Applications of Similarity

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Learning goals

Justify shadow and mirror triangle correspondence from explicit geometric and physical assumptions
Determine inaccessible distances using established sight lines and angle relationships
Convert units before applying map and drawing scale factors in either direction
Derive perimeter scaling by adding corresponding scaled side lengths
Evaluate model assumptions and report measurements with appropriate precision

Begin with the model, then the proportion

A physical situation does not become a pair of similar triangles just because a sketch looks convincing. Identify the straight segments, establish two corresponding angle pairs, and name the matched lengths. Then AA similarity supplies the proportion. Finally check its direction, units, and whether the reported precision fits the measurements.

If a required assumption is not established, the proposed answer is not guaranteed. That does not mean every possible realization is impossible or that the numerical answer must always be wrong. It means the reasoning has not yet justified it.

Shadows compare vertical heights with their own shadows

Use vertical objects on level ground, measured at the same time, with sunrays treated as parallel. Each object makes a right triangle: its vertical height, its own shadow from base to tip, and the ray from its top to that tip. The base angles at the shadow tips agree because the ground is level and the sunrays are parallel. Together with the right angles at the bases, this establishes AA.

Worked example 1 Measure a tree using a calibration pole

Problem. A vertical calibration pole is $1.50$ m tall and casts its own $1.20$ m shadow. At the same time, a vertical tree casts a $16.8$ m shadow on the same level ground. Treat the sunrays as parallel. Estimate the tree’s height to the nearest tenth of a meter.

The separate side views pair height PQ with TU, and each base-to-tip shadow QR with UV.

Solution. The base right angles match and the parallel rays give matching tip angles. Thus $\triangle PQR\sim\triangle TUV$ , with $P$ to $T$ , $Q$ to $U$ , and $R$ to $V$ . Height corresponds to height, and shadow to shadow.

\begin{aligned}\frac{H}{16.8}&=\frac{1.50}{1.20},\\H&=16.8\cdot1.25,\\&=21.0\text{ m}.\end{aligned}

The estimated tree height is approximately $21.0$ m. The tree’s shadow starts at the tree base, not at the calibration pole. A sloping surface or a pole that leans would require a different justification.

Check your understanding

Which facts justify AA for the usual simultaneous shadow model?

A Both objects merely look tall B Vertical objects on level ground give right angles, and parallel sunrays give equal tip angles C The shadows must have the same length D The objects must stand at the same point

A mirror compares ground legs and eye height

Place a small horizontal mirror $M$ on level ground between the observer’s foot $F$ and the object’s base $B$ . Let $E$ be the observer’s eye, vertically above $F$ , and $T$ the object’s top, vertically above $B$ . Align the sight path $T$ to $M$ to $E$ in one vertical plane.

Use the physical reflection assumption: incoming and reflected rays make equal angles with the mirror’s perpendicular. Their acute angles with the horizontal mirror are therefore equal as well. Geometry does not establish this physical law; the model supplies it. Along with the right angles at $F,B$ , it gives $\triangle EFM\sim\triangle TBM$ by AA.

Worked example 2 Use eye height rather than total person height

Problem. Under that horizontal-mirror and equal-angle reflection model on level ground, an observer sees a vertical tower’s top in mirror $M$ . The eye height and the two ground legs are

\begin{aligned}EF&=1.62\text{ m}\\FM&=2.7\text{ m}\\MB&=17.6\text{ m}.\end{aligned}

The mirror lies between $F,B$ . Estimate the tower height $TB$ to the nearest tenth of a meter.

The mirror angles pair; ground FM matches ground MB. Neither slanted sight line is a ground distance.

Solution. The reflection assumption gives the equal acute angles at $M$ ; vertical heights on level ground give right angles at $F,B$ . AA matches $E$ with $T$ , $F$ with $B$ , and $M$ with itself. Therefore eye height matches tower height, and the two ground legs match.

\begin{aligned}\frac{TB}{1.62}&=\frac{17.6}{2.7},\\TB&=10.56\text{ m},\\TB&\approx10.6\text{ m}.\end{aligned}

The estimated tower height is approximately $10.6$ m. The hundredths digit rounds the calculated $10.56$ to $10.6$ at the requested tenth. The approximation sign reflects a measurement-model estimate rather than an exact physical height. Use the eye’s height above the ground, not the observer’s total height, and do not replace $FM$ by slanted $EM$ or whole $FB$ .

Check your understanding

In the mirror model, which proportion correctly pairs the ground and vertical legs?

TB/EF=EM/MT

TB/EF=FB/FM

TB/EF=MB/FM

TB/EF=FM/MB

Check your understanding

The mirror calculation gives 10.56 m, and the task requests the nearest tenth. Which report is appropriate?

A Approximately 10.6 m B Exactly 10.56 m C Approximately 10.5 m D Exactly 11 m

Survey an inaccessible perpendicular distance

In the top view below, $A,B,C$ lie on one straight accessible bank, with $B$ between $A,C$ . Surveying has established $AT\perp AC$ . Here, $T$ is the target across the river. Through accessible $C$ , construct $CD\perp AC$ on dry land. Choose $D$ where that line meets the sight line through $T,B$ , extended past $B$ .

The order on that sight line is $T,B,D$ . The right angles at $A,C$ match, and the angles at $B$ are vertical angles, so

\triangle TAB\sim\triangle DCB.

The correspondence is $T$ to $D$ , $A$ to $C$ , and $B$ to $B$ . Thus inaccessible $AT$ matches accessible $DC$ , while bank interval $AB$ matches bank interval $CB$ .

Worked example 3 Keep every measured segment on dry land

Problem. In a top-view river survey, $A,B,C$ are collinear with $B$ between $A,C$ , and $T,B,D$ are collinear with $B$ between $T,D$ . Surveying establishes

\begin{aligned}AT\perp AC\\CD\perp AC.\end{aligned}

Segment $CD$ is on dry land. The measured lengths are

\begin{aligned}AB&=42.0\text{ m}\\BC&=15.0\text{ m}\\CD&=11.0\text{ m}.\end{aligned}

Estimate the perpendicular river distance $AT$ to the nearest tenth of a meter. Identify why $TB$ is a different distance.

Right angles and crossing sight lines justify AA. AB, BC and CD are all accessible measurements.

Solution. AA was established before setting up a ratio. Match the perpendicular segments and the bank segments in the same order.

\begin{aligned}\frac{AT}{11.0}&=\frac{42.0}{15.0},\\AT&=30.8\text{ m}.\end{aligned}

The estimate is approximately $30.8$ m across the river along $AT$ . Segment $TB$ is a slanted sight line, not the perpendicular crossing. The construction never requires measuring $AT$ or walking to $T$ .

Maps require common units and a named direction

A scale $1:3200$ means drawing length divided by real length is $1/3200$ when both use the same unit. The reciprocal ratio, real length divided by drawing length, is $3200$ . Writing centimeters over meters without conversion does not give that dimensionless factor.

Worked example 4 Convert a map length into a ground distance

Problem. A map uses one uniform scale $1:3200$ . A straight path measures $7.4$ cm on the map. Find its real length in meters, and state the drawing-to-real and real-to-drawing factors.

Solution. The real length is $3200$ times the drawing length when both are in centimeters.

\begin{aligned}L_{\rm real}&=7.4\cdot3200,\\&=23680\text{ cm},\\L_{\rm real}&=236.8\text{ m}.\end{aligned}

The drawing-to-real factor is $3200$ ; the reverse real-to-drawing factor is $1/3200$ . Name the starting figure and target figure when choosing which reciprocal to use.

Check your understanding

A 1:5000 map shows a segment of 3 cm. Which equation uses common units correctly?

L_{\rm real}=15000\text{ cm}

L_{\rm real}=15000\text{ m}

L_{\rm real}=3/5000\text{ m}

L_{\rm real}=5000/3\text{ cm}

If a printer stretches the horizontal and vertical directions by different factors, the print has no single similarity scale. One correct width comparison cannot establish a scale for every route and boundary edge.

Perimeter scales by adding the scaled sides

Why one length factor also scales the perimeter#

For a triangular boundary with sides $5, 7, 9$ cm, the perimeter is $21$ cm. An image with length factor $2.4$ has sides $12, 16.8, 21.6$ cm. Their sum is $50.4$ cm, which is $2.4\cdot 21$ .

Every corresponding boundary edge uses the same factor. Adding those scaled edges gives the scaled perimeter.

In general, for original side lengths $a,b,c$ and a positive scale factor $k$ from the original to the image,

\begin{aligned}P_{\rm image}&=ka+kb+kc,\\&=k(a+b+c),\\&=kP_{\rm original}.\end{aligned}

The same addition works for any matching polygon boundary: add all corresponding sides once. This argument concerns lengths around the boundary.

Worked example 5 Use the boundary length on a scale plan

Problem. A triangular garden plan has boundary side lengths $5.0$ , $6.5$ , and $8.0$ cm. The uniform scale is $1:500$ . Find the planned real perimeter in meters.

The perimeter follows all three boundary sides of the plan, each at the same scale.

Solution. Add the drawing sides, then apply the common length factor to their sum.

\begin{aligned}P_{\rm drawing}&=5.0+6.5+8.0,\\&=19.5\text{ cm},\\P_{\rm real}&=500\cdot19.5,\\&=9750\text{ cm},\\P_{\rm real}&=97.5\text{ m}.\end{aligned}

The three real sides would be $25.0$ , $32.5$ , and $40.0$ m; their sum confirms $97.5$ m. This is the complete boundary length, not a single direct distance across the garden.

Applications of Similarity

Learning goals

Begin with the model, then the proportion

Shadows compare vertical heights with their own shadows

A mirror compares ground legs and eye height

Survey an inaccessible perpendicular distance

Maps require common units and a named direction

Perimeter scales by adding the scaled sides

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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