A lamppost is hard to measure directly, but its shadow lies on the ground. Nearby, you can place a short vertical rod of known height. Why should comparing the shadows tell you anything about the post?
Similar triangles have equal corresponding angles and proportional corresponding sides. Once similarity is established correctly, one unknown length can therefore be found using another, even without reaching the object. You will use triangle congruence and proportional intercepts.
Match first, divide second
Writing ABC∼PQR specifies A↔P, B↔Q, C↔R. Side AB corresponds to PQ, BC to QR, and CA to RP. Matching angles at the endpoints determine this, not which side looks like the “bottom” of a drawing.
Define the scale factor from ABC to PQR as:
k = (PQ)/(AB) = (QR)/(BC) = (RP)/(CA) > 0.
The diagram takes 3–4–5 to 6–8–10, so k=2. In reverse, the factor is 1/2. With k=1 the triangles are congruent. Rotation or reflection does not prevent similarity; stretching only horizontally usually destroys it. AMSI.
ABC ∼ PQR, k = 2
Shadows: equal angles, scale factor 4
You need not check everything
For nondegenerate triangles, any one test suffices:
- AA, two angles. Two angles in one triangle respectively equal two in the other. The third angles then agree because the sum is 180°. Euclid VI.4.
- SAS, two sides and their included angle. Two corresponding side ratios agree, and the angles between those sides are equal. VI.6.
- SSS, three sides. All three ratios of corresponding sides agree. VI.5.
The idea behind AA: on a side of the larger triangle, mark the corresponding side length of the smaller one and draw a parallel to another side. The cut-off triangle is similar to the larger through proportional intercepts and equal angles; it is congruent to the smaller by a side and the adjacent angles. Thus the required proportion applies to all three sides.
Two proportional sides alone do not fix a shape: their included angle can vary. One equal angle is insufficient too. In SAS, you cannot substitute an arbitrary other angle.
What else changes by a factor of k?
Corresponding altitudes, medians, angle bisectors and perimeters scale just like the sides. For example, 3+4+5=12 becomes 6+8+10=24. For altitudes, the corresponding vertex and the perpendicular to the corresponding side form similar right triangles. Angles are not multiplied by k: they remain unchanged.
This statement concerns lengths. Areas follow a different scaling law, which we will study in the area lesson.
Height from a shadow: when the model works
For a rod of height h with shadow s and a post of height H with shadow S, assume:
- Both are vertical on the same local horizontal plane.
- The shadows are measured simultaneously, from each base to its corresponding tip, without overlap from other shadows.
- We treat sunlight rays over this small area as parallel.
Then there are two right triangles with the same acute angle determined by the rays. By AA they are similar, so:
(H)/(h) = (S)/(s), H = h(S)/(s).
With h=1.5 m, s=2 m and S=8 m, we get H=1.5·8/2=6 m. First convert every length to the same units. Parallel light is an approximation, not a claim about every ray from every light source. The argument does not automatically carry over to a nearby lamp. Illustrative Mathematics: assumptions and limits.
An honest answer with imperfect measurements
For this estimate only, take h=1.5 m exactly, the short shadow between 1.95 and 2.05 m, and the long one between 7.9 and 8.1 m. The smallest calculated height uses the shortest long shadow and longest short shadow; the largest uses the reverse:
H_(min) = (1.5 ⋅ 7.9)/(2.05), H_(max) = (1.5 ⋅ 8.1)/(1.95).
Rounding the bounds outward to centimetres gives 5.78–6.24 m. This is a range under the specified errors in two lengths, not a statistical confidence interval. It does not include a tilted post, uneven ground or uncertainty in the rod’s height. A real measurement therefore starts by checking assumptions, before multiplying three numbers.