Last updated .
6.3 — Similar Triangles and Other Figures
In Section 6.2, you used a dilation to produce a new triangle with the same angles and proportionally changed sides. This lesson asks the reverse question: if we know some matching angles or sides, is that enough to conclude that two shapes are similar?
1. In class: A Solidify Understanding Task
What does it mean for figures to be similar?
Congruent figures can be matched through translations, rotations, and reflections: their sizes do not change. Similar figures may additionally use a dilation, so the size can change while the shape stays the same. We write \(\triangle ABC\sim\triangle DEF\) when the named vertices correspond in order: \(A\leftrightarrow D\), \(B\leftrightarrow E\), \(C\leftrightarrow F\).
For polygons, there is an equivalent practical description: every pair of corresponding angles is congruent and every pair of corresponding side lengths is in the same ratio. Both requirements matter. A side-length ratio such as \(\frac{DE}{AB}\) represents image length divided by matching original length; the common value is a scale factor.
- Match the vertices and check corresponding angles.
- Compare corresponding sides in one consistent direction. Every ratio must agree.
Which familiar families of polygons are always similar?
The main task presents five conjectures. We can evaluate each by asking whether angle measures and side ratios are forced to match.
Rectangles: not all similar
All rectangles have four right angles, but their side ratios need not match. An \(8\times4\) rectangle has width-to-height ratio \(8/4=2\), while a \(6\times5\) rectangle has ratio \(6/5=1.2\). No uniform dilation can make both pairs of sides match.
Compare a \(2\times3\) rectangle with a \(2\times5\) rectangle. If the short sides are matched, their scale factor would be \(2/2=1\), but matching long sides would require \(5/3\). Since these disagree, the rectangles are not similar. Equal right angles alone do not settle polygon similarity.
Equilateral triangles: always similar
All equilateral triangles have three \(60^\circ\) angles. Once you match one vertex and align its two sides, one dilation matches both side lengths because all three sides in either triangle are equal. Thus every equilateral triangle is similar to every other.
Isosceles triangles: not all similar
“Isosceles” guarantees two equal sides (and the opposite base angles), but it does not fix the angle sizes. A triangle with angles \(70^\circ,70^\circ,40^\circ\) and one with \(50^\circ,50^\circ,80^\circ\) are both isosceles but not similar.
Rhombuses: not all similar
A rhombus has four equal sides, but the angles may differ. A square is a rhombus with four right angles; a rhombus with angles \(60^\circ\) and \(120^\circ\) cannot be similar to a square.
Squares: always similar
Every square has four right angles and four equal sides. Any square can be aligned to another by rigid motions and then dilated by the ratio of its side length to the target side length.
| Packet conjecture | Always true? | Key reason |
|---|---|---|
| All rectangles are similar | No | Width-to-height ratio varies |
| All equilateral triangles are similar | Yes | All angles are 60° and all sides scale together |
| All isosceles triangles are similar | No | Vertex angles can differ |
| All rhombuses are similar | No | Interior angles can differ |
| All squares are similar | Yes | All angles and side ratios match |
Mason's rotation-and-dilation argument assumes one scale factor will make both the width and height match. That works only if the two rectangles had equal width-to-height ratios to begin with. Aligning one right angle does not guarantee a uniform dilation can match the other vertices.
Why triangles are special: AA, SAS, and SSS similarity
A quadrilateral can change its shape while keeping all four angles fixed. A triangle is more restricted: once its angle measures are fixed, its side proportions are determined.
AA (angle–angle)
If two pairs of corresponding angles are congruent, two triangles are similar. Because every triangle's angles total \(180^\circ\), the third pair also matches. This is often called AAA similarity, but matching just two angles is enough.
Triangle \(ABC\) has angles \(45^\circ,65^\circ,70^\circ\). Triangle \(DEF\) has angles \(45^\circ,65^\circ,70^\circ\) in the same corresponding order. Two angles match, so \(\triangle ABC\sim\triangle DEF\) by AA. The triangles might have different sizes.
SAS (side–angle–side)
Two triangles are similar if two pairs of corresponding sides are proportional and the included angles between those sides are congruent. “Included” means the angle made by the two given sides.
Suppose \(AB=6\), \(AC=9\), \(DE=10\), \(DF=15\), and \(\angle A\cong\angle D\). Then \(DE/AB=10/6=5/3\) and \(DF/AC=15/9=5/3\). Because the included angles match, \(\triangle ABC\sim\triangle DEF\).
SSS (side–side–side)
Two triangles are similar if all three pairs of corresponding sides are proportional.
Triangle \(ABC\) has side lengths \(3,4,5\), and \(DEF\) has matching side lengths \(4.5,6,7.5\). Check: \(4.5/3=6/4=7.5/5=1.5\). Therefore the triangles are similar by SSS.
Why those three similarity tests work
Each test can be understood with the transformations already learned. The basic strategy is to line up a vertex and one side, then use a dilation to make that side the correct length.
Translate and rotate the first triangle until a matching vertex coincides with its partner and one side points along the matching ray. Dilate around the common vertex until that side matches the second triangle. Because both angles at that vertex agree, the second side points along the correct ray. Matching the other given angle forces the remaining vertex to be the same intersection point. Hence the triangles coincide after the similarity transformations.
Align the equal included angles by a rigid motion. The proportional side-length ratios give one common dilation factor \(k\); dilating the first triangle around the angle vertex makes each of the two sides land on its corresponding side. Their endpoints, and therefore the connecting third side, coincide.
Dilate the first triangle so that one side matches its partner. Because all three corresponding side ratios are equal, every side length now matches the second triangle. The triangles are congruent by SSS congruence, so rigid motions finish the alignment.
These tests are specifically tests for triangles. Do not apply AA or SSS to arbitrary polygons without considering their additional freedom to change shape.
2. Ready — Solving proportions (problems 1–9)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
A proportion is an equation between two ratios. If \(a/b=c/d\) and both denominators are nonzero, multiplying both sides by \(bd\) gives \(ad=bc\). You can also scale numerators and denominators or simplify ratios before solving.
Solve \(\frac{3}{4}=\frac{x}{20}\). Cross-multiply: \(3(20)=4x\), so \(60=4x\) and \(x=15\). Check: \(15/20=3/4\).
Solve \(\frac{3}{4}=\frac{b+3}{20}\). Then \(3(20)=4(b+3)\); \(60=4b+12\); \(b=12\). Check: \((12+3)/20=15/20=3/4\).
Solve \(\frac{3}{b+2}=\frac65\). Multiply: \(15=6(b+2)\), giving \(15=6b+12\), then \(b=1/2\). Check the denominator: \(b+2=2.5\ne0\), and \(3/2.5=6/5\).
Solve \(\frac{\sqrt3}{2}=\frac{\sqrt{12}}{c}\). Since \(\sqrt{12}=2\sqrt3\), cross-multiplication gives \(c\sqrt3=4\sqrt3\). Thus \(c=4\). A useful reminder: simplify radicals before solving when possible.
Ready includes simple unknown numerators, unknown denominators, and expressions such as \(b+3\) and \(b+2\). Check each result by substitution into the original proportion, making sure no denominator is zero.
Check your work — Ready 1–9 answer key
- \(x=15\): \(4x=60\).
- \(x=6\): \(21x=126\).
- \(x=16\): \(3x=48\).
- \(c=15\): \(6c=90\).
- \(b=12\): \(4(b+3)=60\).
- \(a=14\): \(12a=168\).
- \(a=\frac{13}{10}\): \(20a=26\).
- \(b=\frac12\): \(6(b+2)=15\).
- \(c=4\): \(c\sqrt3=2\sqrt{12}=4\sqrt3\).
3. Set — Proofs, counterexamples, and transformations (problems 10–17)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
Prove or disprove a universal claim
To prove that all figures in a group share a property, give a reason that works for every member. To disprove it, one valid counterexample is enough.
No. One right triangle can have acute angles \(30^\circ\) and \(60^\circ\), another \(45^\circ\) and \(45^\circ\). Their angle sets differ, so they cannot be similar. Knowing only the right angle is not enough for AA.
Yes. A regular \(n\)-gon has all \(n\) equal side lengths and all interior angles \(\frac{(n-2)180^\circ}{n}\). Two regular polygons with the same \(n\) therefore have equal corresponding angles, and every corresponding side ratio is the constant ratio of their side lengths. Hence they are similar.
Read a grid diagram and justify a mapping
For Set 12–13, match corresponding vertices in order, count the horizontal and vertical grid movements for each side, and check that lengths change by a common factor and angles are preserved. For Set 14–17, name a specific sequence: for example, translate a selected vertex to the target position, rotate or reflect to align matching edges, and dilate by the ratio of their lengths. The original packet diagrams already provide labeled vertices—use them rather than guessing a shape's correspondence from its position.
Let a triangle have vertices \(A=(1,1)\), \(B=(3,1)\), \(C=(1,3)\). Its image has \(A'=(5,2)\), \(B'=(9,2)\), \(C'=(5,6)\). First dilate by factor 2 about \(A\), giving \(A=(1,1)\), \(B=(5,1)\), \(C=(1,5)\). Then translate by vector \((4,1)\). All three resulting vertices equal the image vertices, so the steps are verified—not merely suggested by appearance.
The grid diagrams must be read with their original vertex correspondence. Different valid sequences of transformations can produce the same image; unless the task restricts the order, a correct sequence need not match someone else's. State your steps clearly and check every vertex.
Check your work — Set 10–17 answer key
- “All right triangles are similar” — False. A \(45^\circ,45^\circ,90^\circ\) triangle and a \(30^\circ,60^\circ,90^\circ\) triangle have different angle measures.
- “All regular polygons are similar to other regular polygons with the same number of sides.” — True. Corresponding angles all equal \(\frac{(n-2)180^\circ}{n}\), and every corresponding side ratio equals the ratio of the two polygons’ side lengths.
- “The polygons on the grid below are similar.” — True. A half-size dilation and a \(180^\circ\) rotation, followed by a translation to match \(J\) with \(J\prime\), maps \(I,J,K,G\) to their primed partners. For example, \(IJ=3\) grid units becomes \(I\prime J\prime=\frac32\), and \(JK=2\sqrt2\) becomes \(J\prime K\prime=\sqrt2\).
- “The polygons on the grid below are similar.” — False. On the displayed grid, \(MN=2\) and \(M\prime N\prime=1\), but \(LM=L\prime M\prime=\sqrt5\). The corresponding side ratios are \(\frac12\) and \(1\), so no uniform dilation works.
- Dilate by \(\frac12\) about \(A\), then rotate \(90^\circ\) counterclockwise about \(A\). In grid units relative to \(A\), \(B=(-5,0),C=(-4,-1),D=(-2,2)\) map to \(B\prime\prime=(0,-\frac52),C\prime\prime=(\frac12,-2),D\prime\prime=(-1,-1)\). Plot/construction not shown.
- One intended sequence: translate \(B\) to \(B\prime\prime\); reflect across the line through \(B\prime\prime\) parallel to \(BD\); rotate about \(B\prime\prime\) to align the translated ray \(BD\) with \(B\prime\prime D\prime\prime\); then dilate about \(B\prime\prime\) by \(\frac{B\prime\prime D\prime\prime}{BD}\). This fixes the correspondence and reverses orientation. The drawing is approximate and does not label an exact scale factor or rotation angle; use the stated similarity for this construction. Plot/construction not shown.
- One intended sequence: translate \(L\) to \(L\prime\prime\); reflect across the line through \(L\prime\prime\) parallel to \(LK\); rotate about \(L\prime\prime\) to align the translated ray \(LK\) with \(L\prime\prime K\prime\prime\); then dilate by \(\frac{L\prime\prime K\prime\prime}{LK}\). Corresponding vertices then match under the stated similarity, with reversed orientation. Several image vertices lie between grid intersections; the drawing does not specify an exact numerical rotation angle or scale factor. Plot/construction not shown.
- Reflect across the line of slope \(1\) through \(A\). Relative to \(A\), \(B=(-2,4),C=(0,5)\) map by \((x,y)\mapsto(y,x)\) to \(B\prime\prime=(4,-2),C\prime\prime=(5,0)\). Plot/construction not shown.
4. Go — Ratios in similar polygons (problems 18–21)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
The Go exercises ask for three equivalent ratios of corresponding sides. First match corresponding vertices and keep the same direction (image length divided by original length) for every pair. For the rectangles immediately below, the width changes from \(7\) to \(10.5\), and the height changes from \(4\) to \(6\). Both ratios are \(1.5\), so the same dilation factor works in both directions.
Two similar triangles have corresponding sides \(6,8,10\) and \(3,4,5\). Write three equivalent ratios and explain their order.
Step 1: Match shortest to shortest, medium to medium, longest to longest.
Step 2: Write larger/smaller throughout: \(\frac63=\frac84=\frac{10}5=2\).
Check: Each larger side is twice its partner. If you reverse one fraction, your ratios will disagree.
Two similar quadrilaterals have corresponding sides \(5,7,9,x\) and \(8,11.2,14.4,20.8\). Find \(x\) and three equivalent ratios.
Step 1: The scale factor from first to second is \(8/5=1.6\). Verify \(11.2/7=1.6\) and \(14.4/9=1.6\).
Step 2: Solve \(20.8/x=1.6\). Hence \(x=13\).
Step 3: Three equivalent second/first ratios are \(\frac85=\frac{11.2}{7}=\frac{20.8}{13}=1.6\).
A smaller right triangle has legs 3 and 7. A similar larger triangle has corresponding base 21. Find its height and hypotenuse.
Step 1: The length scale factor is \(21/7=3\).
Step 2: The height is \(3(3)=9\).
Step 3: The smaller hypotenuse is \(\sqrt{3^2+7^2}=\sqrt{58}\), so the larger hypotenuse is \(3\sqrt{58}\).
Ratios: \(\frac{21}{7}=\frac93=\frac{3\sqrt{58}}{\sqrt{58}}=3\). This reinforces the radical lengths shown in the packet's final Go problem.
Check your work — Go 18–21 answer key
- \(\frac84=\frac63=\frac{10}{5}=2\).
- \(\frac a6=\frac84=\frac n4=2\); thus \(a=12,x=4,y=5,n=8\), also \(\frac42=\frac{10}{y}=2\).
- \(\frac3{1.5}=\frac5a=\frac c3=2\); \(a=2.5,c=6\).
- \(\frac{21}{7}=\frac x3=\frac d{\sqrt{58}}=3\); \(x=9,d=3\sqrt{58}\).
5. Additional learning resources
These are optional and most useful after reading the matching portion of this lesson.
| Resource and direct link | What it teaches / why use it | Scope / what to skip |
|---|---|---|
| Math Is Fun — Similar | Visual reading on resizing, rotating, reflecting and moving similar figures; useful for Part 1 and Set | Some extra material. Read the main transformations and similar shapes examples; the circle example is optional. |
| Khan Academy — Triangle similarity review | Short written reference with AA, SAS, and SSS criteria and check questions, matching Part 2 | Close match. Focus on the three similarity tests and their pictured examples. |
| Khan Academy — Intro to triangle similarity (video) | Video explaining the meaning of similar triangles | Close match. Use after the polygon overview, before the triangle tests. |
| Khan Academy — Determine similar triangles: SSS (practice) | Independent practice deciding triangle similarity from corresponding sides | Close match. Focus on identifying corresponding side ratios. |
External links require internet access. A specifically matched Professor Leonard lecture was not verified, so none is assigned here.
6. Teacher's review (teachers and parents)
No serious mathematical errors, ambiguous givens, or incomplete instructions were identified in Section 6.3. For grid transformation questions, multiple valid transformation sequences may exist; students should verify the final vertex mapping.
7. Curriculum coverage
| Packet assignment | Instruction here |
|---|---|
| Main Part 1, 1–7 | Definitions; five polygon conjectures; arguments and counterexamples (In class) |
| Main Part 2, 8–11 | AAA/AA and SAS/SSS triangle similarity with transformation justifications (In class) |
| Ready 1–9 | Solving and checking proportions, variables in numerator/denominator, radicals (Section 2) |
| Set 10–11 | Universal claims: right triangles and regular polygons (Section 3) |
| Set 12–17 | Establish similarity on a grid and describe verified transformation sequences (Section 3) |
| Go 18–21 | Equivalent corresponding-side ratios and additional fully worked practice (Section 4) |
Source: Secondary Math II, Module 6: Similarity & Right Triangle Trigonometry, Mathematics Vision Project (2017), Section 6.3, printed pages 13–19. Independently authored supplemental instruction.
Last updated .