Examine the angles of both triangles:
\[ \angle A = \angle S' \] \[ \angle B = \angle O \]
The triangles satisfy AA (Angle-Angle) similarity criterion.
Set up the proportion of corresponding sides:
\[ \frac{AC}{S'O} = \frac{AB}{SO} = \frac{BC}{OS'} \]
Plug in known values and solve for \( x \):
\[ \frac{8}{4} = \frac{10}{x} \Rightarrow x = 5 \]
The mirror creates similar triangles due to reflection laws.
Set up the proportion:
\[ \frac{\text{Girl's height}}{\text{Distance from mirror}} = \frac{\text{Lamp height}}{\text{Distance to lamp}} \]
Plug in values:
\[ \frac{1.25}{2.5} = \frac{h}{6.6} \]
Cross-multiply:
\[ 1.25 \times 6.6 = 2.5 \times h \]
Solve:
\[ h = \frac{8.25}{2.5} = 3.3 \text{ meters} \]
Convert all measurements to same units (meters):
\[ 400 \text{ cm} = 4 \text{ m} \]
The triangles formed are similar (AA criterion - right angles and common sun angle).
Set up proportion:
\[ \frac{\text{Stick height}}{\text{Stick shadow}} = \frac{\text{Tower height}}{\text{Tower shadow}} \]
Plug in values:
\[ \frac{6}{4} = \frac{h}{28} \]
Solve:
\[ h = \frac{6 \times 28}{4} = 42 \text{ m} \]
From the diagram, triangles \( QPT \) and \( SRT \) are similar (AA criterion).
Write the proportion:
\[ \frac{PT}{ST} = \frac{TQ}{TR} \]
Cross-multiply:
\[ PT \times TR = ST \times TQ \]
Both triangles share angle \( A \) and have right angles (\( \angle C \) and \( \angle DEA \)).
They are similar by AA criterion.
Given \( AB = 13 \) cm, \( AC = 5 \) cm, \( BC = 12 \) cm (Pythagorean triple).
Set up proportions:
\[ \frac{AE}{AC} = \frac{AD}{AB} \Rightarrow \frac{AE}{5} = \frac{4}{13} \Rightarrow AE \approx 1.54 \text{ cm} \]
\[ \frac{DE}{BC} = \frac{AD}{AB} \Rightarrow \frac{DE}{12} = \frac{4}{13} \Rightarrow DE \approx 3.69 \text{ cm} \]
Scale factor \( k = \frac{PQ}{BC} = \frac{4}{8} = 0.5 \).
Find \( CA \):
\[ \frac{AP}{AB} = k \Rightarrow \frac{2.8}{6.5} \approx 0.43 \]
\[ CA = \frac{AP}{k} = \frac{2.8}{0.5} = 5.6 \text{ cm} \]
Find \( AQ \):
\[ AQ = AC - QC = 5.6 - 2.8 = 2.8 \text{ cm} \]
All three triangles are similar by AA criterion (right angles and shared angles).
From similarity \( \Delta QMO \sim \Delta RPN \):
\[ \frac{QM}{RP} = \frac{QO}{RN} \]
Since \( RP = QO = QR \):
\[ QR^2 = MQ \times RN \]
Area ratio \( = \frac{9}{16} \), so scale factor \( k = \sqrt{\frac{9}{16}} = \frac{3}{4} \).
Find \( EF \):
\[ \frac{BC}{EF} = \frac{3}{4} \Rightarrow EF = \frac{4 \times 2.1}{3} = 2.8 \text{ cm} \]
The triangles formed are similar (AA criterion).
Set up proportion for first pole:
\[ \frac{6}{x} = \frac{3}{y} \]
If distance between poles is given as \( d \), then:
\[ y = \frac{x}{2} \]
\[ x + y + d = \text{total shadow length} \]
Draw the original \( \triangle PQR \).
Choose point \( P \) as the center of scaling.
Measure \( PQ \) and mark \( PQ' = \frac{2}{3}PQ \).
Measure \( PR \) and mark \( PR' = \frac{2}{3}PR \).
Connect \( Q'R' \) to complete the similar triangle.
Draw the original \( \triangle LMN \).
Choose point \( L \) as the center of scaling.
Measure \( LM \) and mark \( LM' = \frac{4}{5}LM \).
Measure \( LN \) and mark \( LN' = \frac{4}{5}LN \).
Connect \( M'N' \) to complete the similar triangle.
Draw the original \( \triangle ABC \).
Extend side \( AB \) beyond \( B \).
Mark \( AB' = \frac{6}{5}AB \).
From \( B' \), draw a line parallel to \( BC \) intersecting extended \( AC \) at \( C' \).
\( \triangle AB'C' \) is the required enlarged triangle.
Draw the original \( \triangle PQR \).
Extend side \( PQ \) beyond \( Q \).
Mark \( PQ' = \frac{7}{3}PQ \).
From \( Q' \), draw a line parallel to \( QR \) intersecting extended \( PR \) at \( R' \).
\( \triangle PQ'R' \) is the required enlarged triangle.