Exercises: Use congruence and similarity to solve - Common Core Math - HS Geometry

A

Warm-Up: Review What You Know

These problems review skills you already know.

1.

Two triangles share a side. You know two angles of the first triangle are congruent to two angles of the second triangle. Which congruence criterion applies?

A.

SSS — three pairs of congruent sides

B.

AAS — two angles and a non-included side are congruent

C.

SAS — two sides and the included angle are congruent

D.

The triangles are congruent because they look the same size

2.

Triangle $PQR$ is similar to triangle $XYZ$ with a scale factor of 3. If $PQ = 6$ , what is $XY$ ?

A.

6 — similar triangles are the same size

B.

18

C.

3

D.

2

3.

In the diagram, $DE \parallel BC$ with $D$ on $\overline{AB}$ and $E$ on $\overline{AC}$ . Which similarity criterion proves $\triangle ADE \sim \triangle ABC$ ?

A.

SSS — three pairs of proportional sides

B.

SAS~ — two proportional sides and the included angle

C.

AA — two pairs of congruent angles

D.

SSA — two sides and a non-included angle

B

Fluency Practice

Apply congruence or similarity criteria directly. State the criterion used.

1.

In parallelogram $ABCD$ , diagonal $\overline{AC}$ is drawn. You know $AB \parallel CD$ and $BC \parallel AD$ . Which criterion proves $\triangle ABC \cong \triangle CDA$ ?

A.

SSA — two sides and a non-included angle

B.

SAS — $AC = CA$ , $\angle BAC = \angle DCA$ , $\angle BCA = \angle DAC$

C.

ASA — $\angle BAC \cong \angle DCA$ , $AC \cong CA$ , $\angle BCA \cong \angle DAC$

D.

SSS — three pairs of congruent sides

2.

In rectangle $ABCD$ , $\overline{AB} \cong \overline{DC}$ , $\overline{BC} \cong \overline{BC}$ (reflexive), and $\angle ABC \cong \angle DCB$ (both right angles). A student concludes: "By CPCTC, $\overline{AC} \cong \overline{DB}$ ." Identify the missing step and state the complete proof.

3.

In right triangle $ABC$ with the right angle at $C$ , the altitude from $C$ to hypotenuse $\overline{AB}$ meets $\overline{AB}$ at point $H$ . If $AH = 4$ and $HB = 9$ , find the length of altitude $CH$ .

4.

In $\triangle ABC$ , point $D$ is on $\overline{AB}$ and point $E$ is on $\overline{AC}$ such that $DE \parallel BC$ . Given $AD = 8$ , $DB = 4$ , and $AE = 12$ , find $EC$ .

5.

In isosceles triangle $PQR$ with $PQ \cong PR$ , $M$ is the midpoint of $\overline{QR}$ . Which congruence criterion proves $\triangle PQM \cong \triangle PRM$ ?

A.

AAS — $\angle PQM \cong \angle PRM$ , $\angle QPM \cong \angle RPM$ , $QM \cong RM$

B.

SAS — $PQ \cong PR$ , $\angle QPM \cong \angle RPM$ , $PM \cong PM$

C.

SSS — $PQ \cong PR$ , $QM \cong RM$ , $PM \cong PM$

D.

ASA — $\angle QPM \cong \angle RPM$ , $PM \cong PM$ , $\angle PMQ \cong \angle PMR$

C

Mixed Practice

These problems test the same skills in different formats.

1.

Quadrilateral $ABCD$ has $AB \cong CD$ and $AB \parallel CD$ . Prove that $ABCD$ is a parallelogram by showing that $BC \parallel AD$ .

2.

A flagpole casts a shadow 24 feet long. At the same time, a 5-foot person standing nearby casts a shadow 4 feet long. The sun's rays create similar triangles. Find the height of the flagpole.

3.

$\triangle ABC \sim \triangle DEF$ with $AB = 10$ , $DE = 15$ , and $EF = 12$ . Which proportion correctly finds $BC$ ?

A.

$\dfrac{AB}{EF} = \dfrac{BC}{DE}$

B.

$\dfrac{DE}{AB} = \dfrac{EF}{BC}$

C.

$\dfrac{AB}{DE} = \dfrac{EF}{BC}$

D.

$\dfrac{AB}{BC} = \dfrac{EF}{DE}$

4.

In $\triangle ABC$ , $D$ and $E$ are the midpoints of $\overline{AB}$ and $\overline{AC}$ , respectively. The midsegment $\overline{DE}$ is parallel to $\overline{BC}$ . If $BC = 14$ , find $DE$ .

5.

To prove that the diagonals of a rhombus are perpendicular, a student draws diagonal $\overline{AC}$ and focuses on triangles $\triangle AOB$ and $\triangle COB$ , where $O$ is the intersection point of the diagonals. Which sequence of steps is correct?

A.

$OA \cong OC$ (diagonals of a rhombus bisect each other), $AB \cong CB$ (all sides of rhombus equal), $OB \cong OB$ (reflexive); by SSS, $\triangle AOB \cong \triangle COB$ ; by CPCTC $\angle AOB \cong \angle COB$ ; since they are supplementary, each is 90°.

B.

By CPCTC, $\angle AOB = 90$ ° because the diagonals look perpendicular in the diagram.

C.

$AB \cong CB$ and $\angle OAB \cong \angle OCB$ ; by SSA, $\triangle AOB \cong \triangle COB$ ; by CPCTC the diagonals are perpendicular.

D.

Use AA similarity: both triangles share angle $B$ , so they are similar, so the diagonals are perpendicular.

D

Word Problems

Read each scenario carefully. Identify similar or congruent triangles, then solve.

1.

A city park has a triangular flower bed $\triangle PQR$ . The groundskeeper wants to verify that the two halves of the bed on either side of a center stake $M$ (the midpoint of $\overline{QR}$ ) are mirror images. The two sides of the bed meeting at $P$ are equal ( $PQ = PR$ ).

Explain how to prove that $\triangle PQM \cong \triangle PRM$ , and identify what CPCTC lets you conclude about the angles at $M$ .

2.

A surveyor needs to find the width of a river. She places stakes at points $A$ and $B$ on one bank and at points $C$ and $D$ on the other bank so that $AC \parallel BD$ , $AE = 30$ m, $CE = 20$ m, $DE = 15$ m, and $AC = 45$ m, where $E$ is the intersection of $\overline{AD}$ and $\overline{BC}$ .

1.

Explain why $\triangle AEC \sim \triangle DEB$ and state the similarity criterion used.

2.

Using the similarity from part (a), find $BD$ (the width of the river at that crossing). Express your answer in meters.

3.

On a map drawn to scale, two cities are 4.5 cm apart. The map's scale is 1 cm : 80 km.

Find the actual distance between the two cities in kilometers.

4.

In $\triangle ABC$ , $\overline{DE}$ is a midsegment connecting the midpoints of $\overline{AB}$ and $\overline{AC}$ . A second midsegment $\overline{FG}$ connects the midpoints of $\overline{AB}$ and $\overline{BC}$ .

If $BC = 18$ and $AC = 22$ , find the perimeter of parallelogram $DFGE$ formed by the two midsegments and portions of the sides. (Hint: use the Midsegment Theorem twice.)

E

Error Analysis

Each problem shows a student's incorrect work. Identify the error and explain how to correct it.

1.

Student's work:

Given: $\triangle ABC$ and $\triangle DEF$ where $AB = DE$ , $BC = EF$ , and $\angle A \cong \angle D$ .

$AB = DE$ (given)
$BC = EF$ (given)
$\angle A \cong \angle D$ (given)
By CPCTC, $\angle B \cong \angle E$ .
Therefore $\triangle ABC \cong \triangle DEF$ by ASA.

This student's proof contains two errors. Identify both errors and explain how to correct the proof.

A.

The student used SAS information (two sides and an included angle) but labeled it ASA. Simply change 'ASA' to 'SAS' and the proof is correct.

B.

The student applied CPCTC before proving congruence (Step 4 before Step 5), and $\angle A$ is not the included angle for SAS. The given information is SSA (side-side-non-included angle), which is invalid. The proof cannot proceed as written.

C.

The student forgot to write a reason for Step 3. Adding 'given' to Step 3 fixes the proof.

D.

The proof is correct. ASA and SAS are interchangeable when the side is shared.

2.

Student's work:

Given: $\triangle PQR \sim \triangle XYZ$ with $PQ = 6$ , $QR = 8$ , $XY = 9$ .

Student sets up: $\dfrac{PQ}{QR} = \dfrac{XY}{XZ}$

$\dfrac{6}{8} = \dfrac{9}{XZ}$

$XZ = 12$

Identify the error in the student's proportion and find the correct value of $XZ$ .

A.

The proportion is correct. $XZ = 12$ .

B.

The student compared sides within a triangle ( $PQ$ to $QR$ ) rather than corresponding sides between similar triangles. The correct proportion is $\dfrac{PQ}{XY} = \dfrac{QR}{YZ}$ , giving $YZ = 12$ . For $XZ$ , use $\dfrac{PQ}{XY} = \dfrac{PR}{XZ}$ and the given values of $PR$ (not available from the given information alone).

C.

The student forgot to simplify the ratio. $\dfrac{6}{8} = \dfrac{3}{4}$ , so $XZ = \dfrac{9 \cdot 4}{3} = 12$ .

D.

The proportion should use the scale factor: $6/9 = 2/3$ , so $XZ = QR \cdot (2/3) = 8 \cdot (2/3) \approx 5.3$ .

F

Challenge

These problems require multi-step reasoning. Plan your approach before writing.

1.

In $\triangle ABC$ , $D$ is the midpoint of $\overline{BC}$ . Let $E$ be the midpoint of $\overline{AD}$ . Line $\overline{BE}$ is extended to meet $\overline{AC}$ at point $F$ . Prove that $AF = \dfrac{1}{3} AC$ .

2.