Title: (Canvas-Quiz) Semester 1 Final Review
Course: HBUHSD Geometry Resource Course > Section: Unit 0 Multi Unit Review

Q: Which graph shows a triangle and its reflection image over the x-axis?
A: (a choice with no text)

Q: Name the smallest angle of The diagram is not to scale.
- Two angles are the same size and smaller than the third.
A: (a choice with no text)

Q: If what is the relationship between
- not enough information
A: (a choice with no text)

Q: List the sides in order from shortest to longest. The diagram is not to scale.
A: (a choice with no text)

Q: The vertices of a triangle are P(–8, 6), Q(1, –3),
and R(–6, –3). Name the vertices of after a
reflection over the line y = x.
A: (a choice with no text)

Q: The vertices of a triangle are P(–3, 8), Q(–6, –4),
and R(1, 1). Name the vertices of over the x-axis..
- _
A: _

Q: Name three collinear points.
- P, G, and N
- R, P, and N
- R, P, and G
- R, G, and N
A: R, G, and N
If right: Collinear points are points that lie on the same line. R, G, and N are three collinear points.
If wrong: Collinear points are points that lie on the same line. R, G, and N are three collinear points.

Q: Name a plane that contains .
- plane ACR
- plane WCT
- plane WRT
- plane RCA
A: plane WRT
If right: A plane can be described by any three noncollinear points. Of the choices given, only points W, R, and T are noncollinear. Thus, lies in plane WRT.
If wrong: A plane can be described by any three noncollinear points. Of the choices given, only points W, R, and T are noncollinear. Thus, lies in plane WRT.

Q: D is between C and E. = , = , and DE = 27. Find CE.
- CE = 17.5
- CE = 78
- CE = 105
- CE = 57
A: CE = 105
If right: | Segment Addition Postulate | Substitute for and for . | Simplify. | Subtract from both sides. | Divide both sides by 2. or 17.5 | Simplify.
If wrong: | Segment Addition Postulate | Substitute for and for . | Simplify. | Subtract from both sides. | Divide both sides by 2. or 17.5 | Simplify.

Q: K is the midpoint of . and . Find JK, KL, and JL.
- JK = 1, KL = 1, JL = 2
- JK = 6, KL = 6, JL = 12
- JK = 12, KL = 12, JL = 6
- JK = 18, KL = 18, JL = 36
A: JK = 6, KL = 6, JL = 12
If right: Step 1 Write an equation and solve. | K is the midpoint of . | Substitute for and for . | Subtract from both sides. | Divide both sides by 3. Step 2 Find JK, KL, and JL.
If wrong: Step 1 Write an equation and solve. | K is the midpoint of . | Substitute for and for . | Subtract from both sides. | Divide both sides by 3. Step 2 Find JK, KL, and JL.

Q: Find the measure of .
Then, classify the angle as acute, right, or obtuse.
- m; obtuse
- m; acute
- m; right
- m; obtuse
A: m; right
If right: By the Protractor Postulate, . First, measure and . Thus, is a right angle.
If wrong: By the Protractor Postulate, . First, measure and . Thus, is a right angle.

Q: m and m. Find m.
- m
- m
- m
- m
A: m
If right: m m m | Angle Addition Postulate m | Substitute for m and 20° for m. m | Subtract ° from both sides.
If wrong: m m m | Angle Addition Postulate m | Substitute for m and 20° for m. m | Subtract ° from both sides.

Q: bisects , m, and m. Find m.
- m = 22°
- m = 3°
- m = 40°
- m = 20°
A: m = 20°
If right: Step 1 Solve for x. m m | Definition of angle bisector. | Substitute for and for . | Add 1 to both sides. | Subtract from both sides. | Divide both sides by 3. Step 2 Find m. m
If wrong: Step 1 Solve for x. m m | Definition of angle bisector. | Substitute for and for . | Add 1 to both sides. | Subtract from both sides. | Divide both sides by 3. Step 2 Find m. m

Q: Find the measure of the complement of , where m
A: (a choice with no text)
If right: Subtract from 90° and simplify. 58.9
If wrong: Subtract from 90° and simplify. 58.9

Q: Name all pairs of vertical angles.
- ;
- ;
- ;
- ;
A: ;
If right: The vertical angle pairs are . These angles appear to have the same measure.
If wrong: The vertical angle pairs are . These angles appear to have the same measure.

Q: Find the coordinates of the midpoint of with endpoints C(1, –6) and M(7, 5).
- (, )
- (8, –1)
- (, )
- (, )
A: (, )
If right: (, )
If wrong: (, )

Q: Find CD and EF. Then determine if .
- , ,
- , ,
- , ,
- , ,
A: , ,
If right: Step 1 Find the coordinates of each point. , , , and Step 2 Use the Distance Formula. CD | | EF | Since , .
If wrong: Step 1 Find the coordinates of each point. , , , and Step 2 Use the Distance Formula. CD | | EF | Since , .

Q: A figure has vertices at E(–3, 1), F(1, 1), and G(4, 5). After a transformation, the image of the figure has vertices at E’(–3, –1), F’(1, –1), and G’(4, –5). Draw the preimage and image. Then identify the transformation.
- The transformation is a reflection across the x-axis.
- The transformation is a 180° rotation.
- The transformation is a 90° rotation.
- The transformation is a translation.
A: The transformation is a reflection across the x-axis.
If right: Plot the points. Then use a straightedge to connect the vertices. The transformation is a reflection across the x-axis because each point and its image are the same distance from the x-axis.
If wrong: Plot the points. Then use a straightedge to connect the vertices. The transformation is a reflection across the x-axis because each point and its image are the same distance from the x-axis.

Q: Find the coordinates for the image of EFG after the translation (x, y) (x – 6, y + 2). Draw the image.
A: (a choice with no text)
If right: Step 1 Find the coordinates of EFG. The vertices of EFG are E(3, 0), F(1, –2), and G(5, –4). Step 2 Apply the rule to find the vertices of the image. E'(3 – 6, 0 + 2) = E'(–3, 2) F'(1 – 6, –2 + 2) = F'(–5, 0) G'(5 – 6, –4 + 2) = G'(–1, –2) Step 3 Plot the points. Then finish drawing the image by using a straightedge to connect the vertices.
If wrong: Step 1 Find the coordinates of EFG. The vertices of EFG are E(3, 0), F(1, –2), and G(5, –4). Step 2 Apply the rule to find the vertices of the image. E'(3 – 6, 0 + 2) = E'(–3, 2) F'(1 – 6, –2 + 2) = F'(–5, 0) G'(5 – 6, –4 + 2) = G'(–1, –2) Step 3 Plot the points. Then finish drawing the image by using a straightedge to connect the vertices.

Q: Tell whether the transformation appears
to be a reflection. Explain.
- Yes; the image appears to be flipped across a line.
- No; the image does not appear to be flipped.
A: Yes; the image appears to be flipped across a line.
If right: A reflection is a transformation that moves a figure (the preimage) by flipping it across a line.
If wrong: A reflection is a transformation that moves a figure (the preimage) by flipping it across a line.

Q: Tell whether the transformation appears to be a translation. Explain.
- Yes; all of the points have moved the same distance in the same direction.
- No; not all of the points have moved the same distance.
A: No; not all of the points have moved the same distance.
If right: A translation is a transformation where all the points of a figure are moved the same distance in the same direction. This transformation is not a translation because not all of the points have moved the same distance.
If wrong: A translation is a transformation where all the points of a figure are moved the same distance in the same direction. This transformation is not a translation because not all of the points have moved the same distance.

Q: Rotate with vertices R(4, –1), S(5, 3), and Q(3, 1) by 90° counterclockwise about the origin.
A: (a choice with no text)
If right: The image of (x, y) is (–y, x). R(4, –1) (1, 4) S(5, 3) (–3, 5) Q(3, 1) (–1, 3) Graph the preimage and the image.
If wrong: The image of (x, y) is (–y, x). R(4, –1) (1, 4) S(5, 3) (–3, 5) Q(3, 1) (–1, 3) Graph the preimage and the image.

Q: has vertices A(3, 1), B(4, 5), and C(2, 3). Rotate 90° counterclockwise about the origin and then reflect it across the x-axis.
A: (a choice with no text)
If right: The rotation image of (x, y) 90° counterclockwise is (–y, x), so A(3, 1) (–1, 3), B(4, 5) (–5, 4) and C(2, 3) (–3, 2). The reflection image of (x, y) across the x-axis is (x, –y), so (–1, 3) (–1, –3), (–5, 4) (–5, –4) and (–3, 2) (–3, –2).
If wrong: The rotation image of (x, y) 90° counterclockwise is (–y, x), so A(3, 1) (–1, 3), B(4, 5) (–5, 4) and C(2, 3) (–3, 2). The reflection image of (x, y) across the x-axis is (x, –y), so (–1, 3) (–1, –3), (–5, 4) (–5, –4) and (–3, 2) (–3, –2).

Q: Give an example of corresponding angles.
- and
- and
- and
- and
A: and
If right: Corresponding angles lie on the same side of a transversal, on the same sides of the two lines the transversal crosses. So, and are corresponding angles.
If wrong: Corresponding angles lie on the same side of a transversal, on the same sides of the two lines the transversal crosses. So, and are corresponding angles.

Q: Draw two lines and a transversal such that 1 and 2 are alternate interior angles, 2 and 3 are corresponding angles, and 3 and 4 are alternate exterior angles. What type of angle pair is 1 and 4?
- 1 and 4 are supplementary angles.
- 1 and 4 are corresponding angles.
- 1 and 4 are vertical angles.
- 1 and 4 are alternate exterior angles.
A: 1 and 4 are corresponding angles.
If right: Step 1 Draw two lines m, n, and a transversal p such that 1 and 2 are alternate interior angles. They should lie on opposite sides of the transversal p between lines m and n. | Step 2 2 and 3 are corresponding angles. Corresponding angles lie on the same side of the transversal p and on the same sides of lines m and n. Add 3 to the drawing. Step 3 3 and 4 are alternate exterior angles. They should lie on opposite sides of the transversal p and outside lines m and n. Add 4 to the drawing. | 1 and 4 are corresponding angles. They lie on the same side of the transversal p and on the same sides of lines m and n.
If wrong: Step 1 Draw two lines m, n, and a transversal p such that 1 and 2 are alternate interior angles. They should lie on opposite sides of the transversal p between lines m and n. | Step 2 2 and 3 are corresponding angles. Corresponding angles lie on the same side of the transversal p and on the same sides of lines m and n. Add 3 to the drawing. Step 3 3 and 4 are alternate exterior angles. They should lie on opposite sides of the transversal p and outside lines m and n. Add 4 to the drawing. | 1 and 4 are corresponding angles. They lie on the same side of the transversal p and on the same sides of lines m and n.

Q: Find .
- m = 40°
- m = 45°
- m = 35°
- m = 50°
A: m = 35°
If right: | Corresponding Angles Postulate | Subtract x from both sides. | Add 70 to both sides. | Divide both sides by 2. m | m | Substitute 35 for x. Simplify.
If wrong: | Corresponding Angles Postulate | Subtract x from both sides. | Add 70 to both sides. | Divide both sides by 2. m | m | Substitute 35 for x. Simplify.

Q: Find m.
- m =
- m =
- m =
- m =
A: m =
If right: | Alternate Exterior Angles Theorem | Subtract from both sides. | Divide both sides by . m | Substitute 24 for .
If wrong: | Alternate Exterior Angles Theorem | Subtract from both sides. | Divide both sides by . m | Substitute 24 for .

Q: Use the slope formula to determine the slope of the line containing points A(6, –7) and
B(9, –9).
- 0
- undefined
A: (a choice with no text)
If right: Substitute (6, –7) for () and (9, –9) for () in the slope formula.
If wrong: Substitute (6, –7) for () and (9, –9) for () in the slope formula.

Q: Write the equation of the line with slope 2 through the point (4,7) in point-slope form.
A: (a choice with no text)
If right: First write the point-slope formula. Then substitute 2 for m, 4 for , and 7 for .
If wrong: First write the point-slope formula. Then substitute 2 for m, 4 for , and 7 for .

Q: Determine whether the pair of lines and are parallel, intersect, or coincide.
- intersect
- coincide
- parallel
A: intersect
If right: Solve the first equation for y to find the slope-intercept form. Compare the slopes and y-intercepts of both equations. The slope of the first equation is –4 and the y-intercept is . | The slope of the second equation is 4 and the y-intercept is 1. The lines have different slopes, so they intersect.
If wrong: Solve the first equation for y to find the slope-intercept form. Compare the slopes and y-intercepts of both equations. The slope of the first equation is –4 and the y-intercept is . | The slope of the second equation is 4 and the y-intercept is 1. The lines have different slopes, so they intersect.

Q: Classify by its angle measures, given m, m, and m.
- obtuse triangle
- acute triangle
- right triangle
- equiangular triangle
A: obtuse triangle
If right: and form a linear pair, so they are supplementary. Therefore . By substitution, . So . is an obtuse triangle by definition.
If wrong: and form a linear pair, so they are supplementary. Therefore . By substitution, . So . is an obtuse triangle by definition.

Q: is an isosceles triangle. has length . = and = . Find AB.
- = 110
- =15
- = 43
- = 5
A: =15
If right: Step 1 Find the value of x. | Step 2 Find .
If wrong: Step 1 Find the value of x. | Step 2 Find .

Q: Find and , given , , and .
- m
- ,
- ,
- ,
A: m
If right: | Third Angles Theorem | Definition of congruent angles | Substitute for and for . | Subtract from both sides. | Divide both sides by –3. So . Since , .
If wrong: | Third Angles Theorem | Definition of congruent angles | Substitute for and for . | Subtract from both sides. | Divide both sides by –3. So . Since , .

Q: Find m, given , , and m.
- m
- m
- m
- m
A: m
If right: The Third Angles Theorem states that if two angles of one triangle are congruent to two angles of another triangle, then the third pair of angles are congruent. It is given that and . Therefore, . So, m.
If wrong: The Third Angles Theorem states that if two angles of one triangle are congruent to two angles of another triangle, then the third pair of angles are congruent. It is given that and . Therefore, . So, m.

Q: Given that and m = 27°, find m.
- m = 77°
- m = 63°
- m = 23°
- m = 113°
A: m = 63°
If right: | Triangle Sum Theorem | Substitution. | Simplify. | Subtract 113 from both sides. | Corresponding parts of congruent triangles are congruent. | Definition of congruent angles | Corresponding parts of congruent triangles are congruent.
If wrong: | Triangle Sum Theorem | Substitution. | Simplify. | Subtract 113 from both sides. | Corresponding parts of congruent triangles are congruent. | Definition of congruent angles | Corresponding parts of congruent triangles are congruent.

Q: Find the value of x.
- x = 3
- x = 4
- x = 2
- x = 8
A: x = 3
If right: The triangles can be proved congruent by the SAS Postulate. By CPCTC, . Solve the equation for x.
If wrong: The triangles can be proved congruent by the SAS Postulate. By CPCTC, . Solve the equation for x.

Q: Find m.
- m = 30^ °
- m = 60^ °
- m = 50 °
- m = 75 °
A: m = 75 °
If right: m m | Isosceles Triangle Theorem m m m | Triangle Sum Theorem | Substitute for m and substitute for m and m. | Simplify and subtract 30 from both sides. | Divide both sides by 5. Thus m.
If wrong: m m | Isosceles Triangle Theorem m m m | Triangle Sum Theorem | Substitute for m and substitute for m and m. | Simplify and subtract 30 from both sides. | Divide both sides by 5. Thus m.

Q: Show that GHIJ is a parallelogram for x = 5 and y = 8.
- [1] 15
[2] 15
[3] 24
[4] 24
[5] both sets of opposite sides are congruent.
- [1] 15
[2] 24
[3] 15
[4] 24
[5] one set of opposite sides is parallel and congruent.
- [1] 15
[2] 15
[3] 24
[4] 24
[5] both sets of opposite sides are parallel.
- [1] 24
[2] 24
[3] 15
[4] 15
[5] both sets of opposite angles are congruent.
A: [1] 15
[2] 15
[3] 24
[4] 24
[5] both sets of opposite sides are congruent.
If right: | | Given 15 | GJ = 7(5)20 = 15 | Substitute and simplify. | | Given 24 | 24 | Substitute and simplify. Since HI = GJ and GH = JI, GHIJ is a parallelogram because both sets of opposite sides are congruent.
If wrong: | | Given 15 | GJ = 7(5)20 = 15 | Substitute and simplify. | | Given 24 | 24 | Substitute and simplify. Since HI = GJ and GH = JI, GHIJ is a parallelogram because both sets of opposite sides are congruent.

Q: TRSU is a rhombus. Find .
- = 7
- = 1
- = 5
- = 3
A: = 7
If right: | Definition of a rhombus | Substitute the given values. | Subtract 2x and 2 from both sides. | Divide both sides by 3. | Definition of a rhombus | Substitute for TR. | Substitute 1 for x. | Simplify.
If wrong: | Definition of a rhombus | Substitute the given values. | Subtract 2x and 2 from both sides. | Divide both sides by 3. | Definition of a rhombus | Substitute for TR. | Substitute 1 for x. | Simplify.

Q: Which three lengths could be the lengths of the sides of a triangle?
- 12 cm, 5 cm, 17 cm
- 10 cm, 15 cm, 24 cm
- 9 cm, 22 cm, 11 cm
- 21 cm, 7 cm, 6 cm
A: 10 cm, 15 cm, 24 cm

Q: Identify the lines that are perpendicular:
; ; ;
- and are perpendicular.
- None of the lines are perpendicular.
- and are perpendicular.
- and are perpendicular; and are perpendicular.
A: and are perpendicular; and are perpendicular.
If right: is horizontal, and is vertical. These lines are perpendicular. Check if the product of the slopes of the other two lines is . has slope . has slope . The product of the slopes is , so these two lines are also perpendicular.
If wrong: is horizontal, and is vertical. These lines are perpendicular. Check if the product of the slopes of the other two lines is . has slope . has slope . The product of the slopes is , so these two lines are also perpendicular.

Q: Write an equation in slope-intercept form for the line parallel to y = x – 2 that passes through the point (8, –2).
- y = x – 2
- y = x –
- y = x +
- y = x –
A: y = x –
If right: Parallel lines have the same slope. Since the given line has a slope of , the parallel line has a slope of . y – y_1 = m(x – x_1) | Use the point-slope form to write an equation. y – (–2) = (x – (8)) | Substitute for m and (8, –2) for (x_1, y_1). y = x – | Distribute on the right side. Add –2 to both sides.
If wrong: Parallel lines have the same slope. Since the given line has a slope of , the parallel line has a slope of . y – y_1 = m(x – x_1) | Use the point-slope form to write an equation. y – (–2) = (x – (8)) | Substitute for m and (8, –2) for (x_1, y_1). y = x – | Distribute on the right side. Add –2 to both sides.

Q: Use the Distance Formula and the Pythagorean Theorem to find the distance, to the nearest tenth, from T(4, –2) to U(–2, 3).
- –1.0 units
- 3.4 units
- 0.0 units
- 7.8 units
A: 7.8 units
If right: Method 1 Substitute the values for the coordinates of T and U into the Distance Formula. | Method 2 Use the Pythagorean Theorem. Plot the points on a coordinate plane. Then draw a right triangle. | Count the units for sides a and b. and . Then apply the Pythagorean Theorem.
If wrong: Method 1 Substitute the values for the coordinates of T and U into the Distance Formula. | Method 2 Use the Pythagorean Theorem. Plot the points on a coordinate plane. Then draw a right triangle. | Count the units for sides a and b. and . Then apply the Pythagorean Theorem.

Q: Use the diagonals to determine whether a parallelogram with vertices is a rectangle, rhombus, or square. Give all the names that apply.
- rectangle, rhombus, square
- rectangle, rhombus
- rectangle
- square
A: rectangle, rhombus, square
If right: Step 1 Graph ABCD. Step 2 The diagonals of a rectangle are congruent. Find AC and BD to determine if ABCD is a rectangle. Since , ABCD is a rectangle. Step 3 The diagonals of a rhombus are perpendicular. Find the slopes of and to determine if ABCD is a rhombus. slope of slope of Since , and ABCD is a rhombus. Since ABCD is both a rhombus and a rectangle, it is also a square.
If wrong: Step 1 Graph ABCD. Step 2 The diagonals of a rectangle are congruent. Find AC and BD to determine if ABCD is a rectangle. Since , ABCD is a rectangle. Step 3 The diagonals of a rhombus are perpendicular. Find the slopes of and to determine if ABCD is a rhombus. slope of slope of Since , and ABCD is a rhombus. Since ABCD is both a rhombus and a rectangle, it is also a square.

Q: Which of the following is the best name for figure MNOP with vertices , , , and ?
- parallelogram
- rectangle
- rhombus
- square
A: square
If right: Find the slope of each side to determine if the sides are perpendicular. slope of | slope of slope of | slope of Both sets of opposite sides are parallel, so MNOP is a parallelogram. Rectangle, rhombus, and square are three types of parallelograms. MNOP has perpendicular sides, as the slopes of adjacent sides are negative reciprocals, so MNOP must be a rectangle or square. Find the lengths of consecutive sides to determine if the figure is a square. The sides are congruent, so the figure is a square.
If wrong: Find the slope of each side to determine if the sides are perpendicular. slope of | slope of slope of | slope of Both sets of opposite sides are parallel, so MNOP is a parallelogram. Rectangle, rhombus, and square are three types of parallelograms. MNOP has perpendicular sides, as the slopes of adjacent sides are negative reciprocals, so MNOP must be a rectangle or square. Find the lengths of consecutive sides to determine if the figure is a square. The sides are congruent, so the figure is a square.

Q: Compare m with m.
A: (a choice with no text)
If right: Compare the side lengths in and . By the Converse of the Hinge Theorem, .
If wrong: Compare the side lengths in and . By the Converse of the Hinge Theorem, .

Q: Name the second largest of the four angles named in the figure (not drawn to scale) if the side included by and is 11 cm, the side included by and is 16 cm, and the side included by and is 14 cm.
A: (a choice with no text)

Q: Two sides of a triangle have lengths 5 and 12. Which inequalities represent the possible lengths for the third side, x?
A: (a choice with no text)

Q: What is the range of possible values for x?
The diagram is not to scale.
A: (a choice with no text)

Q: Which labeled angle has the greatest measure? The diagram is not to scale.
- not enough information in the diagram
A: (a choice with no text)