Geometry Midterm 2012 - Weeblywhamrsm.weebly.com/.../8266221/examview_-_geomet… · ·...

Name: ________________________ Class: ___________________ Date: __________ ID: A

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Geometry Midterm 2012

Numeric Response

1. The supplement of an angle is 26 more than five times its complement. Find the measure of the angle.

2. A satellite completely orbits Earth in 216 days. Determine the angle through which the satellite travels over a

period of 12 days.

3. A right triangle is formed by the x-axis, the y-axis and the line y = −2x + 3. Find the length of the hypotenuse.

Round your answer to the nearest hundredth.

4. Find the value of n in the triangle.

5. In parallelogram LMNO, NO = 10.2, and LO = 14.7. What is the perimeter of parallelogram LMNO?

Short Answer

6. D is between C and E. CE = 6x, CD = 4x + 8, and DE = 27. Find CE.

7. K is the midpoint of JL. JK = 6x and KL = 3x + 3. Find JK, KL, and JL.

8. BD→

bisects ∠ABC, m∠ABD = (7x − 1)°, and m∠DBC = (4x + 8)°. Find m∠ABD.

9. Find the measure of the supplement of ∠R, where m∠R = (8z + 10)°

Name: ________________________ ID: A

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10. A billiard ball bounces off the sides of a rectangular billiards table in such a way that ∠1 ≅ ∠3, ∠4 ≅ ∠6, and

∠3 and ∠4 are complementary. If m∠1 = 26.5°, find m∠3, m∠4, and m∠5.

11. The width of a rectangular mirror is 3

4 the measure of the length of the mirror. If the area is 192 in

2, what are

the length and width of the mirror?

12. Use the Distance Formula and the Pythagorean Theorem to find the distance, to the nearest tenth, from T(4,

–2) to U (–2, 3).

13. Use the Converse of the Corresponding Angles Postulate and ∠1 ≅ ∠2 to show that l Ä m.

14. Use the slope formula to determine the slope of the line containing points A(6, –7) and

B(9, –9).

Name: ________________________ ID: A

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15. Write the equation of the line with slope 2 through the point (4, 7) in point-slope form.

16. Graph the line y − 3 = 4(x − 6).

17. Determine whether the pair of lines 12x + 3y = 3 and y = 4x + 1 are parallel, intersect, or coincide.

18. ∆ABC is an isosceles triangle. AB is the longest side with length 4x + 4. BC = 8x + 3 and CA = 7x + 8. Find

AB.

19. One of the acute angles in a right triangle has a measure of 34.6°. What is the measure of the other acute

angle?

20. Find m∠DCB, given ∠A ≅ ∠F, ∠B ≅ ∠E, and m∠CDE = 46°.

Name: ________________________ ID: A

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21. Given: RT⊥SU , ∠SRT ≅ ∠URT, RS ≅ RU . T is the midpoint of SU .

Prove: ∆RTS ≅ ∆RTU

Complete the proof.

Proof:

Statements Reasons

1. RT⊥SU 1. Given

2. ∠RTS and ∠RTU are right angles. 2. [1]

3. ∠RTS ≅ ∠RTU 3. Right Angle Congruence Theorem

4. ∠SRT ≅ ∠URT 4. Given

5. ∠S ≅ ∠U 5. [2]

6. RS ≅ RU 6. Given

7. T is the midpoint of SU . 7. Given

8. ST ≅ UT 8. Definition of midpoint

9. RT ≅ RT 9. [3]

10. ∆RTS ≅ ∆RTU 10. Definition of congruent triangles

22. Given: A(3, –1), B(5, 2), C(–2, 0), P(–3, 4), Q(–5, –3), R(–6, 2)

Prove: ∠ABC ≅ ∠RPQ

Complete the paragraph proof.

AB = RP = 13 , BC = [1] = 53 , and CA = QR = 26 . So AB ≅ [2], BC ≅ PQ, and CA ≅ QR. Therefore

∆ABC ≅ [3] by [4], and ∠ABC ≅ ∠RPQ by [5].

23. Write an equation in point-slope form for the perpendicular bisector of the segment with endpoints A(−2,2)

and B(5,4).

Name: ________________________ ID: A

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24. In ∆ABC, BY = 26.4 and CO = 24. AX BY ,and CZ are medians. Find BO.

25. The size of a TV screen is given by the length of its diagonal. The screen aspect ratio is the ratio of its width

to its height. The screen aspect ratio of a standard TV screen is 4:3. What are the width and height of a 27"

TV screen?

26. Tell if the measures 6, 14, and 13 can be side lengths of a triangle. If so, classify the triangle as acute, right,

or obtuse.

27. An architect designs the front view of a house with a gable roof that has a 45°-45°-90° triangle shape. The

overhangs are 0.5 meter each from the exterior walls, and the width of the house is 16 meters. What should

the side length l of the triangle be? Round your answer to the nearest meter.

28. Find the measure of each interior angle of a regular 45-gon.

29. Find the measure of each exterior angle of a regular decagon.

Name: ________________________ ID: A

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30. The door on a spacecraft is formed with 6 straight panels that overlap to form a regular hexagon. What is the

measure of ∠YXZ?

31. Write a two-column proof.

Given: ABDF and FBCD are parallelograms.

Prove: ∠BCD ≅ ∠ABF

Complete the proof.

Proof:

Statements Reasons

1. ABDF and FBCD are parallelograms. 1. Given

2. ∠BCD ≅ ∠DFB 2. [1]

3. DF Ä AB 3. Opposite sides in a parallelogram are

parallel.

4. ∠DFB ≅ ∠ABF 4. [2]

5. ∠BCD ≅ ∠ABF 5. Substitution

32. Two vertices of a parallelogram are A(2, 3) and B(8, 11), and the intersection of the diagonals is X(7, 6). Find

the coordinates of the other two vertices.

Name: ________________________ ID: A

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33. Show that all four sides of square ABCD are congruent and that AB ⊥ BC.

Name: ________________________ ID: A

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34. Given: ABCD is a rectangle. W, X, Y, and Z are midpoints.

Prove: WXYZ is a rhombus.

Complete the proof.

Proof:

Statements Reasons

1. ABCD is a rectangle.

W, X, Y, and Z are midpoints.

1. Given

2. ∠A, ∠B, ∠C, and ∠D are right angles 2. Definition of a rectangle

3. ∠A ≅ ∠B ≅ ∠C ≅ ∠D 3. Right Angle Theorem

4. AB ≅ CD, BC ≅ AD 4. Theorem: Both pairs of opposite sides are

congruent.

5. AW =1

2AD, WD =

1

2AD,

AX =1

2AB, XB =

1

2AB,

BY =1

2BC, YC =

1

2BC,

CZ =1

2CD, ZD =

1

2CD

5. [1]

6. 1

2AB =

1

2CD,

1

2BC =

1

2AD 6. Division Property of Equality

7. AB = AB, BC = BC

CD = CD, AD = AD

7. Reflexive Property of Equality

8. AW = WD = BY = YC,

AX = XB = CZ = ZD

8. Substitution

9. AW ≅ WD ≅ BY ≅ YC,

AX ≅ XB ≅ CZ ≅ ZD

9. Definition of Congruent Segments

10. ∆AXW ≅ ∆BXY ≅ ∆DZW ≅ ∆CZY 10. [2]

11. WX ≅ XY ≅ YZ ≅ ZW 11. CPCTC

12. WXYZ is a rhombus 12. Definition of a rhombus

Name: ________________________ ID: A

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35. Use the diagonals to determine whether a parallelogram with vertices

A(−1,−2), B(−2,0), C(0,1), and D(1,−1)is a rectangle, rhombus, or square. Give all the names that apply.

36. Which of the following is the best name for figure MNOP with vertices M(−3,5), N(0,9), O(4,6), and P(1,2)?

37. In kite PQRS, m∠QPO = 50° and m∠QRO = 70°. Find m∠PSR.

38. Given isosceles trapezoid ABCD with AB ≅ CD, BY = 10.3, and AC = 17.2. Find YD.

39. QS = 3x + 4 and RT = 8x − 10. Find the value of x so that QRST is isosceles.

Name: ________________________ ID: A

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40. R is the midpoint of AB. T is the midpoint of AC. S is the midpoint of BC. Use the diagram to find the

coordinates of T, the area of ∆RST, and AB. Round your answers to the nearest tenth.

41. Two angles with measures (2x2

+ 3x − 5)° and (x2

+ 11x − 7)° are supplementary. Find the value of x and the

measure of each angle.

42. Two lines intersect to form two pairs of vertical angles. ∠1 with measure (20x + 7)º and ∠3 with measure

(5x + 7y + 49)º are vertical angles. ∠2 with measure (3x − 2y + 30)º and ∠4 are vertical angles. Find the values

x and y and the measures of all four angles.

43. Violin strings are parallel. Viewed from above, a violin bow in two different positions forms two transversals

to the violin strings. Find x and y in the diagram.

Name: ________________________ ID: A

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44. ∆ABF ≅ ∆EDG. ∆ABF and ∆GCF are equilateral. AG = 21 and CG =1

4AB. Find the total distance from A to

B to C to D to E.

45. Find the missing coordinates for the rhombus.

46. Polygon ABCDEFGHIJKL is a regular dodecagon (12-sided polygon). Sides EF and GH are extended so that

they meet at point O in the exterior of the polygon. Find m∠FOG.

47. The perimeter of isosceles trapezoid WXYZ is 55.9. AB is the midsegment of WXYZ. If XY = 3(ZY), find ZW,

WX, XY, and ZY.

ID: A

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Geometry Midterm 2012

Answer Section

NUMERIC RESPONSE

1. ANS: 74

PTS: 1 DIF: Average NAT: 12.2.1.f STA: 6MG2.2

TOP: 1-4 Pairs of Angles KEY: supplementary angles | complementary angles

2. ANS: 20

PTS: 1 DIF: Advanced NAT: 12.2.1.f

TOP: 1-7 Transformations in the Coordinate Plane

KEY: transformations | application | rotations

3. ANS: 3.35

PTS: 1 DIF: Advanced NAT: 12.3.3.d STA: GE15.0

TOP: 3-6 Lines in the Coordinate Plane

4. ANS: 11

PTS: 1 DIF: Average NAT: 12.3.3.f

TOP: 5-4 The Triangle Midsegment Theorem

5. ANS: 49.8

PTS: 1 DIF: Average NAT: 12.2.1.h STA: GE7.0

TOP: 6-2 Properties of Parallelograms

SHORT ANSWER

6. ANS:

CE = 105CE = CD + DE Segment Addition Postulate

6x = 4x + 8( ) + 27 Substitute 6x for CE and 4x + 8 for CD.

6x = 4x + 35 Simplify.2x = 35 Subtract 4x from both sides.

2x

2=

35

2Divide both sides by 2.

x =35

2 or 17.5 Simplify.

CE = 6x = 6 17.5( ) = 105

PTS: 1 DIF: Average REF: Page 15

OBJ: 1-2.3 Using the Segment Addition Postulate NAT: 12.3.5.a

STA: GE1.0 TOP: 1-2 Measuring and Constructing Segments

KEY: segment addition postulate

ID: A

2

7. ANS:

JK = 6, KL = 6, JL = 12

Step 1 Write an equation and solve.JK = KL K is the midpoint of JL.6x = 3x + 3 Substitute 6x for JK and 3x + 3 for KL.3x = 3 Subtract 3x from both sides.x = 1 Divide both sides by 3.

Step 2 Find JK, KL, and JL.

JK = 6x = 6 1( ) = 6

KL = 3x + 3 = 3(1) + 3 = 6

JL = JK + KL = 6 + 6 = 12


OBJ: 1-2.5 Using Midpoints to Find Lengths NAT: 12.2.1.e

STA: GE1.0 TOP: 1-2 Measuring and Constructing Segments

KEY: midpoints | length

8. ANS:

m∠ABD = 20°

Step 1 Solve for x.

m∠ABD = m∠DBC Definition of angle bisector.

(7x − 1)° = (4x + 8)° Substitute 7x − 1 for ∠ABD and 4x + 8 for ∠DBC.

7x = 4x + 9 Add 1 to both sides.3x = 9 Subtract 4x from both sides.

x = 3 Divide both sides by 3.

Step 2 Find m∠ABD.

m∠ABD = 7x − 1 = 7(3) − 1 = 20°


OBJ: 1-3.4 Finding the Measure of an Angle NAT: 12.2.1.f

STA: GE1.0 TOP: 1-3 Measuring and Constructing Angles

KEY: angle bisectors | angle measures

9. ANS:

(170 − 8z)°

Subtract from 180º and simplify.

180° − (8z + 10)° = 180 − 8z − 10 = (170 − 8z)°


OBJ: 1-4.2 Finding the Measures of Complements and Supplements

NAT: 12.3.3.g STA: 6MG2.2 TOP: 1-4 Pairs of Angles

KEY: complementary angles | supplementary angles

ID: A

3

10. ANS:

m∠3 = 26.5°; m∠4 = 63.5°; m∠5 = 53°

Since ∠1 ≅ ∠3, m∠1 ≅ m∠3.

Thus m∠3 = 26.5°.

Since ∠3 and ∠4 are complementary,

m∠4 = 90° − 26.5° = 63.5°.

Since ∠4 ≅ ∠6, m∠4 ≅ m∠6.

Thus m∠6 = 63.5°.

By the Angle Addition Postulate,

180° = m∠4 + m∠5 + m∠6

= 63.5° + m∠5 + 63.5°

Thus, m∠5 = 53°.

PTS: 1 DIF: Average REF: Page 30 OBJ: 1-4.4 Problem-Solving Application

NAT: 12.3.3.g STA: 6MG2.2 TOP: 1-4 Pairs of Angles

KEY: application | complementary angles | supplementary angles

11. ANS:

length = 16 in., width = 12 in.

The area of a rectangle is found by multiplying the length and width. Let l represent the length of the mirror.

Then the width of the mirror is 3

4l.

A = lw

192 = l(3

4l)

192 =3

4l

2

256 = l2

16 = l

The length of the mirror is 16 inches. The width of the mirror is 3

4(16) = 12 inches.

PTS: 1 DIF: Advanced NAT: 12.2.1.h STA: GE8.0

TOP: 1-5 Using Formulas in Geometry KEY: area | rectangles | application

ID: A

4

12. ANS:

7.8 units

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.

TU = x 2 − x 1ÊËÁÁ ˆ

¯˜̃

2+ y 2 − y 1ÊËÁÁ ˆ

¯˜̃

2

= −2 − 4( )2

+ 3 − −2( )2

= −6( ) 2+ 5( ) 2

= 61

≈ 7.8 units

Count the units for sides a and b. a = 6 and

b = 5. Then apply the Pythagorean Theorem.

c2

= a2

+ b2

= 62

+ 52

= 36 + 25 = 61

c ≈ 7.8 units


OBJ: 1-6.4 Finding Distances in the Coordinate Plane NAT: 12.2.1.e

STA: GE15.0 TOP: 1-6 Midpoint and Distance in the Coordinate Plane

KEY: congruent segments | distance formula | Pythagorean Theorem

13. ANS:

∠1 ≅ ∠2 is given. From the diagram, ∠1 and ∠2 are corresponding angles. So by the Converse of the

Corresponding Angles Postulate, l Ä m.

∠1 ≅ ∠2 is given. From the diagram, ∠1 and ∠2 are corresponding angles. So by the Converse of the

Corresponding Angles Postulate, l Ä m.

PTS: 1 DIF: Basic REF: Page 162

OBJ: 3-3.1 Using the converse of the Corresponding Angles Postulate

NAT: 12.3.3.g STA: GE7.0 TOP: 3-3 Proving Lines Parallel

14. ANS:

−2

3

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

m =y 2 − y1

x 2 − x 1

=−9 + 7

9 − 6=

−2

3

PTS: 1 DIF: Average REF: Page 182 OBJ: 3-5.1 Finding the Slope of a Line

NAT: 12.3.5.a STA: 7AF3.3 TOP: 3-5 Slopes of Lines

ID: A

5

15. ANS:

y − 7 = 2(x − 4)

First write the point-slope formula.y − y 1 = m(x − x 1 )

Then substitute 2 for m, 4 for x1 , and 7 for y1 .

y − 7 = 2(x − 4)

PTS: 1 DIF: Average REF: Page 191 OBJ: 3-6.1 Writing Equations of Lines

NAT: 12.3.5.a STA: 1A7.0 TOP: 3-6 Lines in the Coordinate Plane

16. ANS:

The equation is given in point-slope form y − y 1 = m(x − x 1 ).

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.

PTS: 1 DIF: Average REF: Page 191 OBJ: 3-6.2 Graphing Lines


ID: A

6

17. ANS:

intersect

Solve the first equation for y to find the slope-intercept form. Compare the slopes and y-intercepts of both

equations.

12x + 3y = 3

3y = −12x + 3

y = −4x +1

The slope of the first equation is –4 and the

y-intercept is 1.

y = 4x + 1

The slope of the second equation is 4 and the

y-intercept is 1.

The lines have different slopes, so they intersect.

PTS: 1 DIF: Average REF: Page 192 OBJ: 3-6.3 Classifying Pairs of Lines


18. ANS:

AB = 24

Step 1 Find the value of x.

BC = CA

8x + 3 = 7x + 8

x = 5

Step 2 Find AB.

AB = 4x + 4

= 4(5) + 4

= 24

PTS: 1 DIF: Average REF: Page 217 OBJ: 4-1.3 Using Triangle Classification

NAT: 12.3.3.f STA: GE12.0 TOP: 4-1 Classifying Triangles

19. ANS:

55.4°

Let the acute angles be ∠M and ∠N, with m∠M = 34.6°.

m∠M + m∠N = 90° The acute angles of a right triangle are complementary.

34.6° + m∠N = 90° Substitute 34.6° for m∠M .

m∠N = 55.4° Subtract 34.6° from both sides.

PTS: 1 DIF: Basic REF: Page 225

OBJ: 4-2.2 Finding Angle Measures in Right Triangles NAT: 12.3.3.f

STA: GE12.0 TOP: 4-2 Angle Relationships in Triangles

20. ANS:

m∠DCB = 46°

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 ∠A ≅ ∠F and ∠B ≅ ∠E. Therefore, ∠CDE ≅ ∠DCB. So, m∠DCB = 46°.

PTS: 1 DIF: Advanced NAT: 12.3.3.f STA: GE12.0

TOP: 4-2 Angle Relationships in Triangles

ID: A

7

21. ANS:

[1] Definition of perpendicular lines

[2] Third Angles Theorem

[3] Reflexive Property of Congruence

Proof:

Statements Reasons

1. RT⊥SU 1. Given

2. ∠RTS and ∠RTU are right angles. 2. Definition of perpendicular lines

3. ∠RTS ≅ ∠RTU 3. Right Angle Congruence Theorem

4. ∠SRT ≅ ∠URT 4. Given

5. ∠S ≅ ∠U 5. Third Angles Theorem

6. RS ≅ RU 6. Given

7. T is the midpoint of SU . 7. Given

8. ST ≅ UT 8. Definition of midpoint

9. RT ≅ RT 9. Reflexive Property of Congruence

10. ∆RTS ≅ ∆RTU 10. Definition of congruent triangles

PTS: 1 DIF: Average REF: Page 232 OBJ: 4-3.3 Proving Triangles Congruent

NAT: 12.3.5.a STA: GE5.0 TOP: 4-3 Congruent Triangles

ID: A

8

22. ANS:

[1] PQ

[2] RP

[3] ∆RPQ

[4] SSS

[5] CPCTC

Step 1 Plot the points on a coordinate plane.

Step 2 Use the Distance Formula to find the lengths of the sides of each triangle.

D = x 2 − x1ÊËÁÁ ˆ

¯˜̃

2+ y 2 − y1ÊËÁÁ ˆ

¯˜̃

2

AB = 5 − 3( )2

+ 2 − −1( )ÊËÁÁ ˆ

¯˜̃

2

= 4 + 9 = 13

BC = −2 − 5( ) 2+ 0 − 2( ) 2

= 49 + 4 = 53

CA = 3 − −2( )ÊËÁÁ ˆ

¯˜̃

2+ −1 − 0( )

2

= 25 + 1 = 26

RP = −3 − −6( )ÊËÁÁ ˆ

¯˜̃

2+ 4 − 2( )

2

= 9 + 4 = 13

PQ = −5 − −3( )ÊËÁÁ ˆ

¯˜̃

2+ −3 − 4( )

2

= 4 + 49 = 53

QR = −6 − −5( )ÊËÁÁ ˆ

¯˜̃

2+ 2 − −3( )ÊËÁÁ ˆ

¯˜̃

2

= 1 + 25 = 26

AB = RP = 15 , BC = PQ = 53 , and CA = QR = 26 . So AB ≅ RP, BC ≅ PQ, and CA ≅ QR. Therefore

∆ABC ≅ ∆RPQ by SSS, and ∠ABC ≅ ∠RPQ by CPCTC.


OBJ: 4-6.4 Using CPCTC in the Coordinate Plane NAT: 12.2.1.e

STA: GE5.0 TOP: 4-6 Triangle Congruence: CPCTC

ID: A

9

23. ANS:

y − 3 = −7

2(x − 1.5)

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 =−2 + 5

2,2 + 4

2

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃̃˜

= (1.5,3)

Step 3 Find the slope of the perpendicular bisector.

Slope of AB =(4) − (2)

(5) − (−2)=

2

7

Since the slopes of perpendicular lines are opposite reciprocals, the slope of the perpendicular bisector is −7

2.

Step 4 Use point-slope form to write the equation.y − y 1 = m(x − x 1 )

y − 3 = −7

2(x − 1.5)


OBJ: 5-1.4 Writing Equations of Bisectors in the Coordinate Plane

NAT: 12.5.2.c STA: 1A8.0 TOP: 5-1 Perpendicular and Angle Bisectors

24. ANS:

BO = 17.6

BO =2

3BY Centroid Theorem

BO =2

3(26.4) = 17.6 Substitute 3.3 for BY and simplify.


OBJ: 5-3.1 Using the Centroid to Find Segment Lengths NAT: 12.3.3.f

STA: GE1.0 TOP: 5-3 Medians and Altitudes of Triangles

ID: A

10

25. ANS:

width: 21.6 in., height: 16.2 in.

Let 3x be the height in inches. Then 4x is the width of the TV screen.

a2

+ b2

= c2 Pythagorean Theorem

4x( )2

+ 3x( )2

= 272 Substitute 4x for a, 3x for b, and 27 for c.

25x2

= 729 Multiply and combine like terms.

x2

=729

25Divide both sides by 25.

x =729

25= 5.4 in. Find the positive square root.

Width: 4 5.4( ) = 21.6 in.

Height: 3 5.4( ) = 16.2 in.

PTS: 1 DIF: Average REF: Page 349 OBJ: 5-7.2 Application

NAT: 12.3.3.d STA: GE15.0 TOP: 5-7 The Pythagorean Theorem

26. ANS:

Yes; acute triangle

Step 1 Determine if the sides form a triangle.

By the Triangle Inequality Theorem, 6, 14, and 13 can be the sides of a triangle.

Step 2 Classify the triangle.

c2 ? a2 + b2 Compare c2 to a2 + b2.

142

? 62

+ 132 Substitute the longest side length for c.

196 ? 36 + 169 Multiply.

196 < 205 Add and compare.

Since 142

< 62

+ 132, the triangle is acute.

PTS: 1 DIF: Average REF: Page 351 OBJ: 5-7.4 Classifying Triangles

NAT: 12.3.3.d STA: GE6.0 TOP: 5-7 The Pythagorean Theorem

27. ANS:

12 m

The roof is a 45°–45°–90° triangle with a hypotenuse of 17 m.

17 = l 2 Hypotenuse = leg 2

l =17

2≈ 12 m Divide both sides by 2 and round.


NAT: 12.3.3.d STA: GE20.0 TOP: 5-8 Applying Special Right Triangles

ID: A

11

28. ANS:

172°

Step 1 Find the sum of the interior angle measures.

(n – 2)180° Polygon Angle Sum Theorem

= (45 – 2)180° A 45-gon has 45 sides, so substitute 45 for n.

= 7740 Simplify.

Step 2 Find the measure of one interior angle.

7740

45 = 172 The interior angles are ≅, so divide by 45.


OBJ: 6-1.3 Finding Interior Angle Measures and Sums in Polygons

NAT: 12.3.3.f STA: GE12.0 TOP: 6-1 Properties and Attributes of Polygons

29. ANS:

36°

A decagon has 10 sides and 10 vertices.

sum of exterior angle measures = 360° Polygon Exterior Angle Sum Theorem

measure of one exterior angle = 360

10= 36°

A regular decagon has 10 congruent exterior

angles, so divide the sum by 10.

The measure of each exterior angle of a regular decagon is 36°.


OBJ: 6-1.4 Finding Exterior Angle Measures in Polygons NAT: 12.3.3.f

STA: GE12.0 TOP: 6-1 Properties and Attributes of Polygons

30. ANS:

m∠YXZ = 60o

∠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°


NAT: 12.3.3.f STA: GE12.0 TOP: 6-1 Properties and Attributes of Polygons

ID: A

12

31. ANS:

[1] In a parallelogram, opposite angles are congruent.

[2] Alternate Interior Angles Theorem

Proof:

Statements Reasons

1. ABDF and FBCD are parallelograms. 1. Given

2. ∠BCD ≅ ∠DFB 2. In a parallelogram, opposite angles are

congruent.

3. DF Ä AB 3. Opposite sides in a parallelogram are

parallel.

4. ∠DFB ≅ ∠ABF 4. Alternate Interior Angles Theorem

5. ∠BCD ≅ ∠ABF 5. Substitution


OBJ: 6-2.4 Using Properties of Parallelograms in a Proof NAT: 12.3.5.a

STA: GE7.0 TOP: 6-2 Properties of Parallelograms

32. ANS:

(12, 9), (6, 1)

The diagonals of a parallelogram bisect each other. If the vertex opposite A is C, and the vertex opposite B is

D, then X is the midpoint of AC and BD. Use the midpoint formula to find points B and D.

Step 1 Solve for point C. Step 2 Solve for point D.

X = midpoint of AC X = midpoint of BD

X = (7,6) =2 + x

2,

3 + y

2

Ê

ËÁÁÁÁ

ˆ

¯˜̃˜̃ X = (7,6) =

8 + x

2,

11 + y

2

Ê

ËÁÁÁÁ

ˆ

¯˜̃˜̃

7 =2 + x

2, 6 =

3 + y

27 =

8 + x

2, 6 =

11 + y

2

x = 12, y = 9 x = 6, y = 1

C(12, 9) D(6, 1)


TOP: 6-3 Conditions for Parallelograms

ID: A

13

33. ANS:

AB = 3 5 , BC = 3 5 , CD = 3 5 , DA = 3 5 , slope of AB = 2, slope of BC = −1

2. Since the product of

the slopes is −1, AB ⊥ BC.

AB = (−1 − (−4))2

+ (3 − (−3))2

= 3 5

BC = (5 − (−1))2

+ (0 − 3)2

= 3 5

CD = (2 − 5)2

+ (−6 − 0)2

= 3 5

DA = (−4 − 2)2

+ (−3 − (−6))2

= 3 5

slope of AB =3 − (−3)

−1 − (−4)= 2

slope of BC =0 − 3

5 − (−1)= −

1

2

2 ⋅ −1

2

ÊËÁÁÁ

ˆ¯˜̃˜ = −1

Since the product of the slopes is −1, AB ⊥ BC.


OBJ: 6-4.3 Verifying Properties of Squares NAT: 12.3.5.a

STA: GE7.0 TOP: 6-4 Properties of Special Parallelograms

34. ANS:

[1] Midpoint Theorem

[2] SAS

[1] Midpoint Theorem

In Statement 2, each side is treated separately. The Midpoint Theorem is used to show that the measure of

half of a segment is equal to the measure of the segment created by joining an endpoint to its midpoint.

[2] SAS

In Statement 3, four angles are proven congruent. In Statement 9, two sets of four sides are proven congruent.

Therefore, by SAS, four triangles are proven congruent.


OBJ: 6-4.4 Using Properties of Special Parallelograms in Proof

NAT: 12.3.5.a STA: GE7.0 TOP: 6-4 Properties of Special Parallelograms

ID: A

14

35. ANS:

rectangle, rhombus, square

Step 1 Graph ABCD.

Step 2 The diagonals of a rectangle are congruent. Find AC and BD to determine if ABCD is a rectangle.

AC = (0 − (−1))2

+ (1 − (−2))2

= 10

BD = (1 − (−2))2

+ (−1 − 0)2

= 10

Since AC = BD, ABCD is a rectangle.

Step 3 The diagonals of a rhombus are perpendicular. Find the slopes of AC and BD to determine if ABCD is

a rhombus.

slope of AC =1 − (−2)

0 − (−1)= 3

slope of BD =−1 − 0

1 − (−2)= −

1

3

Since (3) −1

3

ÊËÁÁÁ

ˆ¯˜̃˜ = −1, AC ⊥ BD and ABCD is a rhombus.

Since ABCD is both a rhombus and a rectangle, it is also a square.


OBJ: 6-5.3 Identifying Special Parallelograms in the Coordinate Plane

NAT: 12.3.4.d STA: GE7.0 TOP: 6-5 Conditions for Special Parallelograms

ID: A

15

36. ANS:

square

Find the slope of each side to determine if the sides are perpendicular.

slope of MN = (5 − 9)

(−3 − 0)=

−4

−3=

4

3slope of OP =

(6 − 2)

(4 − 1)=

4

3

slope of NO = (6 − 9)

(4 − 0)=

−3

4slope of PM =

(5 − 2)

(−3 − 1)=

3

−4

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.

NO = (9 − 6)2

+ (0 − 4)2

= 9 + 16 = 25 = 5

OP = (2 − 6)2

+ (1 − 4)2

= 16 + 9 = 25 = 5

The sides are congruent, so the figure is a square.


TOP: 6-5 Conditions for Special Parallelograms

37. ANS:

m∠PSR = 60°

Since diagonals of a kite are perpendicular, the four triangles are right triangles.m∠QOR = 90° Diagonals of a kite are

perpendicular.

m∠QOP = 90°

m∠QRO + m∠RQO = 90° The acute angles of a right

triangle are complementary.

m∠QPO + m∠PQO = 90°

70° + m∠RQO = 90° Substitute the given values. 50° + m∠PQO = 90°

m∠RQO = 20° Subtract. m∠PQO = 40°

m∠PQR = m∠PSR Theorem: If a quadrilateral is a kite, then exactly one pair

of opposite angles are congruent.m∠PQO + m∠RQO = m∠PSR Angle Addition Postulate

40° + 20° = 60° = m∠PSR Substitute the given values and simplify.

PTS: 1 DIF: Basic REF: Page 429 OBJ: 6-6.2 Using Properties of Kites

NAT: 12.3.3.f STA: GE12.0 TOP: 6-6 Properties of Kites and Trapezoids

ID: A

16

38. ANS:

YD = 6.9

BD ≅ AC Diagonals of an isosceles trapezoid are congruent.

BD = AC Definition of congruent segmentsBD = 17.2 Substitute 17.2 for AC.BY + YD = BD Segment Addition Postulate10.3 + YD = 17.2 Substitute the given values.YD = 6.9 Subtract 10.3 from both sides.


OBJ: 6-6.3 Using Properties of Isosceles Trapezoids NAT: 12.3.3.f

STA: GE7.0 TOP: 6-6 Properties of Kites and Trapezoids

39. ANS:

x = 2.8

QS ≅ RTTheorem: A trapezoid is isosceles if and only if its diagonals are

congruent.QS = RT Definition of congruent segments

3x + 4 = 8x − 10 Substitute the given values.14 = 5x Subtract 3x from both sides and add 10 to both sides.2.8 = x Divide both sides by 5.


OBJ: 6-6.4 Applying Conditions for Isosceles Trapezoids NAT: 12.3.3.f

STA: GE7.0 TOP: 6-6 Properties of Kites and Trapezoids

40. ANS:

T(3, 1); area of ∆RST = 8; AB ≈ 8.9

Using the given diagram, the coordinates of T are (3, 1).

The area of a triangle is given by A =1

2bh.

From the diagram, the base of the triangle is b = RT = 4.

From the diagram, the height of the triangle is h = 4.

Therefore the area is A =1

2(4)(4) = 8.

To find AB, use the Distance Formula with points A(1,5) and B(−3,−3).

AB = (x 2 − x 1)2

+ (y2 − y 1 )2

= (−3 − 1)2

+ (−3 − 5)2

= 16 + 64 = 80 ≈ 8.9

PTS: 1 DIF: Advanced NAT: 12.2.1.e STA: GE17.0

TOP: 1-6 Midpoint and Distance in the Coordinate Plane KEY: area | distance formula | triangles

ID: A

17

41. ANS:

x = 6; 85°; 95°

Step 1 Create an equation

The angles are supplements and their sum equals 180°.

(2x2

+ 3x − 5) + (x2

+ 11x − 7) = 180

Step 2 Solve the equation

3x2

+ 14x − 12 = 180

3x2

+ 14x − 192 = 0(3x + 32)(x − 6) = 0

x = −32

3or 6.

When x = −32

3, the measurement of the second angle is

x2

+ 11x − 7 = −10.6°.

Angles cannot have negative measurements, so x = 6.

Step 3 Solve for the required values

The measurement of the first angle is 2x2

+ 3x − 5 = 2(6)2

+ 3(6) − 5 = 85°.

The measurement of the second angle is x2

+ 11x − 7 = (6)2

+ 11(6) − 7= 95°.

PTS: 1 DIF: Advanced NAT: 12.2.1.f STA: 6MG2.2

TOP: 2-6 Geometric Proof

ID: A

18

42. ANS:

x = 7; y = 9; 147°; 147°; 33°; 33°

Step 1 Create a system of equations.

m∠1 = m∠320x + 7 = 5x + 7y + 49

15x − 7y = 42

The sum of the measures of supplementary angles equals 180°.

m∠1 + ∠2 = 18020x + 7 + 3x − 2y + 30 = 180

23x − 2y = 143

Create a system of equations.15x − 7y = 42

23x − 2y = 143

Step 2 Solve the system of equations.15x − 7y = 42

23x − 2y = 143

−30x + 14y = −84

161x − 14y = 1001

Multiply the first equation by −2.

Multiply the second equation by 7.

131x = 917 Add the two equations together.x = 7 Divide both sides by 131.

Solve for y.

Substitute x = 7 into 15x − 7y = 42.

15(7) − 7y = 42

y = 9

The values are x = 7 and y = 9.

Step 3 Solve for the four angles.

Angle 1: (20(7) + 7)° = 147°

Angle 2: (3(7) − 2(9) + 30)° = 33°

Angle 3: (5(7) + 7(9) + 49)° = 147°

Angle 4 and angle 2 are vertical and thus have equal measures.

The measurement of angle 4 is 33°.

The measures of all four angles are 147°, 147°, 33°, and 33°.

PTS: 1 DIF: Advanced NAT: 12.2.1.f TOP: 2-7 Flowchart and Paragraph Proofs

ID: A

19

43. ANS:

x = 10, y = 20

By the Corresponding Angles Postulate, (4x + y)° = 60°.

By the Alternate Interior Angle Postulate, (8x + y)° = 100°.

8x + y

−(4x + y)

4x

= 100

= −60

= 40

Subtract the first equation from the second.

x = 10 Divide both sides by 4.8(10) + y = 100 Substitute 10 for x.

y = 20 Simplify.

PTS: 1 DIF: Advanced REF: Page 157 OBJ: 3-2.3 Application

NAT: 12.3.3.g STA: GE7.0 TOP: 3-2 Angles Formed by Parallel Lines and Transversals

44. ANS:

98

∆GCF is equilateral, so CG = GF.

∆ABF is equilateral, so AB = AF.

∆ABF ≅ ∆EDG, so AB = DF, BC = CD, and CG = CF.

The total distance from A to B to C to D to E = AB + BC + CD + DE.

Step 1 Find AB and DE by finding AF.

Since CG =1

4AB, use substitution to get GF =

1

4AF.

AF = AG + GF = AG +1

4AF

3

4AF = AG

3

4AF = 21

AF = 28

Since ∆ABF is equilateral, AB = BF = 28.

Since ∆ABF ≅ ∆EDG, AB = DE = 28.

Step 2 Find BC and CD.

Since ∆GCF is equilateral and CG =1

4AB, CG = CF = 7.

So BC = BF − CF = 28 − 7 = 21.

Since ∆ABF ≅ ∆EDG, DC = BC = 21.

Step 3 Substitute to find the distance from A to B to C to D to E.

AB + BC + CD + DE = 28 + 21 + 21 + 28 = 98.


TOP: 4-3 Congruent Triangles KEY: multi-step

ID: A

20

45. ANS:

(A + C, D)

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.

PTS: 1 DIF: Advanced NAT: 12.2.1.e STA: GE17.0

TOP: 4-7 Introduction to Coordinate Proof

46. ANS:

m∠FOG = 120°

Sketch the relevant sides of the polygon with extended sides meeting at O.

By the Sum of the Exterior Angles of a Polygon Theorem, the sum of the measures of the exterior angles of

the dodecagon is 360º. So, each exterior angle is 360°

12= 30°. Hence, m∠OFG = m∠OGF = 30°.

By the Triangle Sum Theorem, m∠FOG + m∠OFG + m∠OGF = 180°. So, m∠FOG + 30° + 30° = 180°,

which means m∠FOG = 120°.


TOP: 6-1 Properties and Attributes of Polygons

ID: A

21

47. ANS:

ZW = 12.9, WX = 8.6, XY = 25.8, and ZY = 8.61

2(XY + ZW) = 19.35 Trapezoid Midsegment Theorem

XY + ZW = 38.7 Simplify.ZW = 38.7 − XY Solve for ZW.

ZY = WX Isosceles trapezoids have congruent legs.

XY + ZY + WX + ZW = 55.9 Perimeter of the trapezoid3(ZY) + ZY + WX + [38.7 − 3(ZY)] = 55.9 Substitute for XY and ZW.

3(WX) + WX + WX + [38.7 − 3(WX)] = 55.9 Substitute for ZY.

2(WX) = 17.2 Simplify.

WX = 8.6 Solve.

ZW = 12.9, WX = 8.6, XY = 25.8, and ZY = 8.6.

PTS: 1 DIF: Advanced NAT: 12.2.1.h STA: GE7.0

TOP: 6-6 Properties of Kites and Trapezoids

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