Review assignment with sub 1-24

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Review Assignment because the Spring Exam is coming!

Multiple Choice

Identify the choice that best completes the statement or answers the question.

____ 1. Complete the conjecture.

The sum of two odd numbers is _____.

a. even c. sometimes odd, sometimes even

b. odd d. even most of the time

____ 2. Show that the conjecture is false by finding a counterexample.

If ab, then a_b 0.

a. a 11, b  3 c. a3, b11 b. a 11, b 3 d. a 11, b3

____ 3. Write a conditional statement from the statement. A horse has 4 legs.

a. If it has 4 legs then it is a horse. c. If it is a horse then it has 4 legs. b. Every horse has 4 legs. d. It has 4 legs and it is a horse.

____ 4. Give an example of corresponding angles.

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____ 5. Identify the transversal and classify the angle pair 11 and 7.

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____ 6. 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? a.

1 and 4 are supplementary angles. b.

1 and 4 are corresponding angles. c.

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____ 7. Find mABC.

a. mABC = 40° c. mABC = 35° b. mABC = 45° d. mABC = 50°

____ 8. Find mRST.

a. mRST = 108 c. mRST = 156

b. mRST = 24 d. mRST = 72

____ 9. Find m1 in the diagram. (Hint: Draw a line parallel to the given parallel lines.)

a. m195 c. m185

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____ 10. Write and solve an inequality for x.

a. x 2 c. x1

b. x 2 d. x 2

____ 11. Use the slope formula to determine the slope of the line.

a. 0 c. 3₂

b. 2

3 d. undefined

____ 12. Use slopes to determine whether the lines are parallel, perpendicular, or neither.

AB 

andCDforA(3,5),B(2,7),C(10,5), andD(6,15)

a. neither c. parallel

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____ 13. Graph the line y34(x6).

a. c.

b. d.

____ 14. Write an equation for the line parallel to the line shown that passes through the point (–2, 3).

a. y = 3x – 3 c. y = 1₃x + 11₃

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____ 15. Find the missing coordinates for the rhombus.

a. (C,D) c. (A,DC)

b. (AC,D) d. (AC,DC)

____ 16. Find mQ.

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____ 17. Find CA.

a. CA = 10 b. CA = 12 c. CA = 14

d. Not enough information. An equiangular triangle is not necessarily equilateral.

____ 18. Find the measures BC and AC.

a. BC6.4,AC4.6 c. BC6.4,AC2.3 b. BC4.6,AC6.4 d. BC2.3,AC6.4

____ 19. Write an equation in point-slope form for the perpendicular bisector of the segment with endpoints A(2,2) and B(5,4).

a. y37

2(x1.5) c. y1 2

7(x3.5)

b. y3 2

7(x1.5) d. y1

7

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____ 20. Find the circumcenter of ABC with vertices A(2,4),B(2,2), andC(4,2).

a. (1, 1) c. 1₂, 1₂



 _

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____ 21. In ABC, show that midsegment KL is parallel to AB and that KL 1 2AB.

a. K(4,1). L 4 4 2 ,

22 2

 

 _ (4,2).

The slope of KL 1

4. The slope of AB 1

4. The slopes are equal so KL AB.

The length of KL 17. The length of AB2 17. AB 1 2KL.

b. K(4,1). L 4 4 2 ,

22 2

 

 _ (0,0). The slope of KL 1

c. K(4,1). L 4 4 2 ,

22 2

 

 _ (0,0). The slope of KL 1

d. K(4,1). L 4 4 2 ,

22 2

 

 _ (0,0). The slope of KL 1

____ 22. Write the sides of IJK in order from shortest to longest.

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____ 23. The diagram shows the approximate distances from Houston to Dallas and from Austin to Dallas. What is the range of distances, d, from Austin to Houston?

a. 40d 440 c. 200d 240 b. 40d 440 d. 0d440

____ 24. The door on a spacecraft is formed with 6 straight panels that overlap to form a regular hexagon. What is the measure of YXZ?

a. mYXZ = 60o _c. _m_{YXZ = 720}o

b. mYXZ = 120o _{d. m}_{YXZ = 45}o

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____ 26. MNOP is a parallelogram. Find MP.

a. MP = 25 c. MP = 20

b. MP = 30 d. MP = 6

____ 27. Three vertices of parallelogram WXYZ are X(–2,–3), Y(0, 5), and Z(7, 7). Find the coordinates of vertex W.

a. (4, 0) c. (5, 0)

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____ 28. Show that GHIJ is a parallelogram for x = 5 and y = 8.

Complete the explanation.

HI5x10 GJ7x20 Given

HI5(5)10 [1] GJ = 7(5)20 = [2] Substitute and simplify.

GH3y JI5y16 Given

GH3(8) [3] JI5(8)16 [4] Substitute and simplify.

Since HI = GJ and GH = JI, GHIJ is a parallelogram because [5]. a. [1] 15

[2] 15 [3] 24 [4] 24

[5] both sets of opposite sides are congruent.

b. [1] 15 [2] 24 [3] 15 [4] 24

[5] one set of opposite sides is parallel and congruent.

c. [1] 15 [2] 15 [3] 24 [4] 24

[5] both sets of opposite sides are parallel.

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Numeric Response

1. Find the value of x so that mn.

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Review Assignment because the Spring Exam is coming!

Answer Section

MULTIPLE CHOICE 1. A

List some examples and look for a pattern.

358 3710 5712 5914

2. A

Pick values for a and b that follow the condition ab. Then substitute them into the second inequality to see if the conjecture holds.

Values of a and b a > b a_b 0 Conclusion

Let a 4 and b 1. 41 4₁ 0 The conjecture holds.

Let a 11 and b 3. 113 11₃ 0 The conjecture holds.

Let a 11 and b  3. 11 3 11_₃ 0 The conjecture is false.

a 11 and b  3 is a counterexample.

The conjecture is false when a is positive and b is negative. 3. C

Identify the hypothesis and conclusion.

Hypothesis Conclusion

A horse has 4 legs.

If it is a horse, then it has 4 legs.

4. A

Corresponding angles lie on the same side of a transversal, on the same sides of the two lines the transversal crosses. So, 8 and 4 are corresponding angles.

5. A

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6. B

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.

7. C

(x) (3x70) Corresponding Angles Postulate 02x70 Subtract x from both sides.

702x Add 70 to both sides.

35x Divide both sides by 2.

mABC3x70

mABC3(35)7035 Substitute 35 for x. Simplify.

8. D

(3x) (4x24) Alternate Exterior Angles Theorem

x 24 Subtract 4x from both sides. x24 Divide both sides by 1.

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9. C

Step 1 Draw line l parallel to lines m and n. Step 2 Find mx.

m1mxmy Use the Corresponding Angles Postulate with lines m and l. mx35.

Step 3 Find my.

Use the Same-Side Interior Angles Theorem with lines l and n. my 18013050. Step 4 Find m1.

m1mxmy355085

10. A

DADC _DC_{is the shorter segment.}

2x48 Substitute 2x4 for DA and 8 for DC. 2x4 Subtract 4 from both sides.

x2 Divide both sides by 2 and simplify.

11. B

Substitute (6, –7) for (x₁,y₁) and (9, –9) for (x₂,y₂) in the slope formula.

m _xy2y1

2x1

 ₉9_₆7  ₃2

12. A

slope of AB 3 2 57 

5

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13. B

The equation is given in point-slope form yy₁ m(xx₁). The slope is m4 4

1 and the coordinates of a point on the line are (6,3).

Plot the point (6,3) and then rise 4 and run 1 to locate another point. Draw the line connecting the two points.

14. A 15. B

The horizontal sides are parallel, so the y-value is the same as in the point (A,D). The missing y-coordinate is D.

A rhombus has congruent sides, so the x-value is the same horizontal distance from (C, 0) as the point (A,D) is from the point (0, 0). This horizontal distance is A units.

The missing x-coordinate is AC. 16. D

mQ mR2x15 Isosceles Triangle Theorem mP mQ mR180 Triangle Sum Theorem

x2x152x15180 Substitute x for mP and substitute 2x15 for mQ and mR.

5x150 Simplify and subtract 30 from both sides. x30 Divide both sides by 5.

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17. C

ABC is equilateral. Equiangular triangles are equilateral. 2s10s2 Definition of equilateral triangle.

s12 Subtract s and add 10 to both sides of the equation.

AB2s10

AB2 12 10 Substitute 12 for s in the equation for AB.

AB14 Simplify.

CAAB Definition of equilateral triangle. CA14 Substitute 14 for AB.

18. A

BABC Perpendicular Bisector Theorem BC6.4 Substitute 6.4 for BA.

AXXC Given

AX2.3 Substitute.

ACAXXC Segment Addition Postulate AC2.32.3 Substitute.

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19. A

Step 1 Plot AB.

The perpendicular bisector of AB is perpendicular to AB at its midpoint.

Step 2 Find the midpoint of AB. Midpoint of AB 2₂5,24

2       

 (1.5,3)

Step 3 Find the slope of the perpendicular bisector. Slope of AB ₍₅₎(4)_₍(2)_₂₎ 2

7

Since the slopes of perpendicular lines are opposite reciprocals, the slope of the perpendicular bisector is 7₂.

Step 4 Use point-slope form to write the equation. yy₁ m(xx₁)

y37

2(x1.5)

20. A

Step 1 Find equations for two perpendicular bisectors.

Since two sides of the triangle lie along the axes, use the graph to find the perpendicular bisectors of these two sides. The perpendicular bisector of AB is y1, and the perpendicular bisector of BC is x1.

Step 2 Find the intersection of the two equations.

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21. D

Find K by finding the midpoint of AC 4 4 2 ,

2 4 2

 

 _ (4,1)

Find L by finding the midpoint of BC  4 4 2 ,

22 2

 

 _ (0,0). Find the slope of KL 0 1

0 4  1 4

Find the slope of AB 2 4 4 4 

2 8 

1 4.

The slopes are equal so KL AB.

Find the length of KL (0 4)2_₍₀_{ }₁₎2 _ _17.

Find the length of AB (4 4)2_₍_₂_{ }₄₎2 _ ₆₈ __{2 17.}

KL 1 2 AB

22. A

By the Triangle Sum Theorem, mJ 180 (58 62)60. The smallest angle is I, so the shortest side is JK.

The largest angle is K, so the longest side is IJ. From shortest to longest, the sides are JK,IK,IJ. 23. A

d200240 d240200 240200d Triangle Inequality Theorem

d 40 d 40 440d Simplify.

40d440 Combine the inequalities.

The distance from Austin to Houston is greater than 40 miles and less than 440 miles. 24. A

YXZ is an exterior angle of the regular hexagon. All exterior angles add to 360, and for a regular hexagon there are 6 congruent exterior angles, so mYXZ 360

6 60

25. C

To find ST:

ST UR In a parallelogram, opposite sides are congruent. STUR Definition of congruent segments

ST25 Substitute 25 for UR.

To find XT:

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26. B

MP NO Opposite sides of a parallelogram are congruent. MP = NO Definition of congruent segments

5x3x12 Substitute.

x6 Simplify and solve.

MP = 5x = 5(6) = 30 Substitute and solve for entire segment measure.

27. D

Count the units of horizontal and vertical movement required to move from Y to X. This same pattern of motion will occur between Z and W because opposites sides of a parallelogram are parallel and therefore have the same slope.

Step 1 Graph and label the given points.

Step 2 Find the slope of XY by counting the units from X to Y. The rise from –3 to 5 is 8. The run from –2 to 0 is 2.

Step 3 Start at Z and count the same number of units, moving down and to the left to make the slope the same. A rise of 8 to 7 requires a starting point at –1. A run of 2 to 7 requires a starting point at 5. W is at (5, –1).

Step 4 Use the slope formula to verify that XY WZ. slope of XY  _3₂_5₀  _8₂ 4

slope of WZ ₅1_₇7  _8₂ 4

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28. A

HI5x10 GJ7x20 Given

HI5(5)10 15 GJ = 7(5)20 = 15 Substitute and simplify.

GH3y JI5y16 Given

GH3(8) 24 JI5(8)16 24 Substitute and simplify.

Since HI = GJ and GH = JI, GHIJ is a parallelogram because both sets of opposite sides are congruent.

NUMERIC RESPONSE 1. 17