Mathematics

Corrigé

7.7 Shape and Measure

Pack A — Réponses

# Question Réponse
1 A rectangle has perimeter 16 cm. Give two different pairs of whole-number side lengths that would produce this perimeter. Verify each pair by adding. E.g. 6 cm × 2 cm (2×(6+2)=16 ✓) and 7 cm × 1 cm (2×(7+1)=16 ✓)
2 A triangle has sides 5 cm, 7 cm and 9 cm. What is its perimeter? Show the addition. 21 cm; 5 + 7 + 9 = 21
3 A rectangle has area 24 cm² with whole-number sides. Give three different (length, width) pairs. Verify each by multiplying. E.g. (24, 1), (12, 2), (8, 3) — each verified by multiplication
4 Find the area of a rectangle with length 9 cm and width 4 cm. Show the multiplication and write the units. 36 cm²
5 A right-angled triangle has base 6 cm and height 8 cm. Find its area. Explain why you halve the rectangle formula. 24 cm²; a triangle is exactly half a rectangle with the same base and height
6 A shape has 4 sides and 2 lines of symmetry. Name two possibilities and justify each. Rectangle (lines through midpoints of opposite sides) and rhombus (lines through opposite vertices)
7 A cuboid has length 4 cm, width 3 cm and height 2 cm. Find its volume by multiplying length × width × height. Show each step. 24 cm³
8 A regular polygon has perimeter 35 cm and each side is 7 cm. Find the number of sides by dividing. State the name of the polygon. 5 sides (pentagon); 35 ÷ 7 = 5
9 A rectangle has perimeter 26 cm and length 8 cm. Find the width by calculation. Then find the area. Width = 5 cm; Area = 40 cm²
10 A compound shape is made from two rectangles. Rectangle A is 6 cm × 4 cm. Rectangle B is 3 cm × 2 cm. Find the total area by calculating each area separately then adding. 30 cm²
11 Find the area of a parallelogram with base 9 cm and perpendicular height 5 cm. 45 cm²
12 Find the area of a trapezium with parallel sides 5 cm and 9 cm and perpendicular height 4 cm. 28 cm²
13 Find the volume of a cuboid with length 5 cm, width 4 cm and height 3 cm. 60 cm³
14 Find the surface area of a cuboid with length 5 cm, width 3 cm and height 2 cm. 62 cm²
15 A compound shape is made from a rectangle 8 cm by 6 cm with a 2 cm square removed from one corner. Find the area. 44 cm²
16 Find the perimeter of a regular 8-sided polygon with side length 5 cm. 40 cm
17 A net is folded to make a cuboid with dimensions 4 cm × 3 cm × 2 cm. How many faces does it have and what is its surface area? 6 faces; 52 cm²
18 A triangle has base 8 cm and area 28 cm². Find its perpendicular height. 7 cm
19 How many faces, edges and vertices does a triangular prism have? 5 faces, 9 edges, 6 vertices
20 A rectangle has a perimeter of 30 cm and a length of 9 cm. Find its width and hence its area. Width = 6 cm; Area = 54 cm²
21 A rectangle has area 24 cm² and width 3 cm. Find its perimeter. 22 cm
22 A square has side 1.5 cm. Find its area in cm² and then convert it to mm². 2.25 cm² = 225 mm²
23 A trapezium has parallel sides 4 cm and 8 cm. Its area is 30 cm². Find its perpendicular height. 5 cm
24 A compound L-shape is formed by a large rectangle 10 cm by 8 cm with a smaller rectangle 4 cm by 3 cm cut from one corner. Find the area of the L-shape. 68 cm²
25 A rectangle has length $(x + 3)$ cm and width $(2x)$ cm. Write an expression for its perimeter and simplify. $(6x + 6)$ cm
26 A rectangular garden is 3.2 m long and 2.5 m wide. Find its area in m² and convert the answer to cm². 8 m² = 80 000 cm²
27 A square has area 49 cm². Find its perimeter. 28 cm
28 Find the surface area of a cuboid with dimensions 7 cm × 4 cm × 3 cm. 122 cm²
29 A rectangle is 12 cm long and 5 cm wide. What is three-quarters of its area? 45 cm²
30 A triangular prism has a triangular cross-section with base 6 cm and height 4 cm, and a length of 10 cm. Find its volume. 120 cm³
31 A rectangle has length $(2x + 1)$ cm and width $(x - 2)$ cm. Its perimeter is 34 cm. Find $x$ and hence find the area. x = 5; Area = 33 cm²
32 A compound shape is made by joining a rectangle 10 cm × 6 cm and a triangle with base 6 cm and height 4 cm along a shared edge of length 6 cm. Find the total area. 72 cm²
33 A cuboid fish tank is 1.2 m long, 0.5 m wide and 0.4 m deep. It is filled to three-quarters of its height. Find the volume of water in litres. (1 m³ = 1 000 litres.) 180 litres
34 A square tile has perimeter 28 cm. Tiles are arranged in a 4 × 5 grid with no gaps. Find the total area of the tiled surface in cm² and convert to m². 980 cm² = 0.098 m²
35 Two identical trapeziums, each with parallel sides 5 cm and 11 cm and perpendicular height 6 cm, are placed together along their longer parallel side to form a parallelogram. Find the area of the parallelogram. 96 cm²
36 A rectangle has length 10 cm and width 8 cm. Both dimensions are increased by 20%. Find the new area and the percentage increase in area. New area = 115.2 cm²; 44% increase
37 A cuboid has volume 120 cm³. Its length is 6 cm and its width is 4 cm. Find its height and hence its surface area. Height = 5 cm; SA = 148 cm²
38 The area of a square is 196 cm². Express 196 as a product of prime factors, and hence write down the exact side length of the square. $196 = 2^2 \times 7^2$; side = 14 cm
39 A path of uniform width 2 cm runs around the outside of a rectangle 10 cm × 6 cm. Find the area of the path alone. 80 cm²
40 A cube has surface area 216 cm². Find its side length, volume and express the volume in litres. Side = 6 cm; Volume = 216 cm³ = 0.216 litres

Pack B — Réponses

# Question Réponse
1 A rectangle has perimeter 24 cm. Give two different pairs of whole-number side lengths that would produce this perimeter. Verify each pair by adding. E.g. 8 cm × 4 cm (2×(8+4)=24 ✓) and 10 cm × 2 cm (2×(10+2)=24 ✓)
2 A triangle has sides 6 cm, 8 cm and 11 cm. What is its perimeter? Show the addition. 25 cm; 6 + 8 + 11 = 25
3 A rectangle has area 36 cm² with whole-number sides. Give three different (length, width) pairs. Verify each by multiplying. E.g. (36, 1), (18, 2), (12, 3) — each verified by multiplication
4 Find the area of a rectangle with length 7 cm and width 6 cm. Show the multiplication and write the units. 42 cm²
5 A right-angled triangle has base 10 cm and height 5 cm. Find its area. Explain why you halve the rectangle formula. 25 cm²; a triangle is half the enclosing rectangle (10 × 5 = 50, halved = 25)
6 A shape has 4 sides and 2 lines of symmetry. Name two possibilities and justify each. Rectangle (lines through midpoints of opposite sides) and rhombus (lines through opposite vertices)
7 A cuboid has length 5 cm, width 4 cm and height 3 cm. Find its volume by multiplying length × width × height. Show each step. 60 cm³
8 A regular polygon has perimeter 48 cm and each side is 8 cm. Find the number of sides by dividing. State the name of the polygon. 6 sides (hexagon); 48 ÷ 8 = 6
9 A rectangle has perimeter 30 cm and length 11 cm. Find the width by calculation. Then find the area. Width = 4 cm; Area = 44 cm²
10 A compound shape is made from two rectangles. Rectangle A is 8 cm × 5 cm. Rectangle B is 4 cm × 3 cm. Find the total area by calculating each area separately then adding. 52 cm²
11 Find the area of a parallelogram with base 12 cm and perpendicular height 7 cm. 84 cm²
12 Find the area of a trapezium with parallel sides 6 cm and 10 cm and perpendicular height 5 cm. 40 cm²
13 Find the volume of a cuboid with length 8 cm, width 3 cm and height 4 cm. 96 cm³
14 Find the surface area of a cuboid with length 6 cm, width 4 cm and height 3 cm. 108 cm²
15 A compound shape is made from a rectangle 10 cm by 7 cm with a 3 cm square removed from one corner. Find the area. 61 cm²
16 Find the perimeter of a regular 7-sided polygon with side length 6 cm. 42 cm
17 A net is folded to make a cuboid with dimensions 5 cm × 2 cm × 3 cm. How many faces does it have and what is its surface area? 6 faces; 62 cm²
18 A triangle has base 10 cm and area 35 cm². Find its perpendicular height. 7 cm
19 How many faces, edges and vertices does a square-based pyramid have? 5 faces, 8 edges, 5 vertices
20 A rectangle has a perimeter of 40 cm and a length of 13 cm. Find its width and hence its area. Width = 7 cm; Area = 91 cm²
21 A rectangle has area 45 cm² and width 5 cm. Find its perimeter. 28 cm
22 A square has side 2.5 cm. Find its area in cm² and then convert it to mm². 6.25 cm² = 625 mm²
23 A trapezium has parallel sides 5 cm and 11 cm. Its area is 48 cm². Find its perpendicular height. 6 cm
24 A compound L-shape is formed by a large rectangle 12 cm by 7 cm with a smaller rectangle 5 cm by 4 cm cut from one corner. Find the area of the L-shape. 64 cm²
25 A rectangle has length $(x + 5)$ cm and width $(3x)$ cm. Write an expression for its perimeter and simplify. $(8x + 10)$ cm
26 A rectangular garden is 4.5 m long and 1.8 m wide. Find its area in m² and convert the answer to cm². 8.1 m² = 81 000 cm²
27 A square has area 121 cm². Find its perimeter. 44 cm
28 Find the surface area of a cuboid with dimensions 8 cm × 5 cm × 2 cm. 132 cm²
29 A rectangle is 15 cm long and 8 cm wide. What is 40% of its area? 48 cm²
30 A triangular prism has a triangular cross-section with base 8 cm and height 5 cm, and a length of 9 cm. Find its volume. 180 cm³
31 A rectangle has length $(2x + 3)$ cm and width $(x - 1)$ cm. Its perimeter is 40 cm. Find $x$ and hence find the area. x = 6; Area = 75 cm²
32 A compound shape is made by joining a rectangle 12 cm × 8 cm and a triangle with base 8 cm and height 5 cm along a shared edge of length 8 cm. Find the total area. 116 cm²
33 A cuboid fish tank is 1.5 m long, 0.6 m wide and 0.5 m deep. It is filled to 80% of its height. Find the volume of water in litres. (1 m³ = 1 000 litres.) 360 litres
34 A square tile has perimeter 32 cm. Tiles are arranged in a 3 × 6 grid with no gaps. Find the total area of the tiled surface in cm² and convert to m². 1 152 cm² = 0.1152 m²
35 Two identical trapeziums, each with parallel sides 7 cm and 13 cm and perpendicular height 8 cm, are placed together along their longer parallel side to form a parallelogram. Find the area of the parallelogram. 160 cm²
36 A rectangle has length 15 cm and width 6 cm. Both dimensions are increased by 10%. Find the new area and the percentage increase in area. New area = 108.9 cm²; 21% increase
37 A cuboid has volume 210 cm³. Its length is 7 cm and its width is 5 cm. Find its height and hence its surface area. Height = 6 cm; SA = 214 cm²
38 The area of a square is 324 cm². Express 324 as a product of prime factors, and hence write down the exact side length of the square. $324 = 2^2 \times 3^4$; side = 18 cm
39 A path of uniform width 3 cm runs around the outside of a rectangle 14 cm × 8 cm. Find the area of the path alone. 168 cm²
40 A cube has surface area 486 cm². Find its side length, volume and express the volume in litres. Side = 9 cm; Volume = 729 cm³ = 0.729 litres

Problèmes — Solutions détaillées

1

**Fencing a field.** A farmer wants to fence a rectangular field with area 72 m². She has exactly 34 m of fencing to use as the perimeter. Find the length and width of the field. Show all your working.

Réponse

Length = 9 m, width = 8 m (or length = 8 m, width = 9 m)

Let the length be $l$ and width $w$. We have two equations: $lw = 72$ and $2(l + w) = 34$, so $l + w = 17$. This means $w = 17 - l$. Substitute into the area equation: $l(17 - l) = 72$, giving $17l - l^2 = 72$, or $l^2 - 17l + 72 = 0$. Factorise: $(l - 8)(l - 9) = 0$, so $l = 8$ or $l = 9$. If $l = 9$, then $w = 8$. Check: $9 \times 8 = 72$ ✓ and $2(9 + 8) = 34$ ✓. **Length = 9 m, width = 8 m.**
2

**Optimal pen design.** A farmer has exactly 40 m of fencing. She wants to make a rectangular enclosure divided into **three equal pens** by two internal fences parallel to one pair of sides (as shown below): $$\underbrace{\Big[\;\big|\;\big|\;\Big]}_{\text{3 pens}}$$ Let $W$ be the width of the whole enclosure (perpendicular to the dividers) and $L$ be the length. (a) Explain why the total fencing used is $2L + 4W = 40$. (b) Express $L$ in terms of $W$. (c) Write the total area $A$ as a function of $W$ alone, and complete the table: | $W$ (m) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | |----------|---|---|---|---|---|---|---|---|---| | $A$ (m²) | | | | | | | | | | (d) What value of $W$ gives the maximum area? What is that area? (e) At the maximum, what is the ratio $L : W$? Does this surprise you?

Réponse

(d) W = 5 m gives maximum area 50 m². (e) L : W = 10 : 5 = 2 : 1 — the length is always double the width at the optimum.

(a) The outer rectangle uses 2 fences of length $L$ (top and bottom) and 2 fences of length $W$ (left and right outer sides), plus 2 internal fences of length $W$. Total = $2L + 4W = 40$ ✓. (b) $2L = 40 - 4W$, so $L = 20 - 2W$. (c) $A = L \times W = (20 - 2W) \times W = 20W - 2W^2$. | $W$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | |-----|---|---|---|---|---|---|---|---|---| | $A$ | 18 | 32 | 42 | 48 | **50** | 48 | 42 | 32 | 18 | Calculations: $A(1)=20(1)-2(1)=18$; $A(2)=40-8=32$; $A(3)=60-18=42$; $A(4)=80-32=48$; $A(5)=100-50=50$; $A(6)=120-72=48$; etc. (d) Maximum area = **50 m²** at $W = 5$ m, $L = 20 - 10 = 10$ m. (e) $L : W = 10 : 5 = \mathbf{2:1}$. This is slightly surprising — the optimal ratio is not a square but 2:1. In general, whenever the interior fences divide the enclosure into $k$ equal pens parallel to a side, the optimal ratio is always $L : W = k : 1$ (here $k = 2$ for the widths, or $k$ internal widths vs 2 external). It shows that the "square is optimal" rule only applies when there are no interior dividers.
3

**Trapezoidal swimming pool.** A swimming pool has a trapezoidal cross-section: it is 1 m deep at the shallow end and 3 m deep at the deep end. The pool is 25 m long and 12 m wide. (a) Sketch the cross-section and label all dimensions. (b) Find the area of the trapezoidal cross-section. (c) Find the volume of the pool in m³. (d) Convert the volume to litres and find how long (in hours) it takes to fill at 1 000 litres per minute. (e) A second pool is rectangular with the same length, width, and the same volume of water. How deep is the rectangular pool?

Réponse

(b) 50 m² (c) 600 m³ (d) 600 000 litres; 10 hours (e) 2 m deep

(a) Cross-section is a trapezium with parallel sides 1 m (top/shallow) and 3 m (bottom/deep), width 25 m along the length of the pool. (b) Area of trapezium = $\dfrac{1}{2}(a + b) \times h = \dfrac{1}{2}(1 + 3) \times 25 = \dfrac{1}{2} \times 4 \times 25 = \mathbf{50}$ m². (Here the "height" of the trapezium is the 25 m length of the pool — the cross-section is taken along the length.) (c) Volume = cross-section area × width = $50 \times 12 = \mathbf{600}$ m³. (d) $600 \times 1\,000 = \mathbf{600\,000}$ litres. Time = $600\,000 \div 1\,000 = 600$ minutes = **10 hours**. (e) Rectangular pool: $V = \text{length} \times \text{width} \times \text{depth} = 25 \times 12 \times d = 300d$. Set equal to 600: $300d = 600$, so $d = \mathbf{2}$ m. The trapezoidal pool's average depth is $\frac{1+3}{2} = 2$ m — as expected, since the trapezium equals a rectangle with the average height.
4

**Tiling a floor.** A rectangular kitchen floor is 3.6 m long and 2.4 m wide. Square tiles with side length 30 cm are to be laid with no gaps. How many tiles are needed? Show how you convert units consistently.

Réponse

96 tiles

Convert floor dimensions to cm: 3.6 m = 360 cm; 2.4 m = 240 cm. Number of tiles along length = 360 ÷ 30 = 12. Number of tiles along width = 240 ÷ 30 = 8. Total tiles = 12 × 8 = **96 tiles.** Alternatively: floor area = 3.6 × 2.4 = 8.64 m²; tile area = 0.3 × 0.3 = 0.09 m²; tiles = 8.64 ÷ 0.09 = 96.
5

**Shape investigation.** A shape is made from a rectangle and a right-angled triangle. The rectangle is 12 cm long and 5 cm wide. The triangle is attached to one of the shorter ends (5 cm wide), and has a perpendicular height of 8 cm. (a) Find the total area of the compound shape. (b) Find the perimeter of the compound shape. The slant side of the triangle has length 8.5 cm (given).

Réponse

(a) 80 cm² (b) 45.5 cm

**(a)** Rectangle area = 12 × 5 = 60 cm². Triangle area = ½ × 5 × 8 = 20 cm². Total = 60 + 20 = **80 cm².** **(b)** The outer edges are: top of rectangle (12 cm), far short side (5 cm), slant of triangle (8.5 cm), bottom of rectangle (12 cm), and the left short side (5 cm). But the triangle shares the 5 cm side with the rectangle — that edge is interior. Outer perimeter = 12 (bottom) + 5 (left) + 12 (top) + 8.5 (slant) + 8 (triangle height, right side) = **45.5 cm.**
6

**Nets.** A cube has side length 4 cm. (a) Draw a sketch of a valid net for this cube (describe it in words if you cannot draw). (b) Find the total area of the net. (c) Explain why a cross-shaped net with five squares in a column and one square to the right of the second square from the top is NOT a valid net for a cube.

Réponse

(a) Any valid T- or cross-shaped arrangement of six 4 cm × 4 cm squares. (b) 96 cm². (c) That specific arrangement has only 6 squares but when folded, two faces overlap, so it is not a valid net.

**(a)** One valid net: a T-shape — three squares across the top, one square below the middle, two more squares below that, forming a cross. **(b)** A cube has 6 square faces each with area 4² = 16 cm². Total net area = 6 × 16 = **96 cm².** **(c)** When five squares are in a column, folding gives four faces along the "tube" plus top and bottom — however the sixth square position matters: placing it to the right of the second square from the top creates a face that would coincide with another face on folding, so it is invalid. A valid net must allow each of the 6 faces to map to a distinct face of the cube.
7

**Algebra + area.** A rectangle has length $(2x + 4)$ cm and width $(x + 1)$ cm. Its area is 40 cm². (a) Show that $x^2 + 3x - 18 = 0$ and solve it to find $x$. (b) Write down the dimensions of the rectangle and find its perimeter.

Réponse

(a) x = 3 (b) Length = 10 cm, width = 4 cm; Perimeter = 28 cm

**(a)** Area = length × width = $(2x+4)(x+1) = 2x^2 + 2x + 4x + 4 = 2x^2 + 6x + 4$. Set equal to 40: $2x^2 + 6x + 4 = 40$. Divide through by 2: $x^2 + 3x + 2 = 20$, so $x^2 + 3x - 18 = 0$. Factorise: $(x + 6)(x - 3) = 0$, giving $x = -6$ or $x = 3$. Since $x$ must be positive, $x = 3$. **(b)** Length $= 2(3) + 4 = 10$ cm; width $= 3 + 1 = 4$ cm. Check: $10 × 4 = 40$ ✓. Perimeter $= 2(10 + 4) = 2 × 14 = 28$ cm.
8

**Perimeter puzzle.** The perimeter of an equilateral triangle equals the perimeter of a square. The square has side length 9 cm. (a) Find the side length of the triangle. (b) Find the area of the triangle. (Use the formula: area = $\frac{\sqrt{3}}{4} \times \text{side}^2$, or split into two right-angled triangles.) (c) Which shape has the larger area?

Réponse

(a) 12 cm (b) 36√3 cm² ≈ 62.4 cm² (c) Square (area = 81 cm²)

**(a)** Perimeter of square = 4 × 9 = 36 cm. So the triangle's perimeter = 36 cm and each side = 36 ÷ 3 = **12 cm.** **(b)** Split the equilateral triangle in half: each half is a right-angled triangle with hypotenuse 12 cm and base 6 cm. Height = $\sqrt{12^2 - 6^2} = \sqrt{144 - 36} = \sqrt{108} = 6\sqrt{3}$ cm. Area = ½ × 12 × $6\sqrt{3}$ = $36\sqrt{3}$ ≈ 62.4 cm². **(c)** Square area = 81 cm² > 62.4 cm². The **square** has the larger area.
9

**Thinking about 3D shapes.** A toy factory uses cuboid boxes with dimensions 6 cm × 4 cm × 3 cm. (a) Find the volume and surface area of one box. (b) The boxes are packed into a larger cuboid crate. The crate is 24 cm × 20 cm × 12 cm. How many boxes fit in the crate? (c) What fraction of the crate's volume is taken up by the boxes? Simplify your answer.

Réponse

(a) Volume = 72 cm³; SA = 108 cm² (b) 80 boxes (c) 1 (the boxes fill the crate exactly)

**(a)** Volume = 6 × 4 × 3 = 72 cm³. SA = 2(6×4 + 6×3 + 4×3) = 2(24 + 18 + 12) = 2 × 54 = **108 cm².** **(b)** Boxes along 24 cm: 24 ÷ 6 = 4. Along 20 cm: 20 ÷ 4 = 5. Along 12 cm: 12 ÷ 3 = 4. Total = 4 × 5 × 4 = **80 boxes.** **(c)** Crate volume = 24 × 20 × 12 = 5 760 cm³. Total box volume = 80 × 72 = 5 760 cm³. Fraction = 5 760 ÷ 5 760 = **1** — the boxes fill the crate exactly.
10

**Area reasoning.** Two shapes have the same area. - Shape A is a triangle with base 16 cm and height $h$ cm. - Shape B is a trapezium with parallel sides 5 cm and 11 cm, and perpendicular height 8 cm. Find $h$.

Réponse

h = 8 cm

Area of trapezium (Shape B) = ½(5 + 11) × 8 = ½ × 16 × 8 = 64 cm². Set equal to area of triangle: ½ × 16 × h = 64. So 8h = 64, h = **8 cm.**
11

**Staircase border investigation.** A "staircase" pattern is built from unit squares (each 1 cm × 1 cm). The $n$-step staircase has $n$ columns: column 1 has 1 square, column 2 has 2 squares, …, column $n$ has $n$ squares (like a rising staircase from left to right). A border of width 1 cm is painted around the outside of each staircase. (a) Draw (or describe) the 1-step, 2-step, and 3-step staircases. (b) For each of $n = 1, 2, 3$: - Count the number of unit squares in the staircase. - Count the perimeter of the staircase (in cm). - Calculate the area of the 1 cm border painted around it. (c) Complete the table: | $n$ | Squares in staircase | Perimeter (cm) | Border area (cm²) | |-----|----------------------|----------------|-------------------| | 1 | | | | | 2 | | | | | 3 | | | | | 4 | | | | (d) Find a formula for the number of unit squares in the $n$-step staircase. (e) Find a formula for the border area around the $n$-step staircase. (f) The border area of one staircase equals the number of squares in another staircase. Which two values of $n$ satisfy this? (There may be more than one answer.)

Réponse

(d) Squares = n(n+1)/2. (e) Border area = 4(n+1) cm². (f) 4(n+1) = m(m+1)/2 — e.g. n=1: border=8, squares: m(m+1)/2=8 has no integer solution; n=3: border=16=staircase squares for n=5 (5×6/2=15) — not exact; n=7: border=32=staircase for n=7 (7×8/2=28) — not exact. Closest: border at n=1 is 8 = squares when n=3 gives 6 (not 8); when n=4: 4×5/2=10 (not 8). Actually n=3: border=16; staircase squares = 16 when n(n+1)/2=16 → n²+n-32=0 (not integer). Special case: n=1, border=8; no staircase has exactly 8 squares (closest: n=3→6, n=4→10). This part is open-ended investigation.

(a) 1-step: a single 1×1 square. 2-step: column 1 has 2 squares (height 2), column 2 has 1 square (height 1) — an L-shape rotated (or: column 1: 1 square, column 2: 2 squares — a staircase rising right). Using rising-right convention: column $k$ has $k$ squares tall. (b) and (c): **$n = 1$:** 1 square. Perimeter = 4 cm. Border area: the border of width 1 around a shape with perimeter $P$ and convex corners adds $P \times 1 + 4 \times 1^2 = P + 4$ (four quarter-circles at corners become one full square of area 1 each). Border area = $4 + 4 = \mathbf{8}$ cm². **$n = 2$:** Squares = $1 + 2 = 3$. Perimeter: trace the outline — left side (2 units up), bottom (2 right), right side (1 down), step left (1 left), step down (1 down), top-left (1 left) = total $2+2+1+1+1+1 = 8$ cm. Border area = $8 + 4 = \mathbf{12}$ cm². **$n = 3$:** Squares = $1+2+3 = 6$. Perimeter: left (3 up), bottom (3 right), right (1 down), step (1+1), right (1 down), step (1+1), top (1 left) = $3+3+1+1+1+1+1+1 = \mathbf{12}$ cm. Border area = $12+4 = \mathbf{16}$ cm². **$n = 4$:** Squares = $10$. Perimeter = $4 \times 4 = 16$ cm (pattern: perimeter = $4n$). Border area = $16 + 4 = \mathbf{20}$ cm². | $n$ | Squares | Perimeter | Border area | |-----|---------|-----------|-------------| | 1 | 1 | 4 | 8 | | 2 | 3 | 8 | 12 | | 3 | 6 | 12 | 16 | | 4 | 10 | 16 | 20 | (d) Squares $= 1 + 2 + \cdots + n = \dfrac{n(n+1)}{2}$. (e) Perimeter $= 4n$ (each step adds 2 horizontal and 2 vertical edges — two edges are "new" outer boundary per step plus the staircase grows linearly). Border area $= 4n + 4 = 4(n+1)$. (f) We want $4(n+1) = \dfrac{m(m+1)}{2}$, i.e. $8(n+1) = m(m+1)$. Try values: $n=1$: $8(2)=16$, $m(m+1)=16$ → no integer $m$. $n=3$: $8(4)=32$, $m(m+1)=32$ → no integer. $n=4$: $8(5)=40$, $m(m+1)=40$ → $m=5$: $5 \times 6=30$ (no); $m=6$: $42$ (no). $n=9$: $8(10)=80$, $m(m+1)=80$ → $m=8$: $72$ (no), $m=9$: $90$ (no). $n=2$: $8(3)=24$ → $m(m+1)=24$: no integer. $n=5$: $8(6)=48$ → $m=6$: $42$ (no), $m=7$: $56$ (no). This is an open investigation — students discover there may be no whole-number solution, which itself is a meaningful finding.
12

**Open-ended investigation.** A rectangle has a fixed perimeter of 24 cm. Complete the table of possible integer dimensions, calculate each area, and identify which dimensions give the maximum area. | Length (cm) | Width (cm) | Area (cm²) | |-------------|-----------|------------| | 11 | 1 | | | 10 | 2 | | | 9 | 3 | | | 8 | 4 | | | 7 | 5 | | | 6 | 6 | | What do you notice? What happens if the rectangle becomes a square?

Réponse

Maximum area = 36 cm² when the rectangle is a square (6 cm × 6 cm).

Fill the table: 11×1=11; 10×2=20; 9×3=27; 8×4=32; 7×5=35; 6×6=36. Areas increase as dimensions become more equal. The maximum area occurs when the rectangle is a **square** (6 cm × 6 cm), giving 36 cm². This is a specific instance of the general result: for a fixed perimeter, the square maximises area. Students should notice that the areas increase and then reach a peak — a pattern that connects to optimisation.