# Lesson 15

### Problem 1

Select all quadrilaterals that have 180 degree rotational symmetry.

A:

trapezoid

B:

isosceles trapezoid

C:

parallelogram

D:

rhombus

E:

rectangle

F:

square

### Solution

For access, consult one of our IM Certified Partners.

(From Unit 2, Lesson 14.)

### Problem 2

Lin wrote a proof to show that diagonal $$EG$$ is a line of symmetry for rhombus $$EFGH$$. Fill in the blanks to complete her proof.

Because $$EFGH$$ is a rhombus, the distance from $$E$$ to $$\underline{\hspace{.5in}1\hspace{.5in}}$$ is the same as the distance from $$E$$ to $$\underline{\hspace{.5in}2\hspace{.5in}}$$. Since $$E$$ is the same distance from $$\underline{\hspace{.5in}3\hspace{.5in}}$$ as it is from $$\underline{\hspace{.5in}4\hspace{.5in}}$$, it must lie on the perpendicular bisector of segment $$\underline{\hspace{.5in}5\hspace{.5in}}$$. By the same reasoning, $$G$$ must lie on the perpendicular bisector of $$\underline{\hspace{.5in}6\hspace{.5in}}$$. Therefore, line $$\underline{\hspace{.5in}7\hspace{.5in}}$$ is the perpendicular bisector of segment $$FH$$. So reflecting rhombus $$EFGH$$ across line $$\underline{\hspace{.5in}8\hspace{.5in}}$$ will take $$E$$ to $$\underline{\hspace{.5in}9\hspace{.5in}}$$ and $$G$$ to $$\underline{\hspace{.5in}10\hspace{.5in}}$$ (because $$E$$ and $$G$$ are on the line of reflection) and $$F$$ to $$\underline{\hspace{.5in}11\hspace{.5in}}$$and $$H$$ to $$\underline{\hspace{.5in}12\hspace{.5in}}$$ (since $$FH$$ is perpendicular to the line of reflection, and $$F$$ and $$H$$ are the same distance from the line of reflection, on opposite sides). Since the image of rhombus $$EFGH$$ reflected across $$EG$$ is rhombus $$EHGF$$ (the same rhombus!), line $$EG$$ must be a line of symmetry for rhombus $$EFGH$$.

### Solution

For access, consult one of our IM Certified Partners.

(From Unit 2, Lesson 14.)

### Problem 3

In quadrilateral $$ABCD$$, $$AD$$ is congruent to $$BC$$, and $$AD$$ is parallel to $$BC$$. Andre has written a proof to show that $$ABCD$$ is a parallelogram. Fill in the blanks to complete the proof.

Since $$AD$$ is parallel to $$\underline{\hspace{.5in}1\hspace{.5in}}$$, alternate interior angles $$\underline{\hspace{.5in}2\hspace{.5in}}$$ and $$\underline{\hspace{.5in}3\hspace{.5in}}$$ are congruent. $$AC$$ is congruent to $$\underline{\hspace{.5in}4\hspace{.5in}}$$ since segments are congruent to themselves. Along with the given information that $$AD$$ is congruent to $$BC$$, triangle $$ADC$$ is congruent to $$\underline{\hspace{.5in}5\hspace{.5in}}$$by the $$\underline{\hspace{.5in}6\hspace{.5in}}$$ Triangle Congruence. Since the triangles are congruent, all pairs of corresponding angles are congruent, so angle $$DCA$$ is congruent to $$\underline{\hspace{.5in}7\hspace{.5in}}$$. Since those alternate interior angles are congruent, $$AB$$ must be parallel to $$\underline{\hspace{.5in}8\hspace{.5in}}$$. Since we define a parallelogram as a quadrilateral with both pairs of opposite sides parallel, $$ABCD$$ is a parallelogram.

### Solution

For access, consult one of our IM Certified Partners.

(From Unit 2, Lesson 13.)

### Problem 4

Select the statement that must be true.

A:

Parallelograms have at least one right angle.

B:

If a quadrilateral has opposite sides that are both congruent and parallel, then it is a parallelogram.

C:

Parallelograms have congruent diagonals.

D:

The height of a parallelogram is greater than the lengths of the sides.

### Solution

For access, consult one of our IM Certified Partners.

(From Unit 2, Lesson 13.)

### Problem 5

$$EFGH$$ is a parallelogram and angle $$HEF$$ is a right angle. Select all statements that must be true.

A:

$$EFGH$$ is a rectangle.

B:

Triangle $$HEF$$ is congruent to triangle $$GFH$$.

C:

Triangle $$HEF$$ is congruent to triangle $$FGH$$.

D:

$$ED$$ is congruent to $$HD$$, $$DG$$, and $$DF$$.

E:

Triangle $$EDH$$ is congruent to triangle $$HDG$$.

### Solution

For access, consult one of our IM Certified Partners.

(From Unit 2, Lesson 12.)

### Problem 6

Figure $$ABCD$$ is a parallelogram. Is triangle $$ADB$$ congruent to triangle $$CBD$$? Show or explain your reasoning.

### Solution

For access, consult one of our IM Certified Partners.

(From Unit 2, Lesson 11.)

### Problem 7

Figure $$KLMN$$ is a parallelogram. Prove that triangle $$KNL$$ is congruent to triangle $$MLN$$.

### Solution

For access, consult one of our IM Certified Partners.

(From Unit 2, Lesson 7.)