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.
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.
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.
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? ____
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
How did this worksheet feel?
What I'll revisit before next class:
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.