What is Congruence & similarity on the Geometry Regents?

Congruence & similarity is a Triangles topic on the Geometry Regents. RegentsLab has 21 practice questions on it, each with a full worked solution.

How do you study Congruence & similarity for the Geometry Regents?

Work a range of Congruence & similarity question types and read the worked solution every time you miss one — the exam reuses the same ideas year after year. Start with the questions on this page.

Congruence & similarity: Geometry Regents Practice

Practice Congruence & similarity for the Geometry Regents with real-style questions and full worked solutions. Miss one? Read the solution — the exam reuses the same ideas every year.

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Question 6 · 2 pts

Which pair of conditions is NOT sufficient to prove two triangles congruent?

(A) Angle-Side-Angle (ASA)
(B) Side-Side-Side (SSS)
(C) Side-Angle-Side (SAS)
(D) Angle-Angle-Angle (AAA) ✓

Worked solution

$\text{AAA} \Rightarrow \text{same shape}$ Equal angles fix shape, not size.

$\Rightarrow \text{similar, not necessarily congruent}$

Answer: \text{AAA}

AAA guarantees similar triangles (same shape) but not congruent ones; SSS, SAS, ASA, and AAS each prove congruence.

Question 9 · 2 pts

In $\triangle ABC$ and $\triangle DEF$, $\overline{AB} \cong \overline{DE}$, $\angle A \cong \angle D$, and $\overline{AC} \cong \overline{DF}$. Which criterion proves $\triangle ABC \cong \triangle DEF$?

(A) SSS
(B) SAS ✓
(C) AAS
(D) ASA

Worked solution

$\text{Given: } AB \cong DE,\ \angle A \cong \angle D,\ AC \cong DF$

$\angle A \text{ lies between sides } AB \text{ and } AC$ The marked angle is included between the two marked sides.

$\text{two sides and the included angle} \Rightarrow \text{SAS}$

Answer: SAS

SAS congruence applies when two sides and the angle between them are congruent; the deciding feature is that the angle is included between the two sides.

Question 10 · 2 pts

$\triangle ABC \sim \triangle RST$ with $AB = 8$, $RS = 12$, and $BC = 6$. What is the length of $\overline{ST}$?

(A) $ST = 4$
(B) $ST = 10$
(C) $ST = 18$
(D) $ST = 9$ ✓

Worked solution

$\triangle ABC \sim \triangle RST \Rightarrow \frac{AB}{RS} = \frac{BC}{ST}$ Corresponding sides are proportional.

$\frac{8}{12} = \frac{6}{ST}$

$ST = \frac{6 \cdot 12}{8} = 9$

Answer: ST = 9

Corresponding sides of similar triangles are proportional, so match the correct sides and cross-multiply to solve.

Question 15 · 2 pts

Two triangles are similar with a scale factor of $3{:}1$. If a side of the smaller triangle is $5$, what is the corresponding side of the larger triangle?

(A) $8$
(B) $5/3$
(C) $15$ ✓
(D) $45$

Worked solution

$\text{larger} = 3 \times \text{smaller}$ Scale factor 3:1 scales every length by 3.

$= 3(5) = 15$

Answer: 15

In similar figures, corresponding lengths share the scale factor; areas scale by its square.

Question 20 · 2 pts

If $\triangle ABC \cong \triangle DEF$ and $AB=8$, then which segment must also equal $8$?

(A) $DF$
(B) $DE$ ✓
(C) $BC$
(D) $EF$

Worked solution

$A\!\leftrightarrow\!D,\ B\!\leftrightarrow\!E$ Order of vertices gives correspondence.

$AB\leftrightarrow DE \Rightarrow DE=8$

Answer: DE=8

Corresponding parts of congruent triangles are equal (CPCTC); the vertex order pairs AB with DE.

Question 11 · 2 pts

$\triangle ABC \sim \triangle DEF$ with a scale factor of $3$ (ABC to DEF). If $DE = 12$, find $AB$.

(A) $4$ ✓
(B) $9$
(C) $36$
(D) $15$

Worked solution

$\frac{DE}{AB}=3$ DEF is 3× the size of ABC.

$AB=\frac{12}{3}=4$

Answer: AB=4

Corresponding sides of similar triangles share the scale factor; divide the larger side by 3.

Question 11 · 2 pts

The ratio of similarity of $\triangle ABC$ to $\triangle XYZ$ is $2:3$. If the area of $\triangle ABC$ is $36$ square units, what is the area of $\triangle XYZ$?

(A) $81$ ✓
(B) $24$
(C) $16$
(D) $54$

Worked solution

$\text{ratio of similarity} = \frac{2}{3}$

$\text{ratio of areas} = \left(\frac{2}{3}\right)^2 = \frac{4}{9}$ Areas scale by the square of the linear ratio.

$\frac{36}{[\triangle XYZ]} = \frac{4}{9}$

$[\triangle XYZ] = 36 \cdot \frac{9}{4} = 81$

Answer: 81

For similar figures the ratio of areas equals the square of the scale factor, not the scale factor itself.

Question 12 · 2 pts

In $\triangle ABC$, $D$ lies on $\overline{AB}$ and $E$ lies on $\overline{AC}$ so that $\overline{DE} \parallel \overline{BC}$. If $AD = 8$, $DB = 6$, and $AE = 12$, what is the length of $\overline{EC}$?

(A) $EC = 16$
(B) $EC = 6$
(C) $EC = 4$
(D) $EC = 9$ ✓

Worked solution

$\overline{DE} \parallel \overline{BC} \Rightarrow \frac{AD}{DB} = \frac{AE}{EC}$ Side-splitter theorem.

$\frac{8}{6} = \frac{12}{EC}$

$EC = \frac{12 \cdot 6}{8} = 9$

Answer: EC = 9

A line parallel to one side of a triangle divides the other two sides proportionally (the side-splitter theorem).

Question 13 · 2 pts

A person who is $5$ feet tall casts a shadow $8$ feet long. At the same time, a nearby flagpole casts a shadow $40$ feet long. What is the height of the flagpole?

(A) $64$ feet
(B) $25$ feet ✓
(C) $43$ feet
(D) $32$ feet

Worked solution

$\frac{\text{height}}{\text{shadow}}: \quad \frac{5}{8} = \frac{h}{40}$ Same time of day gives similar triangles.

$h = \frac{5 \cdot 40}{8} = 25$

Answer: 25 feet

An object and its shadow at the same time form similar triangles, so heights and shadow lengths are proportional.

Question 14 · 2 pts

Which sequence of transformations always maps a figure onto an image that is congruent to the original?

(A) A rotation followed by a dilation with scale factor $3$
(B) A reflection followed by a translation ✓
(C) A dilation with scale factor $2$ followed by a rotation
(D) A dilation with scale factor $\frac{1}{2}$

Worked solution

$\text{rigid motions: translation, reflection, rotation preserve distance}$

$\text{reflection then translation are both rigid} \Rightarrow \text{congruent image}$

$\text{any dilation with scale factor} \neq \pm 1 \text{ changes size} \Rightarrow \text{similar, not congruent}$

Answer: A reflection followed by a translation

Rigid motions preserve size and shape and produce congruent images, while a dilation with scale factor other than 1 changes size, giving only similarity.

Question 10 · 2 pts

In triangle $ABC$, point $D$ is on side $\overline{AB}$ and point $E$ is on side $\overline{AC}$ such that $\overline{DE} \parallel \overline{BC}$. If $AD = 6$, $DB = 4$, and $AE = 9$, what is the length of $\overline{EC}$?

(A) $15$
(B) $6$ ✓
(C) $7$
(D) $13.5$

Worked solution

$\frac{AD}{DB} = \frac{AE}{EC}$ A line parallel to one side of a triangle divides the other two sides proportionally (side splitter theorem).

$\frac{6}{4} = \frac{9}{EC}$

$6 \cdot EC = 36$

$EC = 6$

Answer: $6$

Because DE is parallel to BC, the side splitter theorem gives AD/DB = AE/EC. Substituting, 6/4 = 9/EC, so EC = 6. The distractor 13.5 comes from inverting the proportion, and 15 is the full length AC rather than EC.

Question 27 · 4 pts · constructed response

In triangle $ABC$, point $D$ lies on side $\overline{AB}$ and point $E$ lies on side $\overline{AC}$ such that $\overline{DE} \parallel \overline{BC}$. If $AD = 6$, $DB = 4$, and $AE = 9$, determine the length of $\overline{EC}$. Justify your answer.

Model answer: $EC = 6$. Since $\overline{DE} \parallel \overline{BC}$, the side splitter theorem gives $\frac{AD}{DB} = \frac{AE}{EC}$, so $\frac{6}{4} = \frac{9}{EC}$ and $EC = 6$.

Worked solution

$\overline{DE} \parallel \overline{BC} \Rightarrow \frac{AD}{DB} = \frac{AE}{EC}$ A line parallel to one side of a triangle divides the other two sides proportionally (side splitter theorem).

$\frac{6}{4} = \frac{9}{EC}$

$6 \cdot EC = 36$

$EC = 6$

Answer: $EC = 6$. Since $\overline{DE} \parallel \overline{BC}$, the side splitter theorem gives $\frac{AD}{DB} = \frac{AE}{EC}$, so $\frac{6}{4} = \frac{9}{EC}$ and $EC = 6$.

A segment parallel to one side of a triangle splits the other two sides proportionally. Setting up the proportion $\frac{AD}{DB} = \frac{AE}{EC}$ with the given lengths yields $EC = 6$.

Question 2 · 2 pts

Triangle $ABC$ is similar to triangle $DEF$, with vertices corresponding in that order. If $AB = 8$, $BC = 12$, and $DE = 6$, what is the length of $EF$?

(A) $16$
(B) $9$ ✓
(C) $8$
(D) $10$

Worked solution

$\frac{AB}{DE} = \frac{BC}{EF}$ Corresponding sides of similar triangles are proportional.

$\frac{8}{6} = \frac{12}{EF}$

$8 \cdot EF = 72$

$EF = 9$

Answer: $9$

Since $\triangle ABC \sim \triangle DEF$, corresponding sides are proportional: $\frac{AB}{DE} = \frac{BC}{EF}$. Solving $\frac{8}{6} = \frac{12}{EF}$ gives $EF = 9$. Choosing $16$ comes from flipping the ratio, and $10$ comes from wrongly subtracting 2 as if the sides differed by a constant.

Question 16 · 2 pts

Triangle $ABC$ is similar to triangle $DEF$. The ratio of the area of triangle $ABC$ to the area of triangle $DEF$ is $16:49$. What is the ratio of the perimeter of triangle $ABC$ to the perimeter of triangle $DEF$?

(A) $16:49$
(B) $2:7$
(C) $8:49$
(D) $4:7$ ✓

Worked solution

$\frac{A_1}{A_2} = k^2 = \frac{16}{49}$ For similar figures, the ratio of areas equals the square of the scale factor $k$.

$k = \sqrt{\frac{16}{49}} = \frac{4}{7}$

$\frac{P_1}{P_2} = k = \frac{4}{7}$ Perimeters of similar figures are in the same ratio as corresponding sides.

Answer: $4:7$

For similar triangles, areas are proportional to the square of the scale factor, while perimeters are proportional to the scale factor itself. Since the area ratio is $16:49$, the scale factor is $\sqrt{16/49} = 4/7$, so the perimeters are in the ratio $4:7$.

Question 31 · 4 pts · constructed response

In triangle $ABC$, point $D$ is on side $\overline{AB}$ and point $E$ is on side $\overline{AC}$ such that $\overline{DE} \parallel \overline{BC}$. If $AD = 4$, $DB = 6$, and $AE = 6$, determine and state the length of $\overline{EC}$. Justify your answer.

Model answer: $EC = 9$. Because $\overline{DE} \parallel \overline{BC}$, the segment divides the sides proportionally: $\frac{AD}{DB} = \frac{AE}{EC}$, so $\frac{4}{6} = \frac{6}{EC}$ gives $EC = 9$.

Worked solution

$\overline{DE} \parallel \overline{BC} \Rightarrow \frac{AD}{DB} = \frac{AE}{EC}$ A line parallel to one side of a triangle divides the other two sides proportionally (side splitter theorem).

$\frac{4}{6} = \frac{6}{EC}$

$4 \cdot EC = 36$

$EC = 9$

Answer: $EC = 9$. Because $\overline{DE} \parallel \overline{BC}$, the segment divides the sides proportionally: $\frac{AD}{DB} = \frac{AE}{EC}$, so $\frac{4}{6} = \frac{6}{EC}$ gives $EC = 9$.

A segment parallel to one side of a triangle splits the other two sides proportionally, since $\triangle ADE \sim \triangle ABC$ by AA similarity. Setting $\frac{AD}{DB} = \frac{AE}{EC}$ gives $\frac{4}{6} = \frac{6}{EC}$, so $EC = 9$.

Question 5 · 2 pts

(A) $4$
(B) $8$
(C) $15$
(D) $9$ ✓

Worked solution

$\text{A segment parallel to one side of a triangle divides the other two sides proportionally.}$ Side splitter theorem.

$\frac{AD}{DB} = \frac{AE}{EC}$

$\frac{4}{6} = \frac{6}{EC}$

$4 \cdot EC = 36 \quad \Rightarrow \quad EC = 9$

Answer: $9$

Since DE is parallel to BC, the side splitter theorem gives AD/DB = AE/EC. Substituting, 4/6 = 6/EC, so EC = 9. The distractor 15 comes from confusing EC with the whole side AC found from the similar-triangle ratio AD/AB = AE/AC.

Question 30 · 4 pts · constructed response

In triangle $ABC$, point $D$ lies on side $\overline{AB}$ and point $E$ lies on side $\overline{AC}$ such that $\overline{DE} \parallel \overline{BC}$. If $AD = 4$, $DB = 6$, and $AE = 6$, determine and state the length of $\overline{EC}$. Justify your answer.

Model answer: $EC = 9$. Because $\overline{DE} \parallel \overline{BC}$, the segment divides the two sides proportionally (side splitter theorem), so $\frac{AD}{DB} = \frac{AE}{EC}$, giving $\frac{4}{6} = \frac{6}{EC}$ and $EC = 9$.

Worked solution

$\overline{DE} \parallel \overline{BC} \Rightarrow \frac{AD}{DB} = \frac{AE}{EC}$ A line parallel to one side of a triangle divides the other two sides proportionally.

$\frac{4}{6} = \frac{6}{EC}$

$4 \cdot EC = 36$

$EC = 9$

Answer: $EC = 9$. Because $\overline{DE} \parallel \overline{BC}$, the segment divides the two sides proportionally (side splitter theorem), so $\frac{AD}{DB} = \frac{AE}{EC}$, giving $\frac{4}{6} = \frac{6}{EC}$ and $EC = 9$.

When a segment is drawn parallel to one side of a triangle and intersects the other two sides, it splits those sides proportionally, since $\triangle ADE \sim \triangle ABC$ by AA. Setting up the proportion $\frac{AD}{DB} = \frac{AE}{EC}$ and cross multiplying gives $EC = 9$.

Question 23 · 2 pts

In $\triangle ABC$, $DE \parallel BC$ with $AD=4$, $DB=6$, and $AE=6$. Find $EC$.

(A) $6$
(B) $9$ ✓
(C) $4$
(D) $10$

Worked solution

$\frac{AD}{DB}=\frac{AE}{EC}$ Side-splitter theorem.

$\frac{4}{6}=\frac{6}{EC} \Rightarrow EC=9$

Answer: EC=9

A line parallel to one side splits the other two proportionally; set up and solve the proportion.

Question 15 · 2 pts

$\triangle ABC \sim \triangle DEF$ with $AB = x - 1$, $DE = 12$, $BC = 5$, and $EF = 15$. What is the value of $x$?

(A) $x = 5$ ✓
(B) $x = 3$
(C) $x = 4$
(D) $x = 37$

Worked solution

$\triangle ABC \sim \triangle DEF \Rightarrow \frac{AB}{DE} = \frac{BC}{EF}$

$\frac{x-1}{12} = \frac{5}{15} = \frac{1}{3}$

$3(x-1) = 12 \Rightarrow x - 1 = 4$

$x = 5$

Answer: x = 5

Set corresponding sides of similar triangles into a proportion and solve, making sure to finish solving for $x$ itself.

Question 16 · 2 pts

In $\triangle ABC$, $M$ and $N$ are the midpoints of $\overline{AB}$ and $\overline{AC}$, respectively, so $\overline{MN}$ is a midsegment. If $MN = x + 3$ and $BC = 3x - 1$, what is the length of $\overline{BC}$?

(A) $BC = 10$
(B) $BC = 20$ ✓
(C) $BC = 7$
(D) $BC = 5$

Worked solution

$MN = \tfrac{1}{2} BC$ A midsegment is half the length of the third side.

$x + 3 = \tfrac{1}{2}(3x - 1)$

$2x + 6 = 3x - 1 \Rightarrow x = 7$

$BC = 3(7) - 1 = 20$

Answer: BC = 20

The midsegment joining two midpoints is parallel to and half the length of the third side; solve the equation for $x$, then compute the requested length.

Question 21 · 2 pts

In right triangle $ABC$, the right angle is at $C$, and altitude $\overline{CD}$ is drawn to hypotenuse $\overline{AB}$. If $AD = 4$ and $DB = 9$, what is the length of $\overline{CD}$?

(A) $\sqrt{13}$
(B) $6$ ✓
(C) $36$
(D) $6.5$

Worked solution

$\frac{AD}{CD} = \frac{CD}{DB}$ The altitude to the hypotenuse is the geometric mean of the two segments of the hypotenuse, because the two smaller triangles are similar.

$CD^2 = AD \cdot DB = 4 \cdot 9 = 36$

$CD = 6$

Answer: $6$

The altitude to the hypotenuse of a right triangle creates two triangles similar to each other, so the altitude is the geometric mean of the two hypotenuse segments: CD squared = 4 times 9 = 36, so CD = 6. The distractor 6.5 is the arithmetic mean of 4 and 9, and 36 is CD squared.

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