LESSON PLAN

Surface Area of Prisms

Apothem Team

Grade 7 · Measurement

LESSON AT A GLANCE

Warm-up

5 min

Explore

15 min

Formalize

10 min

Practice

12 min

Exit ticket

3 min

Warm-up

Hold up a cereal box and a roll of wrapping paper: "Exactly how much paper wraps this box — zero waste?" Proposals arrive; someone says "add up all the faces." Unfold a box (pre-cut along its edges) into its flat net and let the room see the answer IS the net's area.

Surface area named: the total area of every face — the shape's skin, measured in square units. Volume filled the box; today we paint it.

Explore

Net-and-count lab: pairs unfold three prisms (rectangular box, cube, triangular prism) into nets, compute each face's area, and total. The structure to notice: faces come in PAIRS for rectangular prisms (top=bottom, front=back, sides) — so $SA = 2(\ell w + \ell h + wh)$ compresses the six-face count.

For the triangular prism: two triangle ends plus three rectangle walls — and the walls' widths are the triangle's three SIDES (not its height!). One deliberate trap net has the wrong wall width; pairs find why it won't fold shut.

Formalize

Formalize the general principle and the rectangular shortcut:

$SA = \text{sum of all face areas} \qquad SA_{\text{box}} = 2(\ell w + \ell h + wh)$

For any prism: two congruent bases plus a wrap of rectangles whose widths are the base's perimeter unrolled — $SA_{\text{prism}} = 2A_{\text{base}} + P_{\text{base}} \times H$ . The unrolled-label image (soup can label, cracker box sleeve) carries the formula.

Practice

Practice: SA of a cube from its edge; two boxes (one with decimal edges); one triangular prism from a labelled net; one reverse problem (cube of SA 150 cm² — edge?); one "which needs more paint" comparison where the higher-volume solid has LESS surface.

Exit ticket: SA of a $4 \times 3 \times 2$ box, computed by the pairs structure with the three products named. ( $2(12 + 8 + 6) = 52$ cm².)

Exit ticket

Exit ticket: SA of a $4 \times 3 \times 2$ box, computed by the pairs structure with the three products named. ( $2(12 + 8 + 6) = 52$ cm².)

TIP Units are the fastest error-detector in the unit: surface area in cm², volume in cm³. A student who writes cm³ on a surface-area answer has merged the concepts — reteach the skin-versus-filling distinction immediately.

WORKED EXAMPLES

Example 1 — Wrap the gift exactly: a 25 × 20 × 8 cm box

Step 1: The three distinct face-pairs: top/bottom $25 \times 20 = 500$ each; front/back $25 \times 8 = 200$ each; sides $20 \times 8 = 160$ each.

Step 2: Total: $SA = 2(500 + 200 + 160) = 2(860) = 1720$ cm².

Step 3: Reality pass: wrapping paper needs overlap — a practical wrap buys ~10–15% extra: about 1,900–2,000 cm². The exact answer is the floor, not the shopping list; say both.

Step 4: Structure check: the three products used every pair of dimensions once ( $\ell w, \ell h, wh$ ) — a pattern worth noticing, because it means no face was forgotten and none double-counted.

Example 2 — The Toblerone sleeve: triangular prism SA

The package: triangular ends with sides 6, 6, 6 cm (equilateral, height ≈ 5.2 cm), prism length 30 cm.

Step 1: Two triangle ends: $2 \times \frac{1}{2}(6)(5.2) = 31.2$ cm².

Step 2: Three rectangular walls — widths are the triangle's SIDES: each wall $6 \times 30 = 180$ ; three walls: $540$ cm².

Step 3: Total: $31.2 + 540 = 571.2$ cm².

Step 4: Verify with the unrolled-wrap formula: $SA = 2A_{\text{base}} + P \times H = 31.2 + (18)(30) = 31.2 + 540$ ✓ — the perimeter-times-length term IS the three walls unrolled into one long strip.

Step 5: Spot the trap dodged: using the 5.2 height for wall width gives $468$ — a package that won't close around the chocolate. The walls stand on the sides.

Example 3 — Paint economics: two tanks, one budget

The choice: Tank A is a $2 \times 2 \times 2$ m cube; Tank B is $4 \times 2 \times 1$ m. Same volume (8 m³). Paint covers 10 m² per litre. Which tank is cheaper to paint, and by how much paint?

Step 1: SA of A: $6 \times 4 = 24$ m² (six 2×2 faces).

Step 2: SA of B: $2(4\cdot2 + 4\cdot1 + 2\cdot1) = 2(8 + 4 + 2) = 28$ m².

Step 3: Same filling, different skin: the cube wears 4 m² less. Paint: A needs 2.4 L, B needs 2.8 L.

Step 4: The principle, met again from the perimeter/area unit but one dimension up: among boxes of fixed volume, the CUBE minimizes surface — compact shapes are skin-thrifty. Nature concurs (bubbles, cells); so do shipping departments.

Step 5: Exit thought: what shape would beat even the cube if boxes weren't required? (The sphere — the reason bubbles are round, promised for later grades.)

MATERIALS

•Pre-cut fold-up boxes

•Net templates (with one trap)

•Rulers

•Wrapping paper for the finale wrap-off

•Practice set (PDF)

WATCH FOR

!Surface area and volume interchanged. Skin vs filling; paint vs water; cm² vs cm³ — keep all three contrasts alive.

!Triangular prism walls given the triangle's HEIGHT as width. Walls sit on the triangle's SIDES; the trap net teaches it once and well.

!Faces counted twice or skipped in ad-hoc sums. The net (or the paired formula) imposes the accounting.