Same shape, different size — so the sides come in a fixed ratio.
The problem
A 6 m flagpole casts a 4 m shadow. At the same moment a building casts a 26 m shadow. How tall is the building?
Same sun means similar triangles, so height/shadow matches: h26 = 64, giving h = 26 × 1.5 = 39 m. B inverts the ratio. The pole is taller than its own shadow, so the building must beat 26 m as well.
You never need to check all three, or any of the sides.
90° at the ground
same angle of elevation
Two matching angles force the third, since all three must total 180°. That is the AA rule.
Similar is not the same as congruent. Congruent triangles are identical; similar ones are photocopies at a different zoom.
heightshadow : 64 = h26
Writing 64 = 26h mixes the roles and gives 17.3 m — a building shorter than its own shadow when the flagpole is taller than its own. Nonsense, and easy to spot.
4h = 6 × 26
cross-multiply
4h = 156
tidy
h = 39 m
÷ 4
The scale factor is 26 ÷ 4 = 6.5, and 6 × 6.5 = 39. Same answer, different route.
The mirror version. A mirror lies flat on the ground. We stand 2 m from it with our eyes 1.8 m up, and see the top of a building whose base is 15 m beyond the mirror. Light leaves the mirror at the angle it arrives, so the two right triangles are similar: eye height/our distance = building/its distance, 1.82 = h15, and h = 15 × 1.8 ÷ 2 = 13.5 m. Each height sits over the distance on its own side of the mirror.
| Rule | What you need |
|---|---|
| AA | two pairs of equal angles |
| SSS | all three side pairs in the same ratio |
| SAS | two side pairs in ratio, with equal angles between |
AA does almost all the work in exam questions. Look for a shared angle, a right angle, or parallel lines making equal angles.
A line drawn parallel to one side of a triangle cuts off a smaller triangle similar to the original — a setup that appears constantly.
sides
× k
perimeter
× k
area
× k²
sides in the ratio 2 : 3
k = 32
areas in the ratio 4 : 9
the squares
“The sides are twice as long, so the area is twice as big” is wrong — it is four times as big. Perimeter scales by k, area by k².
Backwards: areas 25 and 49 mean sides in the ratio √25 : √49 = 5 : 7. A side of 10 on the smaller triangle matches 10 × 75 = 14 on the larger. Take the square root first.
In △ABC, DE ∥ BC with D on AB and E on AC. AD = 3, DB = 5 and DE = 6. Find BC.
△ADE ~ △ABC
shared ∠A, corresponding angles
AB = 3 + 5 = 8
the big triangle’s side is the whole
AD/AB = DE/BC → 38 = 6BC
small over big, both times
BC = 16
3 · BC = 48
For the two sides that get cut, the pieces work: AD/DB = AE/EC (so with EC = 10, 35 = AE10 and AE = 6). But DE and BC are whole sides, so they must be compared with AD and the whole AB — using 35 = 6BC gives 10, the classic wrong answer.
In △ABC, D is on AB and E is on AC, and ∠ADE = ∠C (DE is not parallel to BC). AD = 4, AE = 5 and AC = 10. Find AB.
△ADE ~ △ACB
shared ∠A, and ∠D matches ∠C
AD/AC = AE/AB
A↔A, D↔C, E↔B
410 = 5AB, 4 · AB = 50
cross-multiply
AB = 12.5
÷ 4
Treating DE as parallel and writing AD/AB = AE/AC gives 4AB = 510 and AB = 8. The equal angle sits at D but matches C, so the small triangle is turned over: AD now pairs with AC. Let the equal angles, not the picture, decide the letters.
△ABC ~ △DEF with AB = 4, DE = 6, BC = x + 1 and EF = 2x − 1. Find BC and EF.
46 = x + 12x − 1
AB ↔ DE, BC ↔ EF
4(2x − 1) = 6(x + 1)
cross-multiply whole expressions
8x − 4 = 6x + 6, x = 5
− 6x, + 4, ÷ 2
BC = 6, EF = 9
69 = 46 ✓
Different: x sits inside two sides. Brackets around each expression before cross-multiplying, and the answer is the sides, not x.
Right △ABC has its right angle at C. The altitude CD meets the hypotenuse at D, with AD = 4 and DB = 9. Find CD.
△ACD ~ △CBD ~ △ABC
each has a right angle and shares one more
AD/CD = CD/DB
short leg over long leg, both small triangles
CD² = 4 × 9 = 36
cross-multiply
CD = 6
the altitude is the mean of the two pieces
The legs follow the same pattern with the whole hypotenuse: AC² = AD × AB = 4 × 13 = 52 and BC² = DB × AB = 9 × 13 = 117. Check: 52 + 117 = 169 = 13² ✓
Similarity hides in three set-ups: a line parallel to one side, a shadow or mirror story, and the altitude to a hypotenuse. Name the two triangles with matching letters in the same order, then write one ratio. For area or volume questions convert the ratio first — square for areas, square root to go back. Ten-second check: the bigger triangle must get the bigger answer.
The problem
Two similar triangles have sides in the ratio 3 : 5. If the smaller has area 18 cm², what is the area of the larger?
Areas scale by the square of the side ratio: (53)² = 259, so 18 × 259 = 50 cm². C applies the side ratio directly — the commonest error in similar figures. Lengths scale by k, areas by k², volumes by k³.
