Mathematics • Year 8 • Unit 3 • Lesson 4

Perimeter of Polygons

Build fluency with perimeter for regular polygons, rectangles and L-shaped composites. One worked hexagon example, one guided L-shape with blanks, then eight problems on regulars, rectangles, working backwards, and composites.

Build · I Do / We Do / You Do

1. I do — fully worked example

Regular polygons make life easy: every side is the same length. Just multiply.

Problem. A regular hexagon has a side length of 8 cm. Find its perimeter.

Step 1 — Identify the number of sides.

A hexagon has n = 6 sides.

Reason: "hex" means six. Memorise: tri = 3, quad = 4, penta = 5, hexa = 6, octa = 8.

Step 2 — Note the side length.

s = 8 cm. Because it's REGULAR, all 6 sides are 8 cm.

Reason: "regular" = all sides equal + all angles equal. No need to add 6 separate side lengths.

Step 3 — Apply the formula.

P = n × s = 6 × 8

Reason: perimeter is the sum of all sides; for n equal sides, that's n × s.

Step 4 — Calculate, state units.

P = 48 cm

Reason: always state units. Perimeter is a length, so use cm (not cm²).

Answer: P = 48 cm.

Stuck? Revisit lesson § Card 5 — for any regular polygon, P = n × s. To find s when P is known: s = P ÷ n.

2. We do — L-shape with blanks

An L-shape always has 6 sides — and 2 of them are usually unlabelled. Find them first, then add. 4 marks

Problem. An L-shape has total width 15 cm and total height 12 cm. A 9 cm × 7 cm rectangular notch is cut from the bottom-right corner. Find the perimeter.

Step 1 — Find the missing horizontal side:

missing horizontal = total width − partial width = 15 − ______ = ______ cm

Step 2 — Find the missing vertical side:

missing vertical = total height − partial height = 12 − ______ = ______ cm

Step 3 — List all 6 sides:

12, 15, ______ (missing v), 9, 7, ______ (missing h) — all in cm.

Step 4 — Add:

P = 12 + 15 + ______ + 9 + 7 + ______ = ______ cm

Stuck? Revisit lesson § Card 7 — the opposite-sides rule says missing horizontal + given horizontal = total width.

3. You do — independent practice

Show all working. The first four are foundation (single-step). The middle two are standard (work backwards). The last two are extension (composite shapes).

Foundation — direct application

3.1 An equilateral triangle has side length s = 7 cm. Find P. 1 mark

3.2 A square has side length 9 cm. Find P. 1 mark

3.3 A regular pentagon has side length 12 cm. Find P. 1 mark

3.4 A rectangle is 15 cm long and 8 cm wide. Find P. 1 mark

Standard — working backwards

3.5 A rectangle has P = 56 cm and length l = 18 cm. Find the width w. 2 marks

3.6 A regular octagon has perimeter 64 cm. Find the side length. 2 marks

Extension — composite shapes

3.7 An L-shape is formed from a 10 cm × 8 cm rectangle with a 4 cm × 3 cm corner cut out. Find the perimeter. (Hint: find the 2 missing sides first, then add all 6.) 2 marks

3.8 A T-shape is formed from a 12 cm × 4 cm horizontal rectangle sitting on top of a 4 cm × 6 cm vertical rectangle (centred). Sketch the shape, label all sides, and find the perimeter. (Hint: the T-shape has 8 sides.) 2 marks

Stuck on 3.8? Trace the outside boundary of the T with your finger. Top bar: 12 across the top, 4 down each end. Stem: each step in/out of the central part is (12 − 4) ÷ 2 = 4 cm horizontal.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Section 2 — We do (L-shape 15 × 12 with 9 × 7 notch)

Step 1: missing horizontal = 15 − 9 = 6 cm.
Step 2: missing vertical = 12 − 7 = 5 cm.
Step 3: sides are 12, 15, 5, 9, 7, 6 cm.
Step 4: P = 12 + 15 + 5 + 9 + 7 + 6 = 54 cm.

3.1 — Equilateral triangle, s = 7

P = 3 × 7 = 21 cm.

3.2 — Square, s = 9

P = 4 × 9 = 36 cm.

3.3 — Regular pentagon, s = 12

P = 5 × 12 = 60 cm.

3.4 — Rectangle 15 × 8

P = 2l + 2w = 2(15) + 2(8) = 30 + 16 = 46 cm.

3.5 — Rectangle P = 56, l = 18

Half-perimeter = 56 ÷ 2 = 28 = l + w, so w = 28 − 18 = 10 cm.

3.6 — Regular octagon, P = 64

s = P ÷ n = 64 ÷ 8 = 8 cm.

3.7 — L-shape 10 × 8 with 4 × 3 notch

Missing horizontal = 10 − 4 = 6 cm. Missing vertical = 8 − 3 = 5 cm.
Sides: 8, 10, 5, 4, 3, 6. P = 8 + 10 + 5 + 4 + 3 + 6 = 36 cm.

3.8 — T-shape 12 × 4 on top of 4 × 6 (centred)

The horizontal "top bar" is 12 wide × 4 tall. The vertical "stem" is 4 wide × 6 tall, centred under the top bar.
Tracing the boundary clockwise from the top-left corner: 12 (top), 4 (right end of bar down), (12−4)÷2 = 4 (step in to the right edge of the stem), 6 (right of stem down), 4 (bottom of stem), 6 (left of stem up), 4 (step out from stem to left edge of bar), 4 (left end of bar up).
P = 12 + 4 + 4 + 6 + 4 + 6 + 4 + 4 = 44 cm.