Semester 2 Final Exam Review Name ___________________________________Quadrilaterals & Polygons (Unit 5)1. Quadrilateral STUV shown below is a parallelogram. SW = 36, VW = 12x + 1, WU = 8y + 4, and WT = 49. Find the values of x and y. Explain your reasoning.

2. Is enough information given in the diagram to show that quadrilateral JKLM is a square? Explain your reasoning.

3. Find the values of x and y if JKLM is a parallelogram.

4. Decide if you are given enough information to show that the quadrilateral is a parallelogram.

a. One pair of opposite sides are congruent.

b. Two pairs of opposite angles are congruent.

c. Diagonals are perpendicular.

d. One angle is a right angle.

e. One pair of consecutive angles are congruent.

f. Diagonals are perpendicular and congruent.

g. One pair of opposite sides is parallel and congruent.

h. All four sides are congruent.

5. Find the values of x and y. Justify your answer.

VW = WT and SW = UW because diagonal of a parallelogram bisect each other.

12x + 1 = 498y + 4 = 36 12x = 48 8y = 32 X = 4 y = 4

Yes, JKLM is a parallelogram because the diagonals bisect each other. JKLM is also a Rectangle because the diagonals are congruent and a Rhombus because the diagonals are perpendicular. JKLM is a square.

Opposite angles are congruent9y + 1 = 10y – 13 y =1 4

Consecutive angles are supplementary7x + 4 + 10y – 13 = 1807x + 4 + 10(14) – 13 = 180 7x + 131 = 180 7x = 49

Not enough information to show the quadrilateral is a

Not enough information to show the quadrilateral is a

Not enough information to show the quadrilateral is a

Yes, this is enough information to show the quadrilateral is a parallelogram

Yes, this is enough information to show the quadrilateral is a parallelogram

Yes, this is enough information to show the quadrilateral is a parallelogram

Opposite angles are congruent in a parallelogram. x = 110Consecutive angles are supplementary in a parallelogram.

y + 110 = 180 y = 70

Not enough information to show the quadrilateral is a

Not enough information to show the quadrilateral is a

6. Use the figure to answer the questions below.

a. Give the most specific name for the figure.

b. Solve for x.

c. Find the length of the longer side.

In 7 & 8, find the indicated measure. Justify your answer.7.

8.

9. Find the values of x and y. Justify your answer.

10. Prove the theorem that states if a quadrilateral is a parallelogram, then opposite angles are congruent.

Given: XYWZ is a parallelogramProve: ÐX@ÐW

Statements Reasons1. XYWZ is a parallelogram

1. Given

2. XY // WZ ,XZ // WY 2. Definition of a Parallelogram

3. ∠XZY ≅∠WYZ 3. AIA Theorem

4. ∠WZY ≅∠XYZ 4. AIA Theorem

5. ZY≅ZY 5. Reflexive Property of Congruence

6. ∆XYZ≅∆WZY 6. ASA Congruence Criterion

7. ∠X ≅∠ W 7. CPCTC11. Prove the Opposite Sides Criterion for a Parallelogram

The figure is a parallelogram because the opposite sides are congruent.

There are many ways to find x and this is just one example. 6x – 4 = 5x + 1 x = 5

12x = 12(5) = 60 units, the length of the longer side

PQRS is a Rhombus and the diagonals and the diagonals bisect the angles in a Rhombus. m∠PSQ = 49° AIA Thm.

VWXY is a Rectangle because it has 4 right angles and the diagonals are congruent in a Rectangle.

The figure is a parallelogram because the opposite sides are congruent. Diagonals bisect each other.

x + 6 = 10 2y = 7 X = 4 y =

Given: AB¿ DC and AD¿ BCProve: ABCD is a parallelogram.

Statements Reasons

1. AB≃DC and AD≃BC 1. Given

2. Draw DB 2. Through any two points there exists exactly one line.

3. DB≃DB 3. Reflective Property of Congruence

4. ∆ABD ≅ ∆CDB 4. SSS Congruence Criterion

5. ∠ ABD ≅ ∠CDB ∠ ADB ≅ ∠CBD

5. CPCTC

6. AB // DC and AD // BC

6. AIA Converse

7. ABCD is a parallelogram

7. Definition of a parallelogram

12. Prove the Bisecting Diagonals Criterion for a ParallelogramGiven: AE¿ EC and DE¿ BEProve: ABCD is a parallelogram.

Statements Reasons

1. AE≃EC and DE≃BE 1. Given

2. ∠ AED ≅ ∠BEC ∠ AEB ≅ ∠DEC

2. Vertical Angles Theorem

3. ∆AED ≅ ∆BEC ∆AEB ≅ ∆CED

3. SAS Congruence Criterion

4. AD≃BC and AB≃DC 4. CPCTC

5. ABCD is a parallelogram

5. Opposite Sides Criterion for a Parallelogram

13. a. Use the given vertices to graph the quadrilateral ABCD.b. Classify the quadrilateral ABCD. Explain your answer with a specific definition, criterion, or theorem.

c. Find the perimeter of quadrilateral ABCD.

The vertices are A(-2, 3), B(3,2), C(2, -1) and D(-3, 0)

14. a. Use the given vertices to graph the quadrilateral QRST.b. Classify the quadrilateral QRST. Explain your answer with a specific definition, criterion, or theorem.c. Find the perimeter of quadrilateral QRST.

The vertices are

Q(-4, -1), R(-1,3), S(4, 3) and T(1, -1)

Unit Circle and Trigonometry (Unit 6)

BD

b. (Sample Answer)Slope AB Slope DC

Slope BC Slope AD

Quadrilateral ABCD is a parallelogram by the definition of a parallelogram. Both pairs of opposite sides are parallel.

C. Length of AB Length of DC AB2 = 52 + 12 DC2 = 52 + 12 AB2 = 26 DC2 = 26AB = DC = Length of BC Length of ADBC2 = 32 + 12 AD2 = 32 + 12

BC2 = 10 AD2 = 10BC = AD = P(ABCD) = + + + P(ABCD) = 16.5 units

1

1

1

13

3

5

5

C

A4 4

3

3

5

5

T

SR

Q

b. (Sample Answer)RS = 5 QT = 5

RQ = 5 ST = 5(Triple 3 -4 -5)

Quadrilateral QRST is a rhombus by the definition of a rhombus. All sides are congruent. C.

P(QRST) = 5 + 5 + 5 + 5P(QRST) = 20 units

In 15 – 17 find the value of x. Write your answer in simplest radical form.

15.

16.

17.

In 18 & 19, use the trigonometric ratios to find the value of x. Round your answer to the nearest tenth. 18.

19.

20. In right triangle ABC, ∠ A and ∠B are acute angles.

a) What is cos B if sin A = 35?

b) What is tan B if tan A = 35

21. In the diagram below, what is the measure of ∠B to the nearest tenth of a degree?

22. In the diagram below, what is the measure of ∠A to the nearest tenth of a degree?

23

11

C

B

A

299

C

B

A

45° - 45° - 90°Hyp = Leg√2X = 6√2

30° - 60° - 90°Short leg = hyp2X = 7

30° - 60° - 90°Hyp = 2 Short legX = 2(8√3)X = 16√3

SOH CAH TOAtan541

= x32

x =32 tan 54x = 44.0 units

SOH CAH TOA

sin 291

= x17

x = 17 sin 29

5

4

3

C A

B

Start by drawing a picture.To find the hypotenuse - use the triple 3-4-5

cosB =35

Start by drawing a picture.

tanB = 53 5

3

C A

B

SOH CAH TOA

Tan B =

m B = Tan-1 m B =64.4©SOH CAH TOA

Sin A =

m A = Sin-1 m A =18.1©

23. A roofer props a ladder against a wall so that the top of the ladder reaches a 30-foot roof that needs repair. If the angle of elevation from the bottom of the ladder to the roof is 55°, how far is the ladder from the base of the wall? Round your answer to the nearest foot.

24. A student can see a water tower from the closest point of the soccer field at San Lobos High School. The edge of the soccer field is about 100 feet from the water tower and the water tower stands at a height of 32.5 feet. What is the angle of elevation if the eye level of the student viewing the tower from the edge of the soccer field is 6 feet above the ground? Round to the nearest tenth degree.

In 25 – 27, solve the right triangle.

25.

26.

27.

Point P is located on the unit circle. State the quadrant and find the angle θ. Then find the exact values of sin θ and cos θ.

28. P 29. PQuadrant ____________ Quadrant ____________

θ ____________________ θ ____________________

sin θ _________________ sin θ _________________

cos θ ________________ cos θ ________________

SOH CAH TOAtan541

= x32

x =32 tan 54x = 44.0 units

SOH CAH TOA

sin 291

= x17

x = 17 sin 29

X ft

30 ft55©

SOH CAH TOAtan551

=30x

x·tan 55 =30

x = x = 21 feet

SOH CAH TOA

Tan x© =

x© = Tan-1 x© = 14.8©

To Solve a right triangle you find all missing angles and sides.

What is missing?: QR, ∠P, ∠R

Find ∠P: SOH CAH TOACos P =4

6

m∠P = Cos-1 ( 46 )

m∠P = 48.2°

Find ∠R: Triangle Summ∠R = 180 –90 – 48.2m∠R = 41.8°

What is missing?: AB, ∠A, ∠B

Find ∠A: SOH CAH TOA

Find QR: Pythagorean TheoremQR2 + 42 = 62

QR2 + 16 = 36 QR2 = 20 QR = 4.5 Units

Find AB: Pythagorean Theorem152 + 102 = AB2

325 = AB2

AB = 18.0 units

Find ∠K: Triangle Sum∠K = 180 –90 – 30∠K = 60°

II

135°

III

√22

−√22

210°−12

−√32

Find the exact value for 22-25.

30. sin 315º ________________

31. cos 180º ________________

32. sin 60º ________________

33. cos (-135º) ________________

Find the area of the triangle. Round to the nearest tenth. 34.

35.

Circles (Unit 7)In 36-38, AD is a diameter and the measure of arc CE is 121˚.

36. Find the measure of DE.

37. Find the measure of AE.

38. Find the measure of ACE.

39. In the figure below, and are tangent to circle C.

a. m BCD = ___________ b. m BD = _________

14

C

BA

11

34°

What is missing?: AB, ∠A, ∠B

Find ∠A: SOH CAH TOA

A = ½ ac Sin BA = ½ (45)(43)sin 88©A = 966.9 units2

A = ½ ac Sin BA = ½ (11)(14)sin 34©A = 43.1 units2

121°-59° = 62°

180°-62° = 118°

180°+62° = 242°

180° -42° 138 138

Central and equals its arc.

c. m BE = ___________ d. m BED = ___________

e. m ABC = ___________ f. m BED = _________

In 40-42, find the measure of each numbered angle.

40.

41.

42.

43. Find m OPN.

44. Find the measure of each numbered angle for each figure. m∠1 = 4x – 7, m∠2 = 2x + 11, m∠3 = 5y – 14, m∠4 = 3y + 8

360° - 138° - 89°

133 69°

90° 222½(138°)

Tangent Radius Theorem

360° - 138°

Opposite Angles of an inscribed Quadrilateral are supplementarym∠1 + 60 = 180m∠1 = 120°

m∠2 + 105 = 180m∠2 = 75°

Inscribed Angle (ON Circle)Angle = ½ arcm∠1 = ½(145°)m∠1 = 72.5°

Central AngleAngle = arcm∠2 = 145°

Inscribed Angle (Diameter)Angle = ½ arcm∠1 = ½(180°)m∠1 = 90°

Inscribed Angle (Diameter)Angle = ½ arcm∠2 = ½(180°)m∠2 = 90°

Iso triangle right triangle - 45° - 45° - 90°

m∠3 = 45°∠M ≅∠L and ∠N ≅∠OBoth pairs of angles have the same intercepted arc.10x + 10 = 8x + 16 2x = 6 x = 3Plug in x = 3 to find m∠L and m∠O m∠L = 40° and m∠O = 15°By Triangle Sum Theorem m∠LPO = 125°By Linear pairs Theorem m∠OPN = 55°

∠1 ≅∠2 and ∠3 ≅∠4Both pairs of angles have the same intercepted arc. m∠1 = m∠24x – 7 = 2x + 11

45. If A is the center of the circle, find the measure of ∠QRP.

46. In circle B, m WX = 104°, m WZ = 88°, and m∠ZWY = 26°. Find the measure of each angle.

a. m∠1

b. m∠2

c. m∠3

d. m∠4

e. m∠5

f. m∠6

47. Quadrilateral RSTU is inscribed in circle P such that m STU = 220° and m∠S = 95°. Find each measure.

a. m∠R

b. m∠T

c. m∠U

d. m SRU

e. m RUT

f. m RST

P

Q

R

A

∠1 ≅∠2 and ∠3 ≅∠4Both pairs of angles have the same intercepted arc. m∠1 = m∠24x – 7 = 2x + 11

Inscribed Angle (Diameter)Angle = ½ arcm∠QRP = ½(180°)m∠QRP = 90°

Inscribed Angle (On Circle)Angle = ½ arcm∠1 = ½(104°)m∠1 = 52°

Given in directionsm∠2 = m∠ZWY m∠2 = 26°

Inscribed Angle (On Circle)Angle = ½ arcm∠4 = ½(88°)m∠1 = 44°∠4 ≅∠2 Both pairs of angles have the same intercepted arc. m∠4 = m∠2 m∠4 = 26©∠6 ≅∠1 Both pairs of angles have the same intercepted arc. m∠6 = m∠1 m∠4 = 52©

Inscribed Angle (On Circle)

m ZY = 2 · m∠2 Angle = ½ arcm ZY = 2 · 26 = 52® m∠3 = ½(116°)

m∠3 = 58©

Inscribed Angle (On Circle)Angle = ½ arcm∠R = ½(220°)m∠R = 110°

Opposite Angles of an inscribed Quadrilateral are supplementarym∠T + m∠R = 180 m∠T + 110© = 180©

m∠T = 70°Opposite Angles of an inscribed Quadrilateral are supplementarym∠U + m∠S = 180 m∠T + 95© = 180©

m∠T = 85°m SRU = 360° - m STUm SRU = 360© - 220©m SRU =140©

Inscribed Angle (On Circle) BackwardsArc = 2 Anglem RUT = 2(95°)m RUT = 190°Inscribed Angle (On Circle)

BackwardsArc = 2 Anglem RST = 2(85°)m RST = 170°

48. The perpendicular bisectors of ∆RST intersect at K. Find KR & identify the point of concurrency.

49. The angle bisectors of ∆XYZ intersect at W. Find WB and identify the point of concurrency.

50. In the diagram below, P is the centroid of ΔRST.

a) If LS=36, find PL and PS.

b) If TP=20, find TJ and PJ.

c) If JR=25, find JS and RS.

51. Construct the circumcircle of the triangle below.

52. Construct the incircle of the triangle below.

Circumcenter – Distance from center to vertices are equal(KR = KT)

KR2 = 122 + 162

KR2 = 400KR = 20

Incenter – Distance from center to side are equal (WB = WA)

WB2 + 82 = 102

WB2 + 64 = 100WB2 = 36WB = 6

2 short = long3 short = median

PL=363

PL= 12

PJ = ½

Median short Long

Long Median

PS= 2(12)Ps= 24

TJ = 3 (10)

53. Construct the centroid of the triangle below.

54. Constructing perpendicular bisectors of the sides of the triangle allows you to draw the ___________________________ because the point of concurrency of the perpendicular bisectors is equidistant from each _________________________.

55. Constructing the angle bisectors of a triangle allows you to draw the ___________________ of the triangle because the point of concurrency of the angle bisectors is equidistant from each ___________________________.

56. Use a compass and a straight edge to construct an inscribed regular hexagon.

57. Use a compass and straight edge to construct a square.

58. Construct tangents lines to circle C through P

PL=363

PL= 12

PJ = ½

PS= 2(12)Ps= 24

TJ = 3 (10)

Circumcent

Angle or Vertex

Incenter

side

59. Use a compass and straight edge to construct an octagon.

Linear and Area Measurement (Unit 9)

60. Find the perimeter of a pentagon with vertices A(-1, -3), B(-1, 1), C(0, 6), D(3, 2), and E(3, -1). Round to the nearest tenth.

P(ABCDE) ≈_____________________________61. The area of the state of Alabama can be approximated using the vertices A(0,0), B(160, 0) C( 160, 330) and D(0, 330) where every unit represents 1 mile. In 2010, the population of Alabama was 4,779,736.

Length of BC BC2 = 52 + 12 BC2 = 26BC =

Length of AFAF2 = 22 + 42

AF2 = 20AF =

P(ABCDE) = 4 + + 5 + 3 + 21.6 units

10

20

30

40

Population Density

52800 mi

Population Density

A

B

C

D

E4

5

1 3

4

4

2

Length of CD CD = 5 3-4-5 TripleLength of ABAB = 4Length of DFDF = 3

a. Find the approximate area of the state of Alabama.

b. Estimate the population density of Alabama in 2010 in people per square mile.

c. Estimate the population density of Alabama in 2010 in people per acre. (1 square mile = 640 acres)

In 62 – 64, Find the area of the shaded region. Round to the nearest tenth, if necessary. 62.

63.

64.

65. A circle has a diameter of 18 millimeters.a. Find the circumference of the circle.

b. Find the area of the circle.

66. Find the length the arc and the area of the shaded regiona. Length of arc PQ.

10 20 30 40A B

CD

A(ABCD) = b·h (Area of a Rectangle)A(ABCD) = 160(330)A(ABCD) = 52,800 miles 2

A = A(Rectangle) – A(Circle)A = b·h – π· r2

A = 8(4) – π( 2)2

A = 19.4 m 2

A = A(Square) – ½ ·A(Circle)A = s2 – ½ π· r2

A = 82 – ½ π( 4)2

A = 38.9 units 2

A = A(Triangle) + A(Rectangle)A = ½ b·h + b·hA = ½ 12(15) + 12(5)A = 150 units 2

C = d·πC = 18 π mmC = 56.5 mm

A = π· r2

A = π( 9)2

A = 81 π mm2

A = 254.5 mm 2

360x = 5510πx = 48.1 ft

b. Area of shaded sector.

In 67 – 69, use the diagram to the right.67. Find the circumference and area of circle R.

68. Find the length of arc AB.

69. Find the area of sector ARB.

In 70 – 75, Find the indicated measure for the circle shown.70. Length of arc AB

71. Circumference of circle F

72. m GH

73. Area of the shaded sector

74. Area of circle N

75. Radius of circle P

360x = 52345πx = 456.8 ft 2

C = 2πrC = 2π(5)C = 10π cm or 31.4 cm

x2π 5

=105 °360°

360x = 3298.67x = 9.2 cm

A = πr2

A = π(5)2

A = 25π cm 2 or 78.54 cm 2

xπ 52 =

105°360°

360x = 8246.68x = 22.9 cm 2

360x = 11535.9x = 32.0 cm

210x = 23040x = 109.7 in

54πx = 12600x = 74.3©

360x = 51270.8x = 142.4 in 2

360© - 105© =

255

68x = 17640x = 259.4 m 2

76. The area of a circle is 529π m2. Find the radius.

77. The circumference of a circle is 64π, what is the area of the circle?

Three-Dimensional Figures and Volume (Unit 10)In 78 – 81, determine the shape resulting from each cross section of the following solids78. A square pyramid with the cross section cut parallel to the base.

79. A cylinder with the cross section cut parallel to the base.

80. The solid is a cube.

81. The solid is a cube.

82. Draw the two-dimensional shape that can be rotated 360° about the given axis to produce this solid.

A = πr2

529 π = πr2

529 = r2

r = 23 m

C = 2πr64π = 2πr64= 2rr = 32 units

A = πr2

A = π(32)2

A = 1024π units 2 or 3216.99 units 2

Square

Circle

Trapezoid

Hexagon

114πr2 = 36.1868x = 6.0 cm 2

83. Multiple Choice: Draw the solid of revolution formed by the shape rotated around the axis given.

a. b.

c. d.

84. Find the volume of the solid generated by rotating the figure around the given axis. Round your answer to the nearest hundredth.

85. Find the volume of the solid generated by rotating the figure around the given axis. Round your answer to the nearest tenth.

86. According to Cavalieri’s principle, under what conditions are the volumes of two solids equal?

87. Find the volume of the right cylinder. Give your answer in term of π.

C

The solid is a cone with a radius 3 and height 5. V = 1/3 π r2 H V = 1/3 π (8)2 5V = 47.12 units 3

The solid is a cylinder with a radius 4 and height 3.

V = π r2 H V = π (4)2 3V = 150.8 units 3

If two solids have the same height and the same cross sectional area at every level, then the two solids have the same volume.

V = π r2 H V = π (6)2 5V = 180π ft 3

88. Find the volume of the right prism.

89. Find the volume of the right cone. Round you result to two decimal places.

90. Find the volume of the cone. Round to the nearest tenth.

91. Find the volume of the pyramid. Round to the nearest tenth

92. What is the diameter of a sphere whose volume is 580π m3?

93. Find the volume of a sphere with radius of 0.5 inch. Round to the nearest hundredth.

94. Find the volume of a rectangular pyramid with length 11 m, width 7 m, and height 8 m. Round to the nearest tenth.

95. Each side of a cube measure 3.9 centimeters. Its mass is 95.8 grams. Find the density of the cube. Round to the nearest tenth.

V = BH (The base is a rectangle) V = b·h·H V = 12(4)(9)V = 432 m 3

H

First, find the length of the height.

H2 + 62 = H2

H2 = 28H =

V = 1/3 π r2 H V = 1/3 π (6)2 ()V = 199.49 in 3

V = 1/3 π r2 H V = 1/3 π (8)2 (15)V = 1005.3 ft 3

V = 1/3 B H V = 1/3 · ½ · a·c·sin A·HV = 1/6(19(19)sin 60(23)V = 1198.4 cm 3

V = π r3 435 = r3

580 π = π r3 r = 7.6 m

580 = r3 d= 15.2 m

V = π r3

V = π (0.5)3 V= 0.52 in 3

V = 1/3 B H V = 1/3 · b·h·HV = 1/3(11)(7)(8)V = 205.3m 3

11 m7 m

8 m

V = BH (The base is a square) V = s2·H V = (3.9)3

V = 59.3 cm3 Density = 1.6 g/cm 3

96. A concrete cylinder has the dimensions shown, and has a mass of 14,500 kg. To the nearest kilogram per cubic meter, what is the density of the concrete in the cylinder?

97. You have 550 cm3 of clay to make a sculpture in the shape of a cylinder whose height is twice the radius of the cylinder. What radius and height will the cylinder have?

V = π r2 H V = π (1)2 2V = 6.3 m3

Density = 2301.6

kg/m 3

V = π r2 H radius = 4.44 cm550 = π r2· 2r height = 8.88 cm550 = 2π r3

87.5 = r3

r = 4.44 cm