Applications

How tall is the lamp-post outside your house? You cannot climb it with a measuring tape. The flagpole at school? Too high. A tall tree in the field? No ladder reaches.

So here is the question. How do you measure a height you cannot touch?

The answer is hiding on the ground. It is the shadow.

Stop scrolling. Look at any tall object near you in sunlight. Picture its shadow stretching across the ground. We are going to turn that shadow into a ruler.

Here is the idea. An upright object and its shadow make a triangle. A taller object's shadow makes a bigger triangle of the same shape. When two triangles have the same shape, their sides keep the same ratio. That single fact lets you find any height from a height you already know.

Let me define the words before we use them.

• Similar (symbol ~): two triangles are similar if they have the same shape — same angles — but maybe different sizes. We write $\triangle ABC \sim \triangle PQR$.
• $\triangle$: the symbol for "triangle".
• $\angle$: the symbol for "angle".
• Corresponding sides: sides that sit in the same position in two similar triangles (the side facing the equal angle in each).
• Proportional: in the same ratio. If $\dfrac{a}{p} = \dfrac{b}{q}$, the sides are proportional.

You can now say what this whole topic is about: using the shape of a shadow to measure a height.

We need one tool first. It is the rule that tells us when two triangles are similar.

The main tool is the AA criterion (Angle-Angle): if two angles of one triangle equal two angles of another triangle, the two triangles are similar. That is enough. You do not need to check the third angle — it follows automatically, because all three angles of a triangle add to $180^\circ$.

There are two other criteria you should know by name:

• SSS (Side-Side-Side): all three pairs of sides in the same ratio → similar.
• SAS (Side-Angle-Side): two pairs of sides in the same ratio, with the angle between them equal → similar.

For shadow problems, AA is the workhorse. Almost everything below uses it.

Stop scrolling. Try it in your head: two triangles each have an angle of $90^\circ$ and an angle of $40^\circ$. Are they similar?

(Yes — two equal angles, so by AA they are similar.)

Example 1 — The pole and its shadow

A pole stands straight up. In the sunlight it casts a shadow on the ground. The pole, the ground, and the slanting ray of sunlight from the top of the pole to the shadow's tip make a triangle.

What do you notice? The pole meets the ground at a right angle, $90^\circ$. So one angle of this triangle is fixed.

Example 2 — Two poles, same sunlight

Now put a second, taller pole nearby, at the same time of day. It also casts a shadow. It also stands at $90^\circ$ to the ground.

Here is the key. The sun is very far away, so its rays come in parallel. The ray hitting the top of the short pole and the ray hitting the top of the tall pole travel at the same slant. So the angle the shadow makes with the ground is the same for both poles.

So both triangles share:

• a right angle ($90^\circ$, the upright meeting the ground), and
• the equal "sun angle" at the shadow tip.

Two equal angles. By AA, the two triangles are similar.

What do you notice? Same shape, different size. So the sides keep the same ratio.

Example 3 — Putting numbers on it (the sun)

A $6$ m pole casts a shadow of $4$ m. At the same time, a tower casts a shadow of $28$ m. How tall is the tower?

Understand it first. What is given: pole height $6$ m, pole shadow $4$ m, tower shadow $28$ m. What is required: tower height. Call it $h$.

Convert words to maths. The two triangles are similar (AA, as we just argued). So corresponding sides are proportional:

$\dfrac{\text{height}}{\text{shadow}} = \dfrac{\text{height}}{\text{shadow}}$

$\dfrac{6}{4} = \dfrac{h}{28}$

Solve step by step.

$h = \dfrac{6}{4} \times 28 = \dfrac{6 \times 28}{4} = \dfrac{168}{4} = 42$

So the tower is $42$ m tall.

Interpret it. Does this make sense? The tower's shadow ($28$ m) is $7$ times the pole's shadow ($4$ m). So the tower should be $7$ times the pole's height: $7 \times 6 = 42$ m. It matches. Good.

What do you notice? You never measured the tower. You measured a pole and two shadows on the ground.

Example 4 — The lamp-post at night (a moving girl)

Sunlight gives parallel rays. A lamp-post is different — its light spreads out from one bulb. But the same idea still works, because the upright girl and the upright lamp-post are both vertical, and they share the angle at the tip of the shadow.

A girl $90$ cm tall walks away from a lamp-post $3.6$ m high. She walks at $1.2$ m/s. After $4$ seconds, how long is her shadow?

Understand it. Picture the lamp-post $AB$ standing tall. The girl $CD$ stands upright somewhere to the side. The light from the top of the post $A$ grazes the top of the girl's head $C$ and lands on the ground at $E$, the tip of her shadow.

• $AB = 3.6$ m (lamp-post)
• $CD = 0.90$ m (girl — convert $90$ cm to $0.9$ m so units match)
• $BD = 1.2 \times 4 = 4.8$ m (how far she has walked from the post)
• $DE = x$ = her shadow length (what we want)

The two triangles are $\triangle ABE$ (big — post and full ground) and $\triangle CDE$ (small — girl and her shadow). Both have a right angle (post and girl are vertical). Both share the angle at $E$. So by AA, $\triangle ABE \sim \triangle CDE$.

Convert to maths. Corresponding sides proportional. The side from the shadow tip along the ground over the upright height:

$\dfrac{BE}{DE} = \dfrac{AB}{CD}$

Note $BE = BD + DE = 4.8 + x$, and $DE = x$:

$\dfrac{4.8 + x}{x} = \dfrac{3.6}{0.9} = 4$

Solve.

$4.8 + x = 4x$ $4.8 = 3x$ $x = 1.6$

So her shadow is $1.6$ m long.

Interpret. The lamp-post is $4$ times the girl's height, so the part of the ground to the post ($4.8 + 1.6 = 6.4$ m) should be $4$ times the shadow ($1.6$ m). And $4 \times 1.6 = 6.4$. It checks out.

This is the part that trips people: $BE$ is the whole base of the big triangle, post to shadow tip — that is $BD + DE$, not just $BD$. Forget the $+x$ and your answer collapses.

You can now set up and solve a shadow problem with either the sun or a lamp-post.

These two look alike but are not the same. Read slowly.

Sun rays vs lamp-post light.

• The sun is so far away its rays arrive parallel. Two poles in sunlight get the equal angle from the parallel rays themselves.
• A lamp-post's light spreads out from one point. The equal angle there is the shared angle at the shadow tip (both the girl's triangle and the post's triangle meet at that one point $E$).

Different reason for the equal angle — but in both cases you still get two equal angles, so AA still applies. The setup of the proportion is what changes.

This is the part that trips people. In the parallel-sun case you compare two separate objects, each with its own shadow. In the lamp-post case the small triangle sits inside the big triangle, so the big base is the small base plus the gap ($BE = BD + DE$). Do not blindly copy one method onto the other.

You can now spot which kind of shadow problem you are facing before you write a single number.

Strip away the story and one clean rule remains.

For two upright objects lit the same way, height divided by shadow is the same number.

$\dfrac{\text{height of object 1}}{\text{shadow of object 1}} = \dfrac{\text{height of object 2}}{\text{shadow of object 2}}$

Or, with symbols, if heights are $h_1, h_2$ and shadows are $s_1, s_2$:

$\dfrac{h_1}{s_1} = \dfrac{h_2}{s_2}$

That common value — height over shadow — is fixed at any one moment. It is the slant of the light turned into a number.

To use it:

1. Identify your known object (height and shadow both known).
2. Identify your unknown (one of height or shadow missing).
3. Write the two fractions equal.
4. Cross-multiply and solve.

The lamp-post version is the same rule, just remember the big base is "gap + shadow" ($BE = BD + DE$).

Stop scrolling. A $2$ m stick casts a $3$ m shadow. A wall's shadow is $15$ m. Find the wall's height before reading on.

($\dfrac{2}{3} = \dfrac{h}{15}$, so $h = \dfrac{2}{3}\times 15 = 10$ m.)

You can now write the height-to-shadow proportion straight from any sunlight problem.

Why should height-over-shadow stay the same? Why not change from pole to pole?

Here is the why. Similar triangles have all corresponding sides in the same ratio. If the big triangle's sides are each $k$ times the small triangle's sides, then:

• the upright side is $k$ times bigger, AND
• the shadow side is $k$ times bigger.

So when you divide upright by shadow, the $k$ cancels:

$\dfrac{k \times (\text{small height})}{k \times (\text{small shadow})} = \dfrac{\text{small height}}{\text{small shadow}}$

The scaling factor disappears. What is left — height over shadow — depends only on the shape (the slant of the light), not the size. That is why every object lit by the same sun, at the same moment, shares one ratio.

Stop scrolling. Try it in your head: if a triangle is scaled to be $3$ times bigger, does its height-to-base ratio change? (No — both height and base triple, so the ratio is unchanged.)

You can now explain why the trick works, not just how to use it.

A few real gotchas. Name them before you meet them.

Units must match. The girl was $90$ cm; the post was $3.6$ m. You must convert to the same unit ($90$ cm $= 0.9$ m) before dividing. Mix cm and m and your ratio is wrong by a factor of $100$.

Same time of day. The height-to-shadow rule only holds for objects measured at the same moment. As the sun moves, shadows lengthen and the ratio changes. A morning shadow and a noon shadow do not compare.

The lamp-post base is gap + shadow. Said again because it is the number-one error: $BE = BD + DE$, not $BD$.

Keep height on top for both. Do not flip one fraction. Height/shadow $=$ height/shadow, never height/shadow $=$ shadow/height.

You can now avoid the four mistakes that ruin most shadow answers.

Look at what you have built. You started with the AA criterion — two equal angles make triangles similar. You saw that an upright object and its shadow form a triangle, and a same-shape object forms a similar triangle. Similar triangles share equal ratios of corresponding sides. That gave you the height-to-shadow rule, and you proved why the ratio is fixed: the scaling factor cancels.

This is the heart of similarity: it lets you measure the unreachable. The same proportion idea returns whenever two figures share a shape — in maps, in scale drawings, in the mirror method of finding heights. You now hold the key tool.

Here is the lamp-post scene as similar triangles. Watch how the small triangle (girl + shadow) sits inside the big one (post + ground).

Spot it around you: next time you pass a pole's shadow at noon, or a tree's shadow stretching across a field, you are looking at one side of a triangle you can now measure.