LESSON PLAN

Circumference and Area of Circles

Apothem Team

Grade 7 · Geometry

LESSON AT A GLANCE

Warm-up

5 min

Explore

15 min

Formalize

10 min

Practice

12 min

Exit ticket

3 min

Warm-up

Distribute string and six circular objects (lids, tins, wheels). Task: measure each circumference with the string and its diameter with a ruler, then compute circumference ÷ diameter. Collect results on the board: 3.09, 3.2, 3.13, 3.1, 3.18…

The reveal lands itself: every circle, any size, gives the SAME ratio — a little more than 3. That constant is $\pi$ , and it was just measured by hand, not announced.

Explore

From the discovery, build both formulas. Circumference: if $\frac{C}{d} = \pi$ always, then $C = \pi d = 2\pi r$ — done, and students own it because the string did it first.

Area by the wedge rearrangement: cut a paper circle into 8, then 16 wedges; lay them alternating point-up/point-down into a near-parallelogram. Its width is half the circumference ( $\pi r$ ), its height nearly $r$ — so the area approaches $\pi r \times r$ . More wedges, straighter rectangle: the limit is exact. (This is the same argument the site's Area of a Circle explainer runs — connect them.)

Formalize

Formalize both, with $\pi \approx 3.14159\ldots$ irrational (the decimal never ends or repeats — the string measured its opening digits):

$C = \pi d = 2\pi r \qquad A = \pi r^2$

The two formulas answer different questions — around (length units) versus inside (square units) — and the classic error is feeding the diameter into the area formula. Ritual: WRITE the radius explicitly before any area computation. Exact answers keep $\pi$ as a symbol ( $25\pi$ cm²); decimal answers say "≈" and name the rounding.

Practice

Practice: circumference and area for four circles (mixed given-radius / given-diameter); one reverse each (find $r$ from $C = 31.4$ cm; from $A = 78.5$ cm²); one composite — a semicircular window's perimeter (curved half PLUS the flat diameter, the classic omission).

Exit ticket: a circle has diameter 10 cm. Exact circumference and area, then decimals to one place. ( $10\pi \approx 31.4$ cm; $25\pi \approx 78.5$ cm².)

Exit ticket

Exit ticket: a circle has diameter 10 cm. Exact circumference and area, then decimals to one place. ( $10\pi \approx 31.4$ cm; $25\pi \approx 78.5$ cm².)

TIP "Exact means

\pi

stays": establish that

25\pi

is a finished, superior answer, not laziness — the decimal is the approximation. This single convention pre-solves years of rounding disputes.

WORKED EXAMPLES

Example 1 — The bike wheel: circumference doing real work

The wheel: diameter 70 cm. How far does the bike travel in one wheel turn — and how many turns cover one kilometre?

Step 1: One turn = one circumference: $C = \pi d = 70\pi \approx 219.9$ cm ≈ 2.2 m.

Step 2: Turns per km: $1000 \div 2.199 \approx 455$ turns.

Step 3: Sanity: a 2-ish metre roll per turn and about 450 turns per km feels right for a bike (a car wheel would take more turns — smaller? no, car wheels are smaller diameter… ≈ 63 cm → yes, slightly more turns). Estimation against experience is part of the answer.

Step 4: Exact-form note: $455$ came from dividing by an approximation; the exact count is $\frac{1000}{0.7\pi} = \frac{10000}{7\pi}$ — kept symbolic until the final rounding, the professional habit.

Example 2 — The pizza economics: area comparisons that surprise

The menu: a 20 cm-diameter pizza costs \$8; a 30 cm costs \$15. Which is more pizza per dollar?

Step 1: Radii first (the ritual): 10 cm and 15 cm.

Step 2: Areas: small — $\pi(10)^2 = 100\pi \approx 314$ cm². Large — $\pi(15)^2 = 225\pi \approx 707$ cm².

Step 3: The shock to slow down for: the 30 cm pizza is not 1.5× the 20 cm — it's 2.25× the area ( $\frac{225}{100}$ ), because area scales with the SQUARE of the size ratio ( $1.5^2 = 2.25$ ).

Step 4: Per dollar: small — $314 \div 8 \approx 39$ cm²/$; large — $707 \div 15 \approx 47$ cm²/$. The large wins comfortably.

Step 5: The generalization worth a star: doubling a circle's diameter quadruples its area. Length ratios square when they become area ratios — the same law as the m²→cm² conversion, now buying dinner.

Example 3 — The running track: a composite perimeter

The track: a rectangle 60 m long with semicircular ends of diameter 30 m. One lap = ?

Step 1: Decompose the boundary: two straight sides of 60 m each, plus two semicircles of diameter 30 — which together make ONE full circle of diameter 30.

Step 2: Curved part: $C = 30\pi \approx 94.2$ m.

Step 3: Lap: $60 + 60 + 94.2 \approx 214.2$ m.

Step 4: The near-miss everyone makes: adding the rectangle's short sides (30 + 30) too — but those edges are INSIDE the track where the semicircles attach, not on the boundary. Trace the lap with a finger before summing; boundary walks are chosen by the finger, not the formula sheet.

Step 5: Extension with meaning: the infield AREA is rectangle ( $60 \times 30 = 1800$ ) plus circle ( $225\pi \approx 707$ ) ≈ 2507 m² — same decomposition thinking, area edition.

MATERIALS

•String, rulers, circular objects

•Paper circles and scissors for wedges

•Compasses

•Practice set (PDF)

WATCH FOR

!Diameter squared in the area formula:

A = \pi d^2

giving quadruple the truth. The write-the-radius ritual intercepts it.

\pi

treated as exactly 3.14 (or as "a number the calculator has"). It's irrational; 3.14 is a convenience with an error attached.

!Semicircle perimeter as half the circumference only — the diameter edge walks too.