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Applications of Similarity: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Calibrate a shadow with its own triangle

A vertical calibration pole is $1.80$ m tall and casts its own $1.40$ m shadow. A vertical tree simultaneously casts its own $12.6$ m shadow on the same level ground. Treat the sunrays as parallel.

Estimate the tree height to the nearest tenth of a meter. Justify the angle correspondence before writing a proportion.

Use each object’s own ground shadow.

Text description of this figure

Separate schematic triangles show the calibration pole and tree, each with its own base-to-tip shadow. Only given measurements are labeled.
1. Hint 1
  Which angles follow from vertical objects and level ground?
2. Hint 2
  Use the parallel ray directions for a second angle pair, then match each height with its own shadow.
Answer

The tree is approximately $16.2$ m tall.

Full solution

Both base angles are right angles.
The parallel sunrays and level ground give equal acute angles at the shadow tips, so AA makes the triangles similar.
The tree height matches the pole height, and the tree's own shadow matches the pole's own shadow.

Use height over own shadow in both triangles.
$\begin{aligned}\frac{H}{12.6}&=\frac{1.80}{1.40},\\H&=16.2\text{ m}.\end{aligned}$

The model estimate is approximately $16.2$ m to the requested tenth.
No distance between the two object bases enters the proportion.

Answer

The tree is approximately $16.2$ m tall.

Key idea

AA must justify the paired height and own-shadow measurements.
Problem 2 Ground legs in a mirror measurement

A small horizontal mirror $M$ lies on level ground between observer foot $F$ and vertical mast base $B$ . The observer sees the mast top $T$ through $M$ , with eye $E$ vertically above $F$ . Assume equal-angle reflection and a common vertical sight plane.

The eye height is
$\begin{aligned}EF&=1.56\text{ m}\\FM&=2.4\text{ m}\\MB&=17.5\text{ m}.\end{aligned}$
Estimate mast height $TB$ to the nearest tenth of a meter. Name the two matching side pairs used.

Eye height corresponds to the mast height.

Text description of this figure

The horizontal mirror M lies between observer foot F and object base B. Given labels mark eye height 1.56 and the two ground legs 2.4, 17.5; no sight-line length is supplied.
1. Hint 1
  Right angles at F,B provide one pair; the reflection model provides the mirror-angle pair.
2. Hint 2
  Compare object height with eye height and object ground leg with observer ground leg.
Answer

$TB\approx 11.4$ m. The pairs are $TB$ with $EF$ , and $MB$ with $FM$ .

Full solution

The right angles at $F,B$ and the equal acute mirror angles establish $\triangle EFM\sim\triangle TBM$ by AA.
The matching is $E$ to $T$ , $F$ to $B$ , and $M$ to $M$ .

Use the object triangle over the observer triangle.
$\begin{aligned}\frac{TB}{1.56}&=\frac{17.5}{2.4},\\TB&=11.375\text{ m}.\end{aligned}$

To the nearest tenth, the mast is approximately $11.4$ m tall.
Slanted $EM$ is not the observer ground leg; $FM$ is.

Answer

$TB\approx 11.4$ m. The pairs are $TB$ with $EF$ , and $MB$ with $FM$ .

Key idea

Mirror similarity pairs vertical heights and ground legs separately.
Problem 3 Measure across the river from the accessible bank

In a top-view survey, $A,B,C$ are collinear in that order on an accessible straight bank. Target $T$ is across the river, with $AT\perp AC$ established. A dry-land perpendicular $CD\perp AC$ meets sight line $TB$ extended, with $T,B,D$ in order.

Given
$\begin{aligned}AB&=38.0\text{ m}\\BC&=16.0\text{ m}\\CD&=12.0\text{ m}\end{aligned}$
, estimate perpendicular crossing $AT$ to the nearest tenth. Give the ordered similarity and explain why none of the measured inputs requires reaching $T$ .

The dry-land triangle carries the inaccessible measurement.

Text description of this figure

The schematic top view supplies both collinearities and the bank perpendiculars. Blue TAB and red dashed DCB encode the stated AA correspondence without showing a calculated crossing length.
1. Hint 1
  The two bank perpendiculars give right angles; identify the crossing-angle pair at B.
2. Hint 2
  AT matches CD, not the slanted line from T to B.
Answer

$\triangle TAB\sim\triangle DCB$ , and $AT\approx 28.5$ m.

Full solution

The right angles at $A,C$ and the vertical angles at $B$ give $\triangle TAB\sim\triangle DCB$ by AA.
Thus $AT$ matches $DC$ and $AB$ matches $CB$ .

The corresponding ratio gives
$\begin{aligned}\frac{AT}{12.0}&=\frac{38.0}{16.0},\\AT&=28.5\text{ m}.\end{aligned}$

All three inputs $AB,BC,CD$ lie on the accessible bank or dry land.
Sight-line alignment locates $D$ without measuring $AT$ or $TB$ .
The requested crossing is the perpendicular $AT$ , not slanted $TB$ .

Answer

$\triangle TAB\sim\triangle DCB$ , and $AT\approx 28.5$ m.

Key idea

A survey must establish the angle relations and measure only accessible lengths.
Problem 4 Follow a map route out and back

A uniform map uses scale $1:24000$ . A three-leg walking route has map lengths $2.0$ , $3.25$ , and $3.5$ cm in order. The return trip follows the same route.

Find the one-way route length and total out-and-back length in kilometers. Explain why a straight line from the start to the end is not the requested measurement.

Add the route legs before scaling.

Text description of this figure

A three-leg schematic route has lengths 2.0, 3.25, 3.5 cm on a uniform 1:24000 map. The drawn endpoint gap is not a route segment.
1. Hint 1
  Add all three mapped route legs before applying the common scale.
2. Hint 2
  Convert the real centimeter total into kilometers, then account for the return trip.
Answer

The one-way route is $2.1$ km; the out-and-back total is $4.2$ km.

Full solution

The route follows all three legs.
$\begin{aligned}L_{\rm map}&=2.0+3.25+3.5,\\&=8.75\text{ cm}.\end{aligned}$

Let $L$ be the real one-way route length.
Use the uniform real-per-drawing factor in common units.
$\begin{aligned}L&=8.75\cdot 24000,\\&=210000\text{ cm},\\&=2.1\text{ km}.\end{aligned}$

The return covers the same route, so the total is $2(2.1)=4.2$ km.
The direct endpoint gap would omit the bends and measure a different path.

Answer

The one-way route is $2.1$ km; the out-and-back total is $4.2$ km.

Key idea

Scale the length of the requested route, including every leg actually traveled.
Problem 5 Recover a real boundary perimeter from a model

A triangular model and a real triangular boundary are similar with a fixed correspondence. A $7.2$ cm model edge matches a $2.4$ m real edge. The model perimeter is $54$ cm.

Find the factor mapping model lengths to real lengths in common units, then find the real perimeter in meters. Explain why the factor applies to the whole boundary sum.

Convert units before finding the factor.

Text description of this figure

Similar schematic triangular boundaries have a model perimeter 54 cm and matched edges 7.2 cm,2.4 m. The displayed size ratio is not measurement evidence.
1. Hint 1
  Convert the real edge into centimeters before comparing it with the model edge.
2. Hint 2
  Every corresponding boundary edge uses the same factor; add those scaled edges.
Answer

The model-to-real factor is $100/3$ ; the real perimeter is $18$ m.

Full solution

Express both matched edges in centimeters.
$\begin{aligned}2.4\text{ m}&=240\text{ cm},\\k&=\frac{240}{7.2},\\&=\frac{100}{3}.\end{aligned}$

Each real boundary side is $k$ times its matching model side.
Adding those edges therefore multiplies the model perimeter by $k$ .
$\begin{aligned}P_{\rm real}&=54\cdot\frac{100}{3},\\&=1800\text{ cm},\\P_{\rm real}&=18\text{ m}.\end{aligned}$

Answer

The model-to-real factor is $100/3$ ; the real perimeter is $18$ m.

Key idea

A common length factor multiplies perimeter because all corresponding sides are added.
Problem 6 Leena’s mirror ground distance

A horizontal mirror $M$ lies on level ground between observer foot $F$ and a vertical tower base $B$ . The eye is vertically above $F$ , the sight path obeys equal-angle reflection, eye height is $1.60$ m, $FM=2.0$ m, and the whole ground separation $FB=18.0$ m.

Leena writes $H/1.60=18.0/2.0$ and reports a tower height of $14.4$ m. Explain the mismatched length, correct the estimate to the nearest tenth, and name the matching ground legs.
1. Hint 1
  M is between the two bases. Which part of FB belongs to the tower triangle?
2. Hint 2
  Subtract the observer ground leg before comparing the object and observer triangles.
Answer

Leena used whole $FB$ instead of $MB$ . The matching ground legs are
$\begin{aligned}MB&=16.0\text{ m}\\FM&=2.0\text{ m}\end{aligned}$
; the corrected height is approximately $12.8$ m.

Full solution

The mirror AA model pairs tower height with eye height and ground $MB$ with ground $FM$ .
The whole separation $FB$ crosses both triangles and is not the tower's ground leg.

Since $F,M,B$ are in order,
$\begin{aligned}MB&=18.0-2.0,\\&=16.0\text{ m}.\end{aligned}$

The corrected proportion is
$\begin{aligned}\frac{H}{1.60}&=\frac{16.0}{2.0},\\H&=12.8\text{ m}.\end{aligned}$

The model assumptions are supplied; the error lies in which measured ground length was assigned to the tower triangle.

Subtract the observer leg before using the tower leg.

Answer

Leena used whole $FB$ instead of $MB$ . The matching ground legs are
$\begin{aligned}MB&=16.0\text{ m}\\FM&=2.0\text{ m}\end{aligned}$
; the corrected height is approximately $12.8$ m.

Key idea

A whole ground separation cannot replace one triangle’s ground leg.
Problem 7 Compare two shadow records

A vertical staff of height $2.00$ m casts a $1.60$ m shadow at noon. A vertical tree's own shadow is measured as $18.0$ m later that afternoon. Both sites are level, but no evidence says the sunray angle remained unchanged.

(a) Is the estimate $22.5$ m from these two records guaranteed? Explain and propose a repair.

(b) When the tree shadow is measured, the staff's simultaneous shadow is actually $2.40$ m. Under the parallel-ray model at that time, estimate the tree height.
1. Hint 1
  The usual AA argument needs the same ray-to-ground angle for the two records.
2. Hint 2
  Use the calibration shadow recorded at the tree measurement time.
Answer

(a) The estimate is not guaranteed; measure the calibration shadow at the same time as the tree's. (b) The corrected estimate is approximately $15.0$ m.

Full solution

Noon and afternoon need not have the same sunray angle.
The original records therefore do not establish the acute-angle pair for AA.
Their arithmetic value $22.5$ m is not guaranteed, though missing evidence alone does not make every possible coincidence impossible.

Repair the procedure by measuring both own shadows simultaneously on level ground with vertical objects.
The staff's current shadow supplies the correct calibration.

The repaired model has right base angles and equal ray-to-ground angles, hence AA.
$\begin{aligned}\frac{H}{18.0}&=\frac{2.00}{2.40},\\H&=15.0\text{ m}.\end{aligned}$

Answer

(a) The estimate is not guaranteed; measure the calibration shadow at the same time as the tree's. (b) The corrected estimate is approximately $15.0$ m.

Key idea

A changed physical angle requires a new simultaneous calibration.
Problem 8 Restore the missing survey angle

In a river survey, $A,B,C$ lie on the straight bank in order and $AT\perp AC$ is verified. A point $D$ is sighted with $T,B,D$ in order. The lengths
$\begin{aligned}AB&=32.0\text{ m}\\BC&=14.0\text{ m}\\CD&=9.0\text{ m}\end{aligned}$
are recorded, but nobody establishes $CD\perp AC$ or $CD\parallel AT$ .

(a) Explain why $AT=CD\cdot AB/BC$ is not yet justified, and describe a dry-land repair.

(b) After constructing the perpendicular through $C$ and finding its new sight-line intersection, the corrected land leg is $10.5$ m. Find the supported estimate of $AT$ .
1. Hint 1
  The crossing angle at B gives just one pair; identify which second pair is missing.
2. Hint 2
  Construct the bank perpendicular through C, then use its intersection with the extended sight line and remeasure that leg.
Answer

(a) The second equal-angle pair is missing; construct the bank perpendicular through $C$ and remeasure to its sight-line intersection. (b) $AT\approx 24.0$ m.

Full solution

The collinearities establish vertical angles at $B$ , but there is no verified right angle at $C$ .
A single equal-angle pair does not justify AA, so inserting the recorded $CD=9.0$ into the usual formula is unsupported.

On dry land, construct a perpendicular to the bank through $C$ .
Let its intersection with sight line $TB$ extended be the corrected point $D$ .
Now the right angles at $A,C$ and vertical angles at $B$ justify AA.

Use the remeasured corrected leg.
$\begin{aligned}\frac{AT}{10.5}&=\frac{32.0}{14.0},\\AT&=24.0\text{ m}.\end{aligned}$

This repair establishes the missing angle without measuring the inaccessible target distance itself.

The right-angle mark at C belongs to the repaired arrangement.

Answer

(a) The second equal-angle pair is missing; construct the bank perpendicular through $C$ and remeasure to its sight-line intersection. (b) $AT\approx 24.0$ m.

Key idea

A sight-line alignment needs a second established angle pair for similarity.
Problem 9 Check a map print

An original map has uniform scale $1:9000$ . A print changes a $5.0$ cm horizontal check segment to $6.0$ cm, but an $8.0$ cm vertical check segment to $8.8$ cm.

(a) Decide whether one new similarity scale describes this print.

(b) The map is reprinted uniformly at $120\%$ in every direction instead. Find the corrected print:real scale and the new length of the original $8.0$ cm vertical segment.
1. Hint 1
  Compare print divided by original in both directions.
2. Hint 2
  For a uniform corrected print, real lengths stay fixed while all drawing lengths receive the same multiplier.
Answer

(a) No; the factors are $1.2$ and $1.1$ . (b) The corrected scale is $1:7500$ , and the vertical segment becomes $9.6$ cm.

Full solution

The measured directional factors differ.
$\begin{aligned}k_{\rm horizontal}&=6.0/5.0,\\&=1.2,\\k_{\rm vertical}&=8.8/8.0,\\&=1.1.\end{aligned}$

Thus the first print has no one similarity factor for all lengths.
A width check alone cannot validate every map direction.

In the corrected print, every drawing length is multiplied by $1.2$ .
$\begin{aligned}\frac{L_{\rm print}}{L_{\rm real}}&=\frac{1.2}{9000},\\&=\frac 1{7500},\\L_{\rm vertical}&=1.2\cdot 8.0,\\&=9.6\text{ cm}.\end{aligned}$

Answer

(a) No; the factors are $1.2$ and $1.1$ . (b) The corrected scale is $1:7500$ , and the vertical segment becomes $9.6$ cm.

Key idea

A usable uniform print scale must agree across directions.
Problem 10 Compare two justified estimates of the same mast

Two methods measure one vertical mast on level ground. For the shadow method, a vertical $1.60$ m staff and the mast are measured simultaneously under parallel sunrays; their own shadows are $1.25$ m and $18.6$ m respectively.

For the mirror method, a small horizontal mirror lies between the observer foot and mast base, a common vertical sight plane and equal-angle reflection apply, eye height is $1.68$ m, and the observer and mast ground legs are $1.8$ m and $25.5$ m respectively.

Find each height estimate to the nearest tenth of a meter. Do the estimates agree at that precision, and what does their agreement say about the measurement check?
1. Hint 1
  Write a separate justified height-to-ground-leg proportion for each model.
2. Hint 2
  Round after each complete calculation, then compare the reported estimates without treating agreement as proof of every assumption.
Answer

The shadow and mirror estimates are both approximately $23.8$ m to the nearest tenth, so they agree at that precision.

Full solution

The shadow triangles have right base angles and equal parallel-ray tip angles, so AA pairs the two heights and their own shadows.
$\begin{aligned}H_{\rm shadow}&=18.6\cdot\frac{1.60}{1.25},\\&=23.808\text{ m}.\end{aligned}$

The mirror triangles have right base angles and equal reflection-derived mirror angles.
AA pairs heights and ground legs.
$\begin{aligned}H_{\rm mirror}&=1.68\cdot\frac{25.5}{1.8},\\&=23.8\text{ m}.\end{aligned}$

Both report approximately $23.8$ m to the nearest tenth.
The agreement is a useful consistency check between the two supplied models and records.
It supports the estimate without turning approximate measurements into an exact height or independently proving every physical premise.

Answer

The shadow and mirror estimates are both approximately $23.8$ m to the nearest tenth, so they agree at that precision.

Key idea

Independent model calculations can check consistency at the stated reporting precision.

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