LESSON PLAN

Scale Diagrams and Similar Figures

Apothem Team

Grade 9 · Geometry

LESSON AT A GLANCE

Warm-up

5 min

Explore

15 min

Formalize

10 min

Practice

12 min

Exit ticket

3 min

Warm-up

Two photos of the same tower — one 4 cm tall on a phone, one 12 cm on a poster. "Same tower? Prove it without recognizing it." The room converges on checking PROPORTIONS: width-to-height matches; every feature scales by the same factor.

Similarity formalized: same shape, different size — equal angles, proportional sides — and Grade 9 learns to certify it, compute with it, and deploy it for inaccessible measurements.

Explore

Similarity certification lab: (1) triangle pairs with measurements — test all three ratios $\frac{a'}{a} = \frac{b'}{b} = \frac{c'}{c}$ (SSS-similarity) or two angles (AA — the shortcut that usually suffices since angles come in matched pairs); (2) the shadow method OUTSIDE if weather allows: a metre stick's shadow vs the flagpole's shadow at the same moment — similar triangles by AA (sun angle shared, right angles shared) → flagpole height from three easy measurements; (3) scale-diagram drafting: the classroom mapped at 1:50 on grid paper, with the furniture.

The shadow method's audit closes it: does the answer survive a second measurement ten minutes later (new sun angle, new shadows, same flagpole)?

Formalize

Formalize similarity and its certification tests:

$\triangle ABC \sim \triangle DEF \iff \text{AA} \;|\; \text{SSS ratios equal} \;|\; \text{SAS ratio-angle-ratio} \qquad \frac{DE}{AB} = \frac{EF}{BC} = \frac{DF}{AC} = k$

Correspondence discipline: $\sim$ statements pair vertices IN ORDER ( $A \leftrightarrow D$ , $B \leftrightarrow E$ …), and every ratio must respect the pairing — most similarity errors are correspondence errors, fixable by re-lettering the smaller triangle to match the larger. Areas of similar figures scale by $k^2$ (the year's fourth sighting).

Practice

Practice: certify (or refuse) three figure pairs; two missing-side computations with correspondence stated; one shadow/mirror indirect measurement; one scale-diagram build with a scale chosen and defended; one $k^2$ area question.

Exit ticket: $\triangle PQR \sim \triangle XYZ$ , $PQ = 6, XY = 9, QR = 8$ . Find $YZ$ , and state $k$ . ( $k = 1.5$ ; $YZ = 12$ .)

Exit ticket

Exit ticket: $\triangle PQR \sim \triangle XYZ$ , $PQ = 6, XY = 9, QR = 8$ . Find $YZ$ , and state $k$ . ( $k = 1.5$ ; $YZ = 12$ .)

TIP Require the correspondence sentence ("

A

pairs with

D

because both sit at the marked angle") before any ratio is written. It feels bureaucratic and prevents the unit's dominant error class entirely.

WORKED EXAMPLES

Example 1 — The flagpole by shadow: indirect measurement certified

The data (same moment): metre stick shadow 1.6 m; flagpole shadow 12.8 m.

Step 1: Certify similarity — AA: both triangles hold a right angle (ground ⊥ upright) and share the sun's ray angle. ✓

Step 2: Correspondence and ratio: $\frac{\text{pole height}}{1} = \frac{12.8}{1.6} = 8$ .

Step 3: Height: 8 m.

Step 4: The re-measure audit: ten minutes later the stick casts 1.9 m and the pole 15.2 m: ratio $\frac{15.2}{1.9} = 8$ ✓ same pole. The sun moved; the SIMILARITY held — which is exactly why the method works at any hour with a shadow.

Step 5: Lineage note for the class: Thales measured the pyramids this way twenty-six centuries ago. Same triangles, same theorem, same three measurements.

Example 2 — The river crossing: similar triangles without shadows

The problem: find the width of a river without crossing it.

Step 1: The construction: sight a tree $T$ directly across the river from point $A$ . Walk 20 m along the bank to $B$ , plant a stake; walk 4 m further to $C$ . Turn inland perpendicular to the bank and walk until the stake at $B$ lines up with the tree — that spot $D$ is 5 m inland.

Step 2: Certify: $\triangle TAB \sim \triangle DCB$ — right angles at $A$ and $C$ , vertical angles at $B$ . AA ✓.

Step 3: Correspondence ratio: $\frac{TA}{DC} = \frac{AB}{CB} \to \frac{TA}{5} = \frac{20}{4} = 5$ .

Step 4: River width: $TA = 25$ m.

Step 5: The design insight: the surveyor CHOSE the 20:4 layout to make $k = 5$ friendly — good indirect measurement engineers its own similar triangles. Every rangefinder and every camera focus system is this diagram in a metal box.

Example 3 — The poster blow-up:

k

and

k^2

billing

The order: a 20 cm × 30 cm photo enlarged so the long side reaches 90 cm. The print shop charges by AREA (\$0.02 per cm²).

Step 1: Scale factor from the long sides: $k = \frac{90}{30} = 3$ .

Step 2: New short side: $20 \times 3 = 60$ cm. (Both dimensions ride the same $k$ — that's what "similar" means; stretching only one side distorts faces, and clients notice.)

Step 3: Areas: original $600$ cm²; enlargement $60 \times 90 = 5400$ cm² — which is $600 \times 9 = 600 \times k^2$ ✓.

Step 4: The bill: $5400 \times 0.02 = \$108$ — NINE times the original's $12, not three. Customers who budget by $k$ get a 3× surprise; the $k^2$ law is a consumer-protection fact.

Step 5: Fourth and final sighting this year (units, pizza, maps, posters): length by $k$ , area by $k^2$ — at this point the class should be finishing the sentence themselves.

MATERIALS

•Triangle pair cards

•Metre sticks for shadows

•Grid paper for scale plans

•Mirrors (alternate indirect method)

•Practice set (PDF)

WATCH FOR

!Ratios formed across mismatched vertices. The correspondence sentence, mandatory.

!Similar and congruent conflated. Congruent = similar with

k = 1

; similarity permits resizing.

!Areas scaled by

k

(the eternal return).

k^2

, with the tile picture.