Mathematics • Year 8 • Unit 3 • Lesson 20
Similar Figures
Build fluency with the scale factor k = image side ÷ object side, finding missing sides in similar figures, and applying the area ratio k². One worked example, one guided fill-in, then eight independent problems.
1. I do, fully worked example
Read every line. Each step has a short reason so you can see why, not just what.
Problem. △ABC ∼ △PQR. Given AB = 4 cm, PQ = 6 cm, and AC = 5 cm. Find the scale factor k and the length PR.
Step 1, Identify the corresponding pair you can use to find k.
AB corresponds to PQ (both come first in the similarity statement).
Reason: vertex order in △ABC ∼ △PQR pairs A↔P, B↔Q, C↔R. So AB↔PQ and AC↔PR.
Step 2, Calculate k (image ÷ object).
k = PQ ÷ AB = 6 ÷ 4 = 1.5
Reason: scale factor is always image side ÷ object side, taken in the same direction for every pair.
Step 3, Apply k to the unknown side.
PR = AC × k = 5 × 1.5 = 7.5 cm
Reason: object side × k = image side. The same k works for every pair of corresponding sides.
Step 4, Sanity check.
7.5 ÷ 5 = 1.5 = k ✓
Reason: the ratio image:object should be the SAME for every corresponding pair.
Answer: k = 1.5, PR = 7.5 cm.
2. We do, fill in the missing steps
Same shape as Section 1, with the working faded. Fill in each blank. 4 marks
Problem. Two similar rectangles. The small rectangle has length 8 cm and area 24 cm². The large rectangle has length 12 cm. Find (i) the scale factor k and (ii) the area of the large rectangle.
Step 1, Identify a corresponding pair of LENGTHS: small length 8 cm ↔ large length ______ cm.
Step 2, Calculate k (image ÷ object):
k = ______ ÷ ______ = ______
Step 3, For area, use the AREA RATIO rule: areas scale by ______, not by k.
Area of large = Area of small × ______ = 24 × ______ = ______ cm²
Step 4, Sanity check: a length-1.5× rectangle should have an area about ______ times bigger (not just 1.5 times).
3. You do, independent practice
Show working. The first four are foundation (just find k or one missing side). The middle two are standard (k with decimals, or area ratio). The last two are extension (reduction, perimeter vs area).
Foundation, find k or a missing side
3.1 △ABC ∼ △PQR. AB = 6 cm, PQ = 9 cm. Find the scale factor k. 1 mark
3.2 Two similar triangles have corresponding sides 4 cm and 10 cm. Another side of the smaller triangle is 7 cm. Find the corresponding side of the larger triangle. 1 mark
3.3 A shape has a side of 10 cm. Its similar image has the corresponding side of 15 cm. Find k. 1 mark
3.4 Two similar figures have k = 3. The smaller figure has perimeter 20 cm. Find the perimeter of the larger figure. 1 mark
Standard, decimals and area ratio
3.5 △ABC ∼ △XYZ. AB = 5 cm, BC = 8 cm, AC = 6 cm, and XY = 7.5 cm. Find k, then find YZ and XZ. 2 marks
3.6 Two similar shapes have a linear scale factor of k = 4. State the ratio of their areas (larger : smaller). 2 marks
Extension, reduction and dimension analysis
3.7 Two similar triangles have scale factor k = 2.5 (large to small). The smaller triangle has area 16 cm². Find the area of the larger triangle, showing your use of the area ratio rule. 2 marks
3.8 A photo is REDUCED so that what was 20 cm wide becomes 8 cm wide. (i) Find k (image ÷ object). (ii) If the original area was 600 cm², find the reduced area. (iii) If the original perimeter was 70 cm, find the reduced perimeter. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2, We do (rectangles fill-in)
Step 1: small 8 cm ↔ large 12 cm.
Step 2: k = 12 ÷ 8 = 1.5.
Step 3: areas scale by k², not by k. Area of large = 24 × k² = 24 × 2.25 = 54 cm².
Step 4: a length-1.5× rectangle has an area about 2.25 times bigger (k² = 1.5² = 2.25).
3.1, AB = 6, PQ = 9
k = PQ ÷ AB = 9 ÷ 6 = k = 3/2 = 1.5.
3.2, sides 4 and 10, plus 7
k = 10 ÷ 4 = 2.5. Missing side = 7 × 2.5 = 17.5 cm.
3.3, sides 10 and 15
k = 15 ÷ 10 = k = 1.5.
3.4, perimeter scaling
Perimeter scales by k (linear measure), so larger perimeter = 20 × 3 = 60 cm.
3.5, △ABC ∼ △XYZ
k = XY ÷ AB = 7.5 ÷ 5 = 1.5.
YZ = BC × k = 8 × 1.5 = 12 cm.
XZ = AC × k = 6 × 1.5 = 9 cm.
3.6, area ratio when k = 4
Area ratio = k² = 4² = 16, so areas are in ratio 16 : 1.
3.7, k = 2.5, small area 16 cm²
Area ratio = k² = 2.5² = 6.25.
Large area = 16 × 6.25 = 100 cm².
3.8, photo reduced 20 → 8
(i) k = 8 ÷ 20 = 0.4 (a reduction, k < 1).
(ii) Area scales by k² = 0.4² = 0.16, so reduced area = 600 × 0.16 = 96 cm².
(iii) Perimeter scales by k = 0.4, so reduced perimeter = 70 × 0.4 = 28 cm.