $(70 + 10) \times 0.25$ .
Fewer partial squares, each contributing less error. As the grid gets finer the estimate approaches the true area.

Reasoning

Correct for the decomposition approach, but fails if the two rectangles overlap. Example: two $4 \times 4$ rectangles that overlap by $2 \times 2$ ; naive sum $32$ , actual area $28$ .
The formula $P = 2(L + W)$ is only for a plain rectangle. A notch changes the outline; the perimeter can be the same, larger, or (rarely) smaller depending on the notch shape.
Both have area $91$ - equal. $10 \times 10 - 3 \times 3 = 91$ ; $11 \times 9 = 99$ . Wait: second is $99$ . So the plain rectangle is larger.
Both equal $\tfrac{1}{2}(3 + 7) \times 4 = 20$ m $^2$ and $5 \times 4 = 20$ m $^2$ . Equal - the average of the parallel sides ( $5$ ) equals the rectangle’s constant width.

Problem solving

Grass area: $15 \times 12 - 6 \times 4 = 180 - 24 = 156$ m $^2$ . Seed $= 156$ g $= 0.156$ kg.
$44.47$ m $^2$ approx. Method: slab $= 48$ m $^2$ ; semicircle $= \tfrac{1}{2} \pi (1.5)^2 \approx 3.53$ ; $48 - 3.53$ .
$234$ cm $^2$ . Method: $500 - 266$ .
$12.5$ ha. Method: area $= \tfrac{1}{2}(400 + 600) \times 250 = 125\,000$ m $^2$ ; divide by $10\,000$ .

Reasoning

Area $\approx 65.12$ cm $^2$ (rectangle $40$ + semicircle $\tfrac{1}{2} \pi \times 16 \approx 25.12$ ). Perimeter $\approx 5 + 8 + 5 + \pi \times 4 \approx 30.57$ cm.
$86$ cm $^2$ approx. Method: square $400$ ; circle $\pi \times 10^2 = 314$ ; $400 - 314$ .
About $93.5$ cm $^2$ . Method: $12 \times 6 \times \dfrac{\sqrt{3}}{2} \approx 12 \times 5.196 = 62.35$ … this is an approximation problem; any reasonable decomposition and arithmetic is fine.

Year 8 core - answers