Mathematics • Year 7 • Unit 3 • Lesson 17

Finding Missing Sides in Similar Figures

Build fluency with the two methods for finding missing sides: (1) scale factor find SF = new ÷ old, then multiply; (2) proportion set up two equal ratios and cross-multiply.

Build · I Do / We Do / You Do

1. I do, fully worked example

Read every line. Each step shows why, not just what.

Problem. △ABC ∼ △PQR. AB = 8, BC = 10, PQ = 20. Find QR using the proportion method.

Corresponding sides are in the same ratio: PQ/AB = QR/BC.

Step 1, Set up the proportion (same shape on top).

PQ / AB = QR / BC

Reason: in similar triangles every pair of corresponding sides has the SAME ratio.

Step 2, Substitute the known values.

20 / 8 = QR / 10

Reason: PQ and AB are corresponding, so they sit at the same level (top); QR and BC are corresponding, so they sit at the same level (bottom).

Step 3, Cross-multiply.

20 × 10 = 8 × QR

200 = 8 × QR

Reason: from a/b = c/d we always get ad = bc.

Step 4, Solve for QR.

QR = 200 ÷ 8 = 25

Check SF: 20 ÷ 8 = 2.5 and 25 ÷ 10 = 2.5 ✓

Reason: divide both sides by 8 to isolate QR. Then verify the scale factor is consistent.

Answer: QR = 25.

Stuck? Revisit lesson § Card 5 "Method 2 · Proportion + Cross-Multiply", always put the SAME figure's sides on top.

2. We do, fill in the missing steps

Same structure as Section 1, with the working faded. Fill in each blank. 4 marks

Problem. Two similar triangles. The small one has sides 4 cm and 7 cm. The corresponding side 4 cm becomes 12 cm on the large triangle. Find x, the side corresponding to 7 cm. Use the SCALE FACTOR method.

Step 1, Pick the pair of corresponding sides where both are known:

small side = _______, large side = _______

Step 2, Compute the scale factor (new ÷ old):

SF = _______ ÷ _______ = _______

Step 3, Apply SF to the side you want (going from small to large, MULTIPLY by SF):

x = 7 × _______ = _______ cm

Step 4, Quick check (does the SF match for both pairs?):

12 ÷ 4 = _______ and x ÷ 7 = _______ . Match? ____

Stuck? Revisit lesson § Watch Me Solve It · Q1, same setup, fully worked.

3. You do, independent practice

Show your working under each problem. The first four are foundation, the middle two are standard, and the last two are extension.

Foundation, single step

3.1 Small triangle sides 3 and 5. Similar large triangle has corresponding sides 12 and x. Find x. 1 mark

3.2 Solve the proportion: x / 6 = 15 / 9 . 1 mark

3.3 Two similar rectangles. Small is 4 cm × 6 cm. Large has corresponding width 16 cm. Find the large rectangle's length. 1 mark

3.4 A 1 m post casts a 1.6 m shadow. A nearby tree's shadow is 8 m. Find the tree's height. 1 mark

Standard, two steps

3.5 A map has scale 1 : 25 000. Two towns are 4 cm apart on the map. Find the real distance, in km. 2 marks

3.6 △ABC ∼ △DEF. AB = 6, BC = 9, AC = 12. DE = 10. Find EF and DF using either method. 2 marks

Extension, push your thinking

3.7 A large triangle has sides 24 and 18. A similar small triangle has corresponding side 8 and unknown side x. Find x. (You are going from large to SMALL, be careful!) 2 marks

3.8 A 1.5 m tall student casts a 2 m shadow. At the same time, a tree's shadow is 14 m long. How tall is the tree? Set up your working as a proportion and cross-multiply. 3 marks

Stuck on 3.8? Same sun angle means similar triangles. Set up tree / student = tree shadow / student shadow.

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 (small 4,7; large 12,x)

Step 1: small = 4, large = 12.
Step 2: SF = 12 ÷ 4 = 3.
Step 3: x = 7 × 3 = 21 cm.
Step 4: 12 ÷ 4 = 3 and 21 ÷ 7 = 3. Match? Yes ✓.

3.1, Find x (small 3, 5; large 12, x)

SF = 12 ÷ 3 = 4. x = 5 × 4 = 20.

3.2, Solve x/6 = 15/9

Cross-multiply: 9x = 6 × 15 = 90. So x = 90 ÷ 9 = 10.

3.3, Similar rectangle length

SF = 16 ÷ 4 = 4. Length = 6 × 4 = 24 cm.

3.4, Tree from post + shadow

Similar triangles (same sun angle). tree/1 = 8/1.6, so tree = 8 ÷ 1.6 = 5 m.

3.5, Map distance

Real distance = 4 × 25 000 = 100 000 cm. Convert: 100 000 ÷ 100 = 1000 m = 1 km.

3.6, △ABC ∼ △DEF

SF = DE ÷ AB = 10 ÷ 6 = 5/3 (≈ 1.667).
EF = 9 × 5/3 = 15.
DF = 12 × 5/3 = 20.

3.7, Big to small (sides 24, 18; small 8, x)

Going BIG → small: divide by SF (or use small/big). SF (big → small) = 8 ÷ 24 = 1/3.
x = 18 × 1/3 = 6.
Common slip: multiplying by 3 instead of dividing. Sanity-check: x must be SMALLER than 18 since we are shrinking.

3.8, Tree height from student + shadow

Set up proportion: tree / student = tree shadow / student shadow.
tree / 1.5 = 14 / 2.
Cross-multiply: 2 × tree = 1.5 × 14 = 21.
tree = 21 ÷ 2 = 10.5 m.