Digital SAT Math · Geometry & Trigonometry
Similarity & congruence
Similar figures share a shape; congruent figures share a shape and a size. The test’s favorite move is to hand you a scale factor that works for lengths and then ask for an area or a volume — and every wrong choice is the power of you used by accident.
On the test
| Domain | Geometry and Trigonometry (score report) |
| What it looks like | A pair of triangles (often labeled, often right-angled) declared similar, or a solid pair declared similar, plus one missing length, area, or volume |
| Often asked | “What is the length of…?”, “What is the area of the larger…?”, “What is the volume of…?”, “Which conclusion must be true?” |
| Format | Multiple choice and student-produced response |
| Calculator | Allowed; the arithmetic is light so the trap is almost always the power of |
| Figures | Triangle pairs with corresponding sides or nested similar triangles; pure-text scale-factor stems also appear |
Recognition cues: similar, congruent, corresponding, scale factor, in the ratio, AA, SAS, SSS, area of the larger, volume of the smaller.
Pattern recognition
Four shapes cover almost every item on this skill.
- Length from similarity — confirm AA (or a stated similarity), match vertices, multiply one side by .
- Area from — same setup, but the asked quantity is an area, so use .
- Volume from — similar solids; use .
- Criterion — which statement proves similarity, which proves congruence, and which proves neither.
Method
- Confirm similarity. AA is the usual route: two pairs of congruent angles (the third follows). A stem that already says “similar” has done this step for you.
- Write the correspondence. Order the vertices so equal angles sit in the same positions: means , , , and sides , , .
- Find from one matched pair of lengths. Decide the direction: from the figure that has the known side to the figure that has the unknown. enlarges; shrinks.
- Apply the right power. Length ; area ; volume . If the stem gives areas and asks for a length, take first.
- State congruence only with a valid criterion. SSS, SAS, ASA, AAS, or HL (right triangles). AAA is similarity only. SSA is not a congruence criterion in general.
| Words in the stem | What they mean |
|---|---|
| similar | same shape; one scale factor for all corresponding lengths |
| congruent | similar with — same shape and same size |
| corresponding sides / angles | the ones that match under the vertex order you wrote |
| scale factor | the single multiplier for lengths; square it for area, cube it for volume |
| AA, SAS, SSS | named criteria — AA → similar; SAS/SSS/ASA/AAS/HL → congruent |
| area of the larger | use , not |
| volume of the solid | use , not |
Worked example 1 — length from AA
Stem. with , , . , , and . What is ?
Step 1 — correspondence. , .
Step 2 — scale factor. From ABC to DEF:
Step 3 — apply to the matched side.
Check. DEF is larger, so . Inverting would give , which is smaller than and impossible for the image on the larger triangle. Answer: 25.
Trap watch. Additive “” and double-application “” (treating a length like an area) are the usual neighbors of the key.
Worked example 2 — area scales by
Stem. Two similar triangles have corresponding sides and . The smaller area is . What is the larger area?
Step 1 — linear .
Step 2 — square it.
Step 3 — scale the area.
Check. would give length and area ; . Answer: 112.5.
Trap watch. is area-by-. is area-by- (the volume mistake on a plane figure).
Worked example 3 — volume scales by
Stem. Two similar cones have heights and . The volume of the smaller cone is . What is the volume of the larger cone?
Step 1 — linear . Heights are corresponding lengths:
Step 2 — cube it.
Step 3 — scale the volume.
Check. Every linear measure doubles, so the product of three of them multiplies by . Answer: 160.
Trap watch. (volume-by-) and (volume-by-, the area mistake on a solid) are the two distractors this stem is built for.
Practice
Answer before opening the explanation. Two items are student-produced response (type the number, no choices). Four stems carry a triangle-pair figure. Every wrong choice below is a named error — log the name when you miss one.
Question 1 Warm-up
Triangle $ABC$ is similar to triangle $DEF$, with $A$ corresponding to $D$, $B$ to $E$, and $C$ to $F$. If $AB = 4$, $DE = 10$, and $BC = 6$, what is the length of $EF$?
Show the answer Choice C
Why it is right
Similarity means corresponding sides share one scale factor. Match the named vertices: AB corresponds to DE, and BC corresponds to EF. The scale factor from ABC to DEF is $k = DE/AB = 10/4 = 5/2$. Multiply the matched side by that factor: $EF = BC \cdot k = 6 \cdot 5/2 = 15$. Check direction: DEF is larger than ABC, so EF must exceed 6.
Why each other choice fails
- Choice A
- Inverted scale factor: $4/10 \times 6 = 2.4$. That answers the length of a side on the smaller triangle when the larger side is 6, not the asked length on DEF.
- Choice B
- Additive instead of multiplicative: $6 + (10 - 4) = 12$, or a guessed $k = 2$. Corresponding sides scale by a common factor, not by a shared difference.
- Choice D
- Halves the correct answer, as if $k = 5/4$ were applied or the mean of 6 and 9 were taken. The matched ratio is $10/4 = 2.5$, not $1.25$.
Question 2 Standard
In the figure, right triangle $ABC$ is similar to right triangle $DEF$, with right angles at $A$ and $D$. If $AB = 8$, $DE = 12$, and $AC = 6$, what is the length of $DF$?
Show the answer Choice B
Why it is right
AA similarity holds: both triangles are right-angled, and the right angles correspond at $A$ and $D$. Corresponding legs are $AB$ with $DE$ and $AC$ with $DF$. Scale factor from ABC to DEF is $k = 12/8 = 3/2$. Then $DF = AC \cdot k = 6 \cdot 3/2 = 9$. The third sides would be $10$ and $15$, which keep the same ratio $3/2$.
Why each other choice fails
- Choice A
- Inverted $k$: $6 \cdot (8/12) = 4$. That is the image of a length-6 side when shrinking DEF onto ABC, not the asked side on the larger triangle.
- Choice C
- Copies the hypotenuse of the smaller 6-8-10 triangle. The question asks for the leg corresponding to AC, not for BC.
- Choice D
- Applies $k = 3/2$ twice, or multiplies 9 by another $3/2$: $6 \cdot (3/2)^2 = 13.5$. Lengths scale by $k$, not by $k^2$.
Question 3 Standard
Two similar triangles have corresponding side lengths $5$ and $15$. The area of the smaller triangle is $20$. What is the area of the larger triangle?
Show the answer Choice A
Why it is right
The linear scale factor from the smaller triangle to the larger is $k = 15/5 = 3$. Areas of similar figures scale by $k^2$, not by $k$: $k^2 = 9$, so the larger area is $20 \times 9 = 180$. Check: every length triples, so every product of two lengths (an area) multiplies by $9$.
Why each other choice fails
- Choice B
- Scales the area by $k$ instead of $k^2$: $20 \times 3 = 60$. That would be correct for a length, never for an area.
- Choice C
- Scales by $k^3$: $20 \times 27 = 540$. Cubing is the volume rule for similar solids, not the area rule for plane figures.
- Choice D
- Doubles the area as if $k = 2$, or adds $20$ to itself. The side ratio is $3$, not $2$.
Question 4 Standard
In the figure, $D$ lies on $\overline{AB}$ and $E$ lies on $\overline{AC}$, and triangle $ADE$ is similar to triangle $ABC$. If $AD = 6$, $AB = 15$, and $DE = 10$, what is the length of $BC$?
Show the answer Choice D
Why it is right
Shared angle at $A$ plus corresponding right angles (or the proportional legs on the figure) give AA similarity: $\triangle ADE \sim \triangle ABC$. Vertices match $A\leftrightarrow A$, $D\leftrightarrow B$, $E\leftrightarrow C$, so $DE$ corresponds to $BC$. Scale factor from ADE to ABC is $k = AB/AD = 15/6 = 5/2$. Then $BC = DE \cdot k = 10 \cdot 5/2 = 25$. The other leg check: $AC = AE \cdot k$ would use $AE = 8$ from the figure, giving $20$, consistent with a 15-20-25 triangle.
Why each other choice fails
- Choice A
- Inverted $k$: $10 \cdot (6/15) = 4$. That shrinks $BC$ onto $DE$ instead of enlarging $DE$ onto $BC$.
- Choice B
- Uses $k = 8/5$ or adds: $10 + 6 = 16$. Neither matches the side ratio $15/6$.
- Choice C
- Adds the given lengths: $15 + 6 = 21$. Similarity multiplies; it does not add the segments of a side.
Question 5 Standard
Which of the following conditions is sufficient to conclude that two triangles are congruent?
Show the answer Choice B
Why it is right
SAS is a standard congruence criterion: two sides and the included angle determine a triangle uniquely, so any second triangle with the same three measures is congruent to the first. AAA only forces equal angles — the triangles can still differ by a scale factor — so AAA is a similarity criterion, not a congruence criterion. SSA is not a congruence criterion in general (the ambiguous case).
Why each other choice fails
- Choice A
- AAA proves similarity, not congruence. The triangles have the same shape but need not be the same size; a scale factor other than $1$ is still allowed.
- Choice C
- SSA is not a congruence criterion in general. Two sides and a non-included angle can produce zero, one, or two distinct triangles (the ambiguous case).
- Choice D
- Restates AAA without a side. Equal angles alone never force equal sides; congruence needs a valid side-involving criterion (SSS, SAS, ASA, AAS, or HL).
Question 6 Harder
Two similar triangles have corresponding side lengths $8$ and $12$. The area of the smaller triangle is $48$. What is the area of the larger triangle?
Show the answer Choice B
Why it is right
Linear scale factor $k = 12/8 = 3/2$. Areas scale by $k^2 = (3/2)^2 = 9/4$. Larger area $= 48 \times 9/4 = 108$. Equivalently, the area ratio is the square of the side ratio: $(12/8)^2 = 9/4$, and $48 \cdot 9/4 = 108$.
Why each other choice fails
- Choice A
- Scales the area by $k$ only: $48 \times 3/2 = 72$. That treats area like a length.
- Choice C
- Uses $k = 3/2$ cubed, or multiplies by $27/8$: $48 \times 27/8 = 162$. Cubing is for volumes of similar solids.
- Choice D
- Inverts the area factor: $48 \times (8/12)^2 = 48 \times 4/9 \approx 21.3$ is the true small-from-large path; $32$ is $48 \times (2/3)$, as if the area scaled by $k$ inverted.
Question 7 Harder
In the figure, $D$ lies on $\overline{AB}$ and $E$ lies on $\overline{AC}$ so that triangle $ADE$ is similar to triangle $ABC$. If $AD = 8$, $AB = 12$, and $BC = 15$, what is the length of $DE$?
Show the answer Choice C
Why it is right
AA similarity: shared $\angle A$ and corresponding right angles at $A$ force $\triangle ADE \sim \triangle ABC$ with $D\leftrightarrow B$ and $E\leftrightarrow C$. Scale factor from ABC down to ADE is $k = AD/AB = 8/12 = 2/3$. Side $DE$ matches $BC$, so $DE = 15 \cdot 2/3 = 10$. Check with the other leg: $AE = 9 \cdot 2/3 = 6$, and the 6-8-10 triple on ADE matches the 9-12-15 triple on ABC.
Why each other choice fails
- Choice A
- Applies $k = 2/5$ or uses $AD$ alone: $15 \cdot 8/20$, or reports $AE$ from the figure. The matched ratio on the sides that include $D$ and $B$ is $8/12$, not $8/20$.
- Choice B
- Uses $k = 3/4$ (perhaps $9/12$ from the other leg only, misread): $15 \times 0.75 = 11.25$. Both legs must give the same $k$; here both give $2/3$.
- Choice D
- Inverts $k$: $15 \cdot (12/8) = 22.5$. That enlarges the already-large side as if ADE were the outer triangle.
Question 8 Harder Student-produced response
Triangle $PQR$ is similar to triangle $XYZ$, with $P$ corresponding to $X$, $Q$ to $Y$, and $R$ to $Z$. If $PQ = 9$, $XY = 15$, and $PR = 12$, what is the length of $XZ$?
Show the answer 20
Why it is right
Corresponding sides: $PQ \leftrightarrow XY$ and $PR \leftrightarrow XZ$. Scale factor from PQR to XYZ is $k = 15/9 = 5/3$. Then $XZ = PR \cdot k = 12 \cdot 5/3 = 20$. Direction check: XYZ is larger, so $XZ > 12$.
Answers students type instead
- 16
- Additive: $12 + (15 - 9) = 18$ is one additive form; $12 + 4 = 16$ treats the $9$-to-$15$ gap as if $4$ of something transferred. Sides scale by a factor.
- 18
- Uses $k = 2$ or $k = 1.5$ inconsistently: $12 \times 1.5 = 18$. The true ratio is $15/9 = 5/3 \approx 1.667$, not $3/2$.
- 7.2
- Inverted $k$: $12 \cdot (9/15) = 7.2$. That is the image of a length-12 side when mapping the larger triangle onto the smaller one.
Question 9 Harder
Two rectangular prisms are similar. The linear scale factor from the smaller prism to the larger prism is $2$. If the volume of the smaller prism is $40$, what is the volume of the larger prism?
Show the answer Choice D
Why it is right
Volumes of similar solids scale by $k^3$, not by $k$ or by $k^2$. The linear scale factor from the smaller prism to the larger is given as $k = 2$, so $k^3 = 8$. Multiply the smaller volume by that factor: $40 \times 8 = 320$. Each of the three edge lengths doubles, and volume is the product of three lengths, which is why the factor is eight rather than two or four.
Why each other choice fails
- Choice A
- Scales volume by $k$: $40 \times 2 = 80$. That is the length rule applied to a volume.
- Choice B
- Scales volume by $k^2$: $40 \times 4 = 160$. Squaring is the area (or surface-area) rule, not the volume rule.
- Choice C
- Uses a made-up factor of $6$: $40 \times 6 = 240$. No power of $k = 2$ equals $6$.
Question 10 Hardest
In triangles $ABC$ and $DEF$, $\angle A \cong \angle D$, $\angle B \cong \angle E$, and $AB = 2 \cdot DE$. Which conclusion must be true?
Show the answer Choice D
Why it is right
Two pairs of congruent angles give AA similarity (the third angles match automatically). Corresponding vertices are $A\leftrightarrow D$ and $B\leftrightarrow E$, so $AB$ matches $DE$. The given $AB = 2 \cdot DE$ means the scale factor from DEF to ABC is $2$. Congruence would require that factor to be $1$; here it is not, so the triangles are similar but not congruent.
Why each other choice fails
- Choice A
- AAA is a similarity criterion only. Equal angles do not force equal sides; with $k = 2$ the triangles are different sizes and not congruent.
- Choice B
- SAS needs two sides and the included angle. Only one pair of corresponding sides is given, so SAS does not apply — and $k \neq 1$ already rules out congruence.
- Choice C
- Reverses the scale-factor direction. From ABC to DEF the factor is $DE/AB = 1/2$, not $2$. Writing "$AB = 2 \cdot DE$" already points the enlargement from the smaller triangle DEF to the larger ABC.
Question 11 Hardest Student-produced response
Two similar polygons have areas $36$ and $100$. A side of the smaller polygon has length $9$. What is the length of the corresponding side of the larger polygon?
Show the answer 15
Why it is right
Area scale factor from smaller to larger is $100/36 = 25/9$. Linear scale factor is the positive square root: $k = 5/3$. Corresponding side on the larger polygon is $9 \cdot 5/3 = 15$. Check: $15/9 = 5/3$ and $(5/3)^2 = 25/9 = 100/36$.
Answers students type instead
- 20
- Uses $k = \sqrt{100}/\sqrt{36}$ correctly as $10/6 = 5/3$ but then multiplies wrong: e.g. $12 \cdot 5/3$ from a misread side, or $9 + 100/9$. The matched computation is $9 \times 5/3 = 15$.
- 25
- Uses the area factor as if it were the length factor: $9 \cdot (100/36)$ is not $25$, but students often jump to $25$ from $\sqrt{100}\cdot\mathrm{something}$ or report the numerator of the area ratio. Lengths use $\sqrt{\text{area ratio}}$, not the area ratio itself.
- 5.4
- Inverts $k$: $9 \cdot (3/5) = 5.4$. That is the side on the smaller polygon when the larger side is 9.
Question 12 Hardest
In the figure, triangle $ABC$ is similar to triangle $DEF$. The legs of $\triangle ABC$ measure $8$ and $6$, and the corresponding legs of $\triangle DEF$ measure $16$ and $12$. If the area of $\triangle ABC$ is $24$, what is the area of $\triangle DEF$?
Show the answer Choice A
Why it is right
Corresponding legs give $k = 16/8 = 12/6 = 2$ from ABC to DEF. Areas scale by $k^2 = 4$, so the larger area is $24 \times 4 = 96$. Direct check: right triangle with legs $16$ and $12$ has area $\frac{1}{2}\cdot 16\cdot 12 = 96$.
Why each other choice fails
- Choice B
- Scales area by $k$ only: $24 \times 2 = 48$. Lengths double; area must quadruple.
- Choice C
- Halves the given area, as if mapping large onto small without flipping the question. DEF is the larger triangle.
- Choice D
- Scales by $k^3$: $24 \times 8 = 192$. Cubing is for volumes of similar solids, not for triangle areas.
Common mistakes
- Area by instead of — sides double so the area is doubled. Areas are products of two lengths; both scale.
- Volume by or by — the solid’s three dimensions each contribute a factor of , so the volume factor is .
- Inverted — written when the unknown sits on the larger figure. Write “from small to large” next to the number.
- AAA called congruence — three equal angles make the same shape, not the same size.
- SSA treated as a congruence criterion — the ambiguous case; two sides and a non-included angle do not pin down a unique triangle in general.
- Correspondence written out of order — forces ; matching to builds a proportion from the wrong sides.
- Squaring a length (or not squaring an area ratio) — using when the stem asked for a side, or using the area ratio as if it were .
- Additive scaling — “the large side is 4 more, so add 4 to the other side.” Similarity multiplies.
- Stopping at — computing the scale factor carefully and gridding it in when the stem asked for a side, an area, or a volume.
FAQ
Is every pair of equiangular triangles congruent? No. Equal angles give similarity. Congruence needs a side that forces .
If the area ratio is , what is the side ratio? Take square roots: . The volume ratio would then be .
Do I need to memorize all five congruence criteria? Know that SSS, SAS, ASA, AAS, and HL work, and that AAA and SSA do not. The test is more interested in whether you reach for a side than in naming the criterion out loud.
How is this different from the triangles skill? Triangles owns matching vertices and solving a plain proportion. This page owns the power of (area and volume) and the line between similarity criteria and congruence criteria.