Measure the height of a barn without a ladder — using its shadow.
Ancient Greek mathematicians measured the height of the Great Pyramid by comparing its shadow to the shadow of a stick. No climbing required — just proportional reasoning and a sunny day. The same trick works on trees, barns, and flagpoles. In this lesson you'll use similar triangles and the law of reflection to solve problems that look impossible at first glance.
Two key rules for light:
1. Shadows and similar triangles: On a sunny day, all vertical objects cast shadows at the same angle. So:
$$\frac{\text{height of object A}}{\text{shadow of A}} = \frac{\text{height of object B}}{\text{shadow of B}}$$
2. Law of reflection (mirrors): The angle at which light hits a mirror equals the angle at which it bounces off — both measured from the normal (the perpendicular to the surface):
$$\theta_\text{incidence} = \theta_\text{reflection}$$
3. Mirror + similar triangles: Place a mirror flat on the ground. Your eye, the mirror, and the top of a distant object form two similar right triangles. Use the same proportion as above.
Math connection — Lesson 15 (Similar Triangles): This entire lesson is Lesson 15 in disguise! The sun ray, the object, and its shadow form a right triangle. Any other object under the same sun creates a similar triangle — same angles, proportional sides. The scale factor $k$ equals the ratio of the heights (or the ratio of the shadows).
Worked Example
Worked Example — Measuring a tree
At 11 am, a 1.5 m fence post casts a 2 m shadow. At the same moment, a tree casts a 14 m shadow. How tall is the tree?
A 2 m fence post casts a 3 m shadow. At the same time, a barn casts an 18 m shadow. How tall is the barn?
$\dfrac{2}{3} = \dfrac{h_\text{barn}}{18}$
cross multiply → divide →
Problem 2
The same barn from Problem 1 is 12 m tall. Later in the day, its shadow is now 9 m long. How long is the shadow of a 1.5 m tall person standing next to it?
$\dfrac{12}{9} = \dfrac{1.5}{s_\text{person}}$
rearrange → solve →
Problem 3
A mirror is placed flat on the ground. Mia stands 2 m from the mirror and sees the top of a flagpole reflected in it. The flagpole is 6 m from the mirror. Mia's eye height is 1.6 m. How tall is the flagpole? Hint: the reflection creates two similar triangles: (eye height, Mia's distance) and (pole height, pole's distance).
$\dfrac{1.6}{2} = \dfrac{h_\text{pole}}{6}$
set up proportion → solve →
Core Problems
Problem 4
A light ray hits a flat mirror at 42° to the normal (the perpendicular to the mirror surface). (a) At what angle does it reflect? (b) Draw a diagram showing the incoming ray, the normal, and the reflected ray. (c) If the mirror is tilted 10° clockwise, what is the new reflection angle?
law of reflection → diagram → tilted mirror: angles shift by 2× the tilt →
Problem 5
At noon the sun is almost directly overhead and a horse (height 1.6 m at the shoulder) casts a 0.8 m shadow. Three hours later the same horse casts a 4.8 m shadow. (a) Calculate the ratio height ÷ shadow at noon. What angle does this suggest for the sun? (b) Calculate the same ratio at 3 pm. (c) Has the sun's angle increased or decreased? What does this tell you about where the sun is in the sky?
ratio at noon → ratio at 3pm → interpret →
Problem 6
Mia wants to find the height of an oak tree on her family's property. She places a 1 m stick vertically in the ground. At 10 am the stick's shadow is 1.4 m. The tree's shadow falls partly on a slope, so she can only measure the horizontal shadow as 18 m. Use the stick's ratio to estimate the tree's height.
stick ratio = 1 ÷ 1.4 → set up proportion → solve →
Problem 7 Challenge
Two mirrors are placed at right angles to each other (like the corner of a room). A light ray hits mirror 1 at 35° to the normal. (a) Find the reflection angle off mirror 1. (b) This reflected ray then hits mirror 2. What angle does it make with mirror 2's normal? (Draw the geometry — the two normals are perpendicular.) (c) What direction does the final ray travel relative to the incoming ray? (d) This is how a retroreflector works — can you explain why it's useful for bicycle reflectors and satellite ranging?
reflect off M1 → find angle at M2 (angles in triangle sum to 180°) → reflect off M2 → describe final direction →
Problem 8 Open
Design a "shadow clock" (a sundial). Describe: what object you would use as the gnomon (shadow-caster), how you would mark the hours on the ground, and how the shadow's length and direction change from morning to afternoon. Sketch the positions at 8 am, 12 noon, and 4 pm. Where on Earth would this design work best — and where would it fail? (Hint: think about what happens at the equator or the poles.)
$s_\text{person} = \frac{1.5 \times 9}{12} = 1.125$ m
Problem 3
$h_\text{pole} = \frac{1.6 \times 6}{2} = 4.8$ m
Problem 4
(a) 42° (law of reflection) · (b) Diagram: incoming and reflected rays make equal angles with the normal · (c) Tilting the mirror 10° rotates the normal by 10°, so the reflected beam moves by 20° — new reflection angle = 42° (unchanged from normal) but beam direction shifts 20°.
Problem 5
(a) $1.6 \div 0.8 = 2$ — sun is high (≈ 63° above horizon) · (b) $1.6 \div 4.8 = 0.33$ — sun is low (≈ 18°) · (c) Ratio decreased — sun is lower at 3 pm than noon. The sun moves across the sky, lower in the afternoon and evening.
Problem 6
Ratio: $1 \div 1.4 = \tfrac{5}{7}$. Tree height $= 18 \times \tfrac{5}{7} \approx 12.9$ m.
Problem 7
(a) 35° (law of reflection) · (b) The angle between the two mirrors is 90°. In the triangle formed, angles sum to 180°: $90° + 35° + \theta_2 = 180°$ → $\theta_2 = 55°$ · (c) The exit ray is parallel to the incoming ray but travels in the opposite direction — it "bounces straight back" · (d) Retroreflectors return light to its source regardless of angle — ideal for safety reflectors and laser ranging (e.g. moon reflectors left by Apollo).
Science arc 1 complete! 🎉
You've connected real-world science to the maths you already know: speed–distance–time (slope and linear equations), pressure (proportionality), density and volume (area formulas), and light (similar triangles and angles). Every science formula is a maths formula with a story attached.