The Central Idea
Two geometric figures are similar when one has the same shape as the other, possibly at a different size.
Similarity preserves angle measure and multiplies every corresponding length by the same positive number.
\[ \boxed{\text{same shape}+\text{one common length scale factor}=\text{similarity}} \]If the common scale factor is exactly \(1\), the figures have the same size as well as the same shape. In that special case, the figures are congruent.
Similar Figures
For polygons, similarity requires two coordinated conditions:
- corresponding angles have equal measure;
- corresponding side lengths are proportional by one common factor.
If polygon \(P\) is similar to polygon \(Q\), then the geometry of \(P\) can be enlarged or reduced uniformly to match the geometry of \(Q\), possibly after a translation, rotation, or reflection.
Similarity Notation
The symbol \(\sim\) is used to state that two figures are similar.
For triangles, a statement such as
\[ \triangle ABC\sim\triangle DEF \]is read as “triangle \(ABC\) is similar to triangle \(DEF\).”
The order of the letters is important because it identifies corresponding vertices.
Correspondence Order
From
\[ \triangle ABC\sim\triangle DEF, \]the correspondence is
\[ A\leftrightarrow D,\qquad B\leftrightarrow E,\qquad C\leftrightarrow F. \]Therefore, the corresponding angles are
\[ \angle A\leftrightarrow\angle D, \qquad \angle B\leftrightarrow\angle E, \qquad \angle C\leftrightarrow\angle F, \]and the corresponding sides are
\[ AB\leftrightarrow DE, \qquad BC\leftrightarrow EF, \qquad AC\leftrightarrow DF. \]Corresponding Angles
Similar figures preserve angle measure. Thus, if
\[ \triangle ABC\sim\triangle DEF, \]then
\[ \boxed{ m\angle A=m\angle D,\qquad m\angle B=m\angle E,\qquad m\angle C=m\angle F }. \]Enlarging or reducing a figure changes lengths, but it does not open or close its angles.
Corresponding Sides Are Proportional
The corresponding side lengths of similar figures all have the same ratio. If
\[ \triangle ABC\sim\triangle DEF, \]then one consistent way to write the proportionality is
\[ \boxed{ \frac{DE}{AB}=\frac{EF}{BC}=\frac{DF}{AC} }. \]Every numerator in this version comes from the second triangle, and every denominator comes from the corresponding side of the first triangle.
The Scale Factor
The common ratio between corresponding lengths is the scale factor.
If a figure is transformed from an original figure to an image and
\[ \frac{\text{image length}}{\text{original length}}=k, \]then \(k\) is the scale factor from the original to the image.
\[ \boxed{k=\frac{\text{new length}}{\text{original length}}} \]Enlargements and Reductions
The size of the scale factor tells us how the image compares with the original.
- If \(k>1\), the image is an enlargement.
- If \(0<k<1\), the image is a reduction.
- If \(k=1\), the image has the same size as the original.
In ordinary Euclidean similarity, the length scale factor is taken to be positive. Reflections can reverse orientation, but lengths themselves remain nonnegative.
Example: Finding a Scale Factor
Suppose a side of length \(8\) in the original figure corresponds to a side of length \(20\) in the image.
The scale factor from the original to the image is
\[ k=\frac{20}{8}=\frac{5}{2}=2.5. \]Therefore, every image length is \(2.5\) times the corresponding original length.
Direction Matters
Scale factor depends on the direction of comparison.
If the scale factor from figure \(A\) to figure \(B\) is
\[ \frac{5}{2}, \]then the scale factor from figure \(B\) back to figure \(A\) is the reciprocal:
\[ \boxed{\frac{2}{5}}. \]Always identify which figure is the original and which is the image before constructing a ratio.
Finding an Unknown Corresponding Length
Suppose two similar figures have scale factor
\[ k=\frac{3}{2} \]from the original to the image. If an original side has length \(14\), then its corresponding image side has length
\[ 14\left(\frac{3}{2}\right)=21. \]Thus,
\[ \boxed{\text{corresponding image length}=21}. \]Using a Proportion
Unknown lengths can also be found by setting corresponding ratios equal.
Suppose
\[ \triangle ABC\sim\triangle DEF, \]with
\[ AB=6,\qquad DE=15,\qquad BC=10, \]and let \(EF=x\). Then
\[ \frac{DE}{AB}=\frac{EF}{BC}. \]Substitute:
\[ \frac{15}{6}=\frac{x}{10}. \]Therefore,
\[ x=25. \] \[ \boxed{EF=25}. \]Keep Corresponding Parts Aligned
A proportion is useful only when corresponding quantities occupy matching positions.
If
\[ AB\leftrightarrow DE \]and
\[ BC\leftrightarrow EF, \]then a valid proportion is
\[ \frac{AB}{DE}=\frac{BC}{EF}. \]Switching one ratio but not the other destroys the correspondence.
Similarity versus Congruence
Congruence and similarity both preserve shape, but they differ in what they require about size.
- Congruent figures have the same shape and the same size.
- Similar figures have the same shape but may have different sizes.
Congruence is therefore the scale-factor-\(1\) case of similarity.
\[ \boxed{\text{congruence}=\text{similarity with }k=1} \]Dilations
A dilation is a transformation that changes every distance from a fixed center by the same scale factor.
If \(O\) is the center of dilation and point \(P\) maps to point \(P'\), then
\[ OP'=k\,OP. \]The point \(P'\) lies on the ray beginning at \(O\) and passing through \(P\) when \(k>0\).
What a Dilation Preserves
A dilation does not preserve ordinary distance unless \(k=1\), but it does preserve the shape structure needed for similarity.
Under a dilation:
- angle measures are preserved;
- collinearity is preserved;
- betweenness is preserved for positive scale factors;
- parallel lines remain parallel;
- perpendicular lines remain perpendicular;
- all lengths are multiplied by \(k\).
Similarity Transformations
A similarity transformation can be built from a dilation together with rigid motions such as translations, rotations, and reflections.
Rigid motions preserve size and shape. A dilation changes size uniformly while preserving shape. Their combination therefore produces similar figures.
\[ \boxed{\text{rigid motions}+\text{dilation}=\text{similarity transformation}} \]Orientation Does Not Determine Similarity
Similar figures do not need to face the same direction.
One figure may be translated, rotated, or reflected relative to the other. The important questions are whether corresponding angles match and whether corresponding lengths share one common scale factor.
A mirror image can therefore still be similar to the original figure.
Why Triangles Are Special
For general polygons, proving similarity can require checking both angles and side ratios. Triangles are more rigid: a small amount of information is enough to force the entire shape.
Three standard triangle-similarity criteria are:
- AA: two corresponding angles are equal;
- SAS Similarity: two pairs of corresponding sides are proportional and the included angles are equal;
- SSS Similarity: all three pairs of corresponding sides are proportional.
AA Similarity
If two angles of one triangle are equal to two corresponding angles of another triangle, then the triangles are similar.
For example, if
\[ m\angle A=m\angle D \]and
\[ m\angle B=m\angle E, \]then
\[ \boxed{\triangle ABC\sim\triangle DEF}. \]This is called Angle-Angle Similarity, or AA.
Why Two Angles Are Enough
Every Euclidean triangle has angle sum \(180^\circ\). Therefore, if two pairs of corresponding angles are equal, the third pair must also be equal.
If
\[ A+B+C=180^\circ \]and
\[ D+E+F=180^\circ, \]with \(A=D\) and \(B=E\), then necessarily
\[ C=F. \]This is why a separate AAA criterion is unnecessary: AAA reduces to AA.
Example: AA Similarity
Suppose one triangle has angles
\[ 40^\circ,\qquad 65^\circ,\qquad 75^\circ, \]and another triangle also has angles
\[ 40^\circ,\qquad 65^\circ,\qquad 75^\circ. \]The triangles have the same angular shape, so they are similar even if their side lengths are different.
SAS Similarity
Two triangles are similar by SAS Similarity when two pairs of corresponding sides are proportional and the angles included between those sides are equal.
For example, if
\[ \frac{AB}{DE}=\frac{AC}{DF} \]and
\[ m\angle A=m\angle D, \]where \(\angle A\) and \(\angle D\) are the included angles, then
\[ \boxed{\triangle ABC\sim\triangle DEF}. \]The Included Angle Matters
In SAS Similarity, the equal angle must lie between the two side pairs being compared.
Knowing two proportional sides together with some unrelated nonincluded angle does not generally force a unique triangle shape. This is the same structural issue behind the ambiguous side-side-angle situation.
Example: SAS Similarity
Suppose two triangles have side pairs
\[ 6\text{ and }9 \]and
\[ 10\text{ and }15, \]surrounding equal included angles.
The side ratios are
\[ \frac{9}{6}=\frac{15}{10}=\frac{3}{2}. \]Because the included angles are also equal, the triangles are similar by SAS Similarity.
SSS Similarity
Two triangles are similar by SSS Similarity when all three pairs of corresponding side lengths are proportional.
If
\[ \frac{DE}{AB}=\frac{EF}{BC}=\frac{DF}{AC}=k, \]then
\[ \boxed{\triangle ABC\sim\triangle DEF}. \]Example: SSS Similarity
Compare triangles with side lengths
\[ 4,\qquad 6,\qquad 8 \]and
\[ 10,\qquad 15,\qquad 20. \]Matching shortest to shortest, middle to middle, and longest to longest gives
\[ \frac{10}{4}=\frac{15}{6}=\frac{20}{8}=\frac{5}{2}. \]Therefore, the triangles are similar by SSS Similarity.
Similarity Criteria Are Sufficient Tests
The triangle-similarity criteria are powerful because they tell us that we do not need to verify every corresponding angle and side ratio separately.
- AA fixes the angular shape.
- SAS fixes the shape using two proportional sides and their included angle.
- SSS fixes the shape using three proportional sides.
Once similarity is proved, all remaining corresponding angles and side ratios follow automatically.
Equal Angles Alone Do Not Prove General Polygons Similar
The AA shortcut is special to triangles.
For example, every rectangle has four right angles, but a \(2\)-by-\(3\) rectangle and a \(2\)-by-\(5\) rectangle are not similar because their side ratios differ.
Thus, equal corresponding angles are not enough by themselves for arbitrary polygons.
Proportional Side Lengths Alone Do Not Prove General Polygons Similar
The SSS shortcut is also special to triangles.
Two quadrilaterals can have proportional corresponding side lengths while having different angle measures. For instance, rhombi can share the same side lengths but have different acute and obtuse angles.
General polygon similarity requires the full shape relationship, not merely a list of proportional side lengths.
Scale Factor and Perimeter
Perimeter is a one-dimensional measurement built by adding lengths. Therefore, if corresponding lengths scale by \(k\), perimeter also scales by \(k\).
\[ \boxed{P_{\text{image}}=kP_{\text{original}}}. \]A figure enlarged by a factor of \(3\) has three times the perimeter of the original.
Scale Factor and Area
Area is two-dimensional. If every length is multiplied by \(k\), area is multiplied by \(k^2\).
\[ \boxed{A_{\text{image}}=k^2A_{\text{original}}}. \]For example, if \(k=3\), then
\[ k^2=9. \]The enlarged figure has nine times the area of the original, not three times the area.
Scale Factor and Volume
For similar three-dimensional figures, volume scales with the cube of the linear scale factor.
\[ \boxed{V_{\text{image}}=k^3V_{\text{original}}}. \]If \(k=2\), then corresponding lengths double, corresponding areas become four times as large, and corresponding volumes become eight times as large.
The Dimensional Scaling Pattern
Similarity gives a compact hierarchy that connects geometric dimension to scaling.
- length scales by \(k\);
- area scales by \(k^2\);
- volume scales by \(k^3\).
This pattern was developed earlier through dimensional scaling laws. In similarity problems, it becomes a direct consequence of one geometric scale factor acting uniformly on the figure.
Example: Perimeter and Area Scaling
A polygon has perimeter \(18\) and area \(24\). A similar polygon is produced with scale factor \(k=4\).
The new perimeter is
\[ 4(18)=72. \]The new area is
\[ 4^2(24)=16(24)=384. \]Therefore,
\[ \boxed{P=72,\qquad A=384}. \]Recovering a Linear Scale Factor from Areas
Sometimes an area ratio is known instead of a length ratio.
If two similar figures have area ratio
\[ \frac{A_2}{A_1}=25, \]then
\[ k^2=25. \]Since a geometric length scale factor is positive,
\[ \boxed{k=5}. \]Recovering a Linear Scale Factor from Volumes
If two similar solids have volume ratio
\[ \frac{V_2}{V_1}=64, \]then
\[ k^3=64. \]Therefore,
\[ \boxed{k=4}. \]Scale Drawings
A scale drawing represents an object using a consistent ratio between drawing lengths and actual lengths.
A scale written as
\[ 1\text{ cm}:5\text{ m} \]means that every \(1\) centimeter measured on the drawing represents \(5\) meters in the actual object.
The units differ, so unit conversion must be handled carefully before a dimensionless numerical scale factor is formed.
Example: Reading a Scale Drawing
A plan uses the scale
\[ 1\text{ cm}:4\text{ m}. \]If a wall measures \(7.5\) centimeters on the plan, its actual length is
\[ 7.5(4)=30\text{ m}. \]Thus,
\[ \boxed{\text{actual wall length}=30\text{ m}}. \]Maps and Models
Maps and physical models use the same similarity principle as scale drawings.
- A map reduces large real-world distances to manageable page distances.
- An architectural model reduces a building while preserving proportional dimensions.
- A scientific model may enlarge a small object so that its structure is easier to inspect.
In each case, the model is useful only if the relevant dimensions are scaled consistently.
Indirect Measurement
Similarity can determine a distance that is difficult or impossible to measure directly.
If a measurable triangle can be shown similar to a triangle containing the unknown distance, corresponding side ratios create an equation for that distance.
This method appears in surveying, shadow measurements, navigation, optics, and many geometric constructions.
Example: Measuring with Shadows
At the same moment, a \(1.8\)-meter vertical pole casts a \(2.4\)-meter shadow while a vertical tree casts a \(10\)-meter shadow.
Under the same sunlight, the pole-and-shadow triangle and tree-and-shadow triangle have equal corresponding angles and are similar.
Let the tree height be \(h\). Then
\[ \frac{h}{10}=\frac{1.8}{2.4}. \]Therefore,
\[ h=10\left(\frac{1.8}{2.4}\right)=7.5. \] \[ \boxed{h=7.5\text{ m}}. \]Parallel Lines Often Create Similar Triangles
A line parallel to one side of a triangle frequently creates a smaller triangle similar to the original.
Suppose \(D\) lies on \(\overline{AB}\), \(E\) lies on \(\overline{AC}\), and
\[ DE\parallel BC. \]Corresponding and alternate angle relationships give two pairs of equal angles, so
\[ \boxed{\triangle ADE\sim\triangle ABC} \]by AA Similarity.
Proportional Segments from Parallel Lines
Once the triangles in the previous configuration are known to be similar, corresponding sides are proportional:
\[ \frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}. \]Related segment ratios can then be derived from the same structure. This is one of the most common ways that parallel-line geometry produces proportional lengths.
Coordinate Dilations about the Origin
A dilation centered at the origin with scale factor \(k\) maps
\[ (x,y) \]to
\[ \boxed{(kx,ky)}. \]For example, a dilation by \(k=3\) maps
\[ (2,-4) \]to
\[ \boxed{(6,-12)}. \]Coordinate Example: Similar Triangles
Let
\[ A=(0,0),\qquad B=(2,0),\qquad C=(0,3). \]Dilate the triangle about the origin by \(k=4\). The image vertices are
\[ A'=(0,0),\qquad B'=(8,0),\qquad C'=(0,12). \]Every coordinate displacement, and therefore every side length, is multiplied by \(4\). The image triangle is similar to the original triangle.
Similarity Is Transitive
Similarity behaves as an equivalence relationship among geometric figures. In particular, if
\[ A\sim B \]and
\[ B\sim C, \]then
\[ \boxed{A\sim C}. \]If the scale factor from \(A\) to \(B\) is \(k_1\) and from \(B\) to \(C\) is \(k_2\), then the scale factor from \(A\) to \(C\) is
\[ \boxed{k_1k_2}. \]Composing Scale Factors
Suppose a figure is first enlarged by a factor of \(3\) and then reduced by a factor of \(\tfrac{1}{2}\).
The overall scale factor is
\[ 3\left(\frac{1}{2}\right)=\frac{3}{2}. \]Thus, the final figure is \(1.5\) times the size of the original in every corresponding linear dimension.
Scale and Units
Ratios of corresponding lengths become pure numerical scale factors only after the lengths are expressed in compatible units.
For example,
\[ 2\text{ cm}:1\text{ m} \]becomes
\[ 2\text{ cm}:100\text{ cm}. \]Therefore, the drawing-to-actual linear scale factor is
\[ \boxed{\frac{2}{100}=\frac{1}{50}}. \]Scale Does Not Mean Add the Same Amount
Uniform scaling is multiplicative, not additive.
If side lengths
\[ 3,\qquad 5,\qquad 7 \]are enlarged by scale factor \(2\), the new lengths are
\[ 6,\qquad 10,\qquad 14. \]Adding \(3\) to every side would instead give
\[ 6,\qquad 8,\qquad 10, \]which does not preserve the original side ratios and therefore does not represent a uniform scaling of the figure.
Common Error: Matching the Wrong Sides
Suppose
\[ \triangle ABC\sim\triangle DEF. \]Then \(AB\) corresponds to \(DE\), not automatically to whichever side of the second triangle appears visually closest on a diagram.
Use the similarity statement, angle information, or structural markings to establish correspondence before writing proportions.
Common Error: Reversing Only One Ratio
If
\[ \frac{AB}{DE}=\frac{BC}{EF}, \]then reversing both ratios gives another valid equation:
\[ \frac{DE}{AB}=\frac{EF}{BC}. \]But reversing only one ratio creates an inconsistent comparison.
Common Error: Using the Linear Factor for Area
If two similar figures have linear scale factor \(5\), their area ratio is not \(5\).
It is
\[ 5^2=25. \]Always match the exponent of the scale factor to the dimension of the measurement.
Common Error: Assuming Equal Angles Make Any Polygons Similar
Equal corresponding angles are sufficient for triangle similarity through AA, but not for general polygons.
Rectangles provide the simplest counterexample: all rectangles have four right angles, yet rectangles with different length-to-width ratios are not similar.
Common Error: Assuming Every Proportion Proves Similarity
A single matching side ratio does not establish similarity.
Similarity must be supported by a valid criterion or by enough information to verify the required angle and side relationships. One numerical ratio by itself usually says too little about shape.
Common Error: Ignoring Units in a Scale
A scale such as
\[ 1\text{ cm}:2\text{ m} \]is not the numerical ratio \(1:2\). Because
\[ 2\text{ m}=200\text{ cm}, \]the dimensionless drawing-to-actual ratio is
\[ \boxed{1:200}. \]Common Error: Trusting the Picture
Two figures may look similar without actually satisfying the mathematical conditions for similarity. Conversely, similar figures can look different because one has been rotated or reflected.
Use measurements, markings, coordinates, stated relationships, or a valid theorem rather than appearance alone.
A Reliable Procedure for Similar-Figure Problems
- Identify the two figures being compared.
- Determine the correspondence between vertices, angles, and sides.
- Decide whether similarity is given or must be proved.
- If proving triangle similarity, test AA, SAS Similarity, or SSS Similarity.
- Choose two corresponding lengths to determine the scale factor when needed.
- Keep every ratio in the same direction.
- Use the scale factor or a proportion to find unknown lengths.
- Use \(k^2\) for area comparisons and \(k^3\) for volume comparisons.
- Check that units are compatible.
- Verify that the final value is reasonable for an enlargement or reduction.
A Reliable Procedure for Proving Triangle Similarity
- Mark or calculate known equal angles.
- Identify any parallel lines that create corresponding or alternate angle relationships.
- Calculate side ratios only between genuinely corresponding sides.
- Test AA first when two angle pairs are available.
- Use SAS Similarity when two proportional side pairs surround equal included angles.
- Use SSS Similarity when all three corresponding side ratios are equal.
- Write the similarity statement in correct correspondence order.
- Only after similarity is established, use remaining corresponding parts proportionally.
Example: Full Similarity Workflow
Suppose
\[ \triangle ABC\sim\triangle DEF \]with
\[ AB=12,\qquad BC=18,\qquad AC=20, \]and
\[ DE=30. \]Because \(AB\leftrightarrow DE\), the scale factor from \(\triangle ABC\) to \(\triangle DEF\) is
\[ k=\frac{30}{12}=\frac{5}{2}. \]Therefore,
\[ EF=18\left(\frac{5}{2}\right)=45 \]and
\[ DF=20\left(\frac{5}{2}\right)=50. \]Thus,
\[ \boxed{EF=45,\qquad DF=50}. \]Example: Proving Similarity from Parallel Lines
In \(\triangle ABC\), let \(D\) lie on \(\overline{AB}\) and \(E\) lie on \(\overline{AC}\), with
\[ DE\parallel BC. \]Because the lines are parallel,
\[ \angle ADE\cong\angle ABC \]and
\[ \angle AED\cong\angle ACB. \]Therefore,
\[ \boxed{\triangle ADE\sim\triangle ABC} \]by AA Similarity.
Practice 1: Scale Factor
A side of length \(9\) corresponds to a side of length \(24\). Find the scale factor from the first figure to the second.
Practice 2: Unknown Side
Two similar triangles have scale factor \(\tfrac{7}{3}\) from the first triangle to the second. A side of the first triangle has length \(12\). Find the corresponding side of the second triangle.
Practice 3: Proportion
Suppose
\[ \triangle ABC\sim\triangle DEF, \]with \(AB=8\), \(DE=20\), and \(BC=14\). Find \(EF\).
Practice 4: AA Similarity
Triangle \(PQR\) has angles \(48^\circ\), \(62^\circ\), and \(70^\circ\). Triangle \(XYZ\) has angles \(62^\circ\), \(70^\circ\), and \(48^\circ\). Are the triangles similar? State why.
Practice 5: SSS Similarity
One triangle has side lengths \(6\), \(9\), and \(12\). Another has side lengths \(10\), \(15\), and \(20\). Determine whether the triangles are similar.
Practice 6: SAS Similarity
In two triangles, one pair of sides surrounding an equal included angle has lengths \(8\) and \(12\), while the corresponding pair in the second triangle has lengths \(14\) and \(21\). Are the side conditions consistent with SAS Similarity?
Practice 7: Perimeter Scaling
A polygon with perimeter \(34\) is enlarged by scale factor \(2.5\). Find the perimeter of the image.
Practice 8: Area Scaling
A figure has area \(18\). A similar figure is produced with linear scale factor \(4\). Find the area of the image.
Practice 9: Recover the Scale Factor
Two similar figures have areas \(32\) and \(200\). Find the linear scale factor from the first figure to the second.
Practice 10: Scale Drawing
A map uses the scale \(1\text{ cm}:6\text{ km}\). Two locations are \(8.5\) centimeters apart on the map. Find their actual separation.
Practice 11: Coordinate Dilation
Under a dilation centered at the origin with scale factor \(\tfrac{3}{2}\), find the image of the point
\[ (-4,6). \]Practice 12: Identify the Error
A student says, “Two rectangles are always similar because all four angles in both rectangles are \(90^\circ\).” Explain the error.
Practice Answers
Practice 1
\[ k=\frac{24}{9}=\boxed{\frac{8}{3}}. \]Practice 2
\[ 12\left(\frac{7}{3}\right)=28. \]Therefore,
\[ \boxed{28}. \]Practice 3
Since \(AB\leftrightarrow DE\) and \(BC\leftrightarrow EF\),
\[ \frac{DE}{AB}=\frac{EF}{BC}. \] \[ \frac{20}{8}=\frac{EF}{14}. \] \[ EF=14\left(\frac{20}{8}\right)=35. \] \[ \boxed{EF=35}. \]Practice 4
Yes. The triangles contain the same three angle measures. Two matching angle pairs are already sufficient, so
\[ \boxed{\text{the triangles are similar by AA}}. \]Practice 5
Compare corresponding sides from shortest to shortest:
\[ \frac{10}{6}=\frac{15}{9}=\frac{20}{12}=\frac{5}{3}. \]Therefore,
\[ \boxed{\text{the triangles are similar by SSS}}. \]Practice 6
\[ \frac{14}{8}=\frac{7}{4} \]and
\[ \frac{21}{12}=\frac{7}{4}. \]The two side pairs are proportional. Since the problem also states that the included angles are equal,
\[ \boxed{\text{the triangles satisfy SAS Similarity}}. \]Practice 7
Perimeter scales linearly:
\[ 34(2.5)=\boxed{85}. \]Practice 8
Area scales by \(k^2\):
\[ 18(4^2)=18(16)=\boxed{288}. \]Practice 9
The area ratio is
\[ \frac{200}{32}=\frac{25}{4}. \]Therefore,
\[ k^2=\frac{25}{4} \]and
\[ \boxed{k=\frac{5}{2}}. \]Practice 10
\[ 8.5(6)=51. \]Therefore,
\[ \boxed{51\text{ km}}. \]Practice 11
Multiply each coordinate by \(\tfrac{3}{2}\):
\[ \left(-4\cdot\frac{3}{2},\ 6\cdot\frac{3}{2}\right) =\boxed{(-6,9)}. \]Practice 12
Equal corresponding angles alone do not prove similarity for general quadrilaterals. Rectangles can have different length-to-width ratios.
For example, a \(2\)-by-\(3\) rectangle has aspect ratio \(\tfrac{3}{2}\), while a \(2\)-by-\(5\) rectangle has aspect ratio \(\tfrac{5}{2}\). Because those side ratios do not match, the rectangles are not similar even though all corresponding angles are equal.
What to Remember
- Similar figures have the same shape, though they may have different sizes.
- Corresponding angles of similar figures have equal measure.
- Corresponding lengths differ by one common positive scale factor.
- The order of vertices in a similarity statement records correspondence.
- The scale factor from original to image is \(k=\dfrac{\text{image length}}{\text{original length}}\).
- Scale factors greater than \(1\) enlarge; scale factors between \(0\) and \(1\) reduce.
- Congruence is the special case of similarity with \(k=1\).
- Dilations produce similar figures by scaling all lengths uniformly.
- Triangles can be proved similar by AA, SAS Similarity, or SSS Similarity.
- AA and SSS shortcuts are special to triangles and do not transfer unchanged to arbitrary polygons.
- Perimeter scales by \(k\), area by \(k^2\), and volume by \(k^3\).
- Scale drawings, maps, models, and indirect measurement all rely on consistent proportional scaling.
- Parallel lines often create similar triangles through angle relationships.
- Use compatible units and maintain correspondence when writing proportions.
The central scaling relationships are:
\[ \boxed{L_2=kL_1}, \] \[ \boxed{A_2=k^2A_1}, \] \[ \boxed{V_2=k^3V_1}. \]Where We Go Next
Similarity compares figures through proportional shape. The next major Euclidean relationship connects the side lengths of right triangles through one of geometry's most important equations.
The next lesson, The Pythagorean Relationship, develops the relationship among the legs and hypotenuse of a right triangle, examines its geometric meaning, uses it to find unknown lengths and distances, and distinguishes the theorem itself from its converse and related distance formulas.