Geometry Indicators p. 1 Indicators for Geometry An indicator is a measure to determine mastery of a concept, procedure, or application within a specific objective or objectives. These indicators are presented to provide a more in-depth explanation of the objectives in the Geometry Standard Course of Study. Most of the indicators address more than one objective and the answer key/ commentaries will highlight those additional objectives. Many of the indicators are summative in nature and are intended to show a more complete understanding of the mathematics stated in the objective after extended instruction and practice. The following objectives do not have indicators written to them because they are best addressed in the context of other objectives. For instance, 2.05 is addressed in every indicator. The answer key/ commentaries will, for almost all the indicators, present a justification for a solution. It is recommended that if students are assigned a problem, they similarly justify their solution in an orderly manner (addressing 2.07 in the process). 1.01 Select appropriate operations and solve a variety of application problems using real numbers. 2.01 Identify, name, and draw sets of points, such as line, ray, segment, and plane. 2.02 Identify the coordinates of a point in a plane or in space. 2.04 Use inductive reasoning and the tools of construction to reach conclusions. 2.05 Use the structure (definitions, postulates, theorems, properties of equality and inequality) of deductive reasoning to solve problems. 2.07 Write direct (two-column, paragraph, or flow) and indirect proofs. 2.11 Use coordinate geometry to confirm properties of polygons.
Transcript
Geometry Indicators p. 1
Indicators for Geometry
An indicator is a measure to determine mastery of a concept,procedure, or application within a specific objective or objectives.These indicators are presented to provide a more in-depth explanationof the objectives in the Geometry Standard Course of Study. Most ofthe indicators address more than one objective and the answer key/commentaries will highlight those additional objectives. Many of theindicators are summative in nature and are intended to show a morecomplete understanding of the mathematics stated in the objectiveafter extended instruction and practice.
The following objectives do not have indicators written to thembecause they are best addressed in the context of other objectives.For instance, 2.05 is addressed in every indicator. The answer key/commentaries will, for almost all the indicators, present ajustification for a solution. It is recommended that if students areassigned a problem, they similarly justify their solution in an orderlymanner (addressing 2.07 in the process).
1.01 Select appropriate operations and solve a variety of application problems using realnumbers.
2.01 Identify, name, and draw sets of points, such as line, ray, segment, and plane.
2.02 Identify the coordinates of a point in a plane or in space.
2.04 Use inductive reasoning and the tools of construction to reach conclusions.
2.05 Use the structure (definitions, postulates, theorems, properties of equality and inequality)of deductive reasoning to solve problems.
2.07 Write direct (two-column, paragraph, or flow) and indirect proofs.
2.11 Use coordinate geometry to confirm properties of polygons.
Geometry Indicators p. 2
Spatial Sense, Measurement, and Geometry
2.03 Find the length and the midpoint of a segment in two or three dimensions to solveproblems.
A. Parallelogram ABCD hasvertices (8, 9), (9, 3), (2, 5),and (1, 11). What are thecoordinates of theintersection of thediagonals?What is the perimeter ofABCD?
B. Connect the midpoints ofthe sides of ABCDconsecutively to form a newquadrilateral. Which specialquadrilateral is it? Justify.
A
B
C
D
C. If the coordinates ofquadrilateral MNOP areM(7, 6), N(-6, 1), 0(-4, -3),and P(9, 2), what type ofquadrilateral is MNOP?Justify.
D. Find the center of prismABCDEFGH with verticesA(3, 10, 8), B(10, 10, 8),C(10, 6, 8), D(3, 6, 8),E( 1, 6, 3), F(1, 10, 3),G(8, 10, 3), and H(8, 6, 3).
A B
CD
E
FG
H
Geometry Indicators p. 3
2.06 Write and interpret conditional statements including the converse, inverse, andcontrapositive.
A. When the statement “If A,then B” is true, whichstatement must also be true?
(a) If B, then A.(b) If not A, then B.(c) If not B, then A.(d) If not B, then not A.
B. Either I go to camp or I geta summer job.If I get a summer job, then Iwill earn money.If I earn money, then I willbuy new shoes.I do not buy new shoes.
Prove: I go to camp.
C. If Shane is an athlete and heis salaried, then Shane is aprofessional.Shane is not a professional.Shane is an athlete.Which statement must betrue?
a) Shane is an athlete and heis salaried.b) Shane is a professional orhe is salaried.c) Shane is not salaried.d) Shane is not an athlete.
D. If I receive a check for$500, then we will go on atrip.If the car breaks down, thenwe will not go on the trip.Either I receive a check for$500 or we will not buysouvenirs.The car breaks down.
Prove: We will not buysouvenirs.
Geometry Indicators p. 4
E. Which is the converse of thestatement "If today isThanksgiving, then there isno school"?
a) If there is school, thentoday is not Thanksgiving.b) If there is no school, thentoday is Thanksgiving.c) If today is Thanksgiving,then there is school.d) If today is notThanksgiving, then there isschool.
F. If Sue goes out on Fridaynight and not on Saturdaynight, then she does notstudy.If Sue does not failmathematics, then shestudies.Sue does not failmathematics.If Sue does not go out onFriday night, then shewatches a movie.Sue does not watch a movie.
Prove: Sue goes out onSaturday night.
G. Which statement is the con-verse of "If two sides of atriangle are congruent, thenthe triangle is isosceles"?
a) If a triangle is notisosceles, then two sides ofthe triangle are notcongruent.b) If two sides of a triangleare not congruent, then thetriangle is not isosceles.c) If a triangle is isosceles,then two sides of thetriangle are congruent.d) If two sides of a triangleare not congruent, then thetriangle is isosceles.
Geometry Indicators p. 5
2.08 Use properties, definitions, and theorems of angles and lines to solve problems and writeproofs, related to:a) Adjacent, vertical, linear pair, complementary and supplementary angles.b) The segment addition postulate and the angle addition postulate.c) Angle bisectors, segment bisectors, and perpendicular bisectors.d) Special pairs of angles formed by parallel lines and a transversal.e) Skew, parallel, and perpendicular lines.
A. BD bisects ∠ ABC.If m∠ ABD = (4x + 19) andm∠ DBC = (x + 58), findm∠ ABC.
B. CE bisects ∠ ACB, BEbisects ∠ ABD, andm∠ A = 80. Find m∠ E.
A
B CD
E
C. In the diagram, AB and CD
intersect at E, AC isparallel to DB, m∠ A = 41,and m∠ D = 56.Find m∠ AEC.
A
B
C
D
E41o
56o
Geometry Indicators p. 6
D. ABC is shown with BEbisecting ∠ DBC. Ifm∠ ABD = 8x andm∠ DBE = (2x + 15),find m∠ EBC.
A B C
D E
E. In the diagram, RS
intersects parallel lines MN
and PQ at A and B,respectively.If m∠ RAN = (3x + 24) andm∠ RBQ = (7x - 16), findm∠ ABP.
M A
R
N
P B
S
Q
F. Find m∠ FGJ, m∠ KHI, andm∠ CAD to the nearesthundredth.
AB
C
D
EF
G
H
I
J
K
G. In the figure shown, lines mand l are parallel and line pis perpendicular to line l. Ifx = y, what is the value ofx?
p
m
l
yo
xo
Geometry Indicators p. 7
x
55o
L1
L2
H. In the figure shown, if L1 is
parallel to L2, then x =?
I. Express m∠ c in terms ofm∠ a and m∠ e.
a b
c
d e
Geometry Indicators p. 8
2.09 Use properties, definitions, and theorems of polygons to solve problems related to:a) Modeling and describing polygons (convex, concave, regular, nonregular).b) The interior and exterior angles of a convex polygon.c) Congruent and similar polygons.
A
B
C
D
E
A. Relocate vertex B so thatABCDE is convex and allsides remain the samelength.
A
B
C
D
EF
B. Find the measures of all theinterior angles of ABCDEF.Find the measures of theexterior angles at each vertex.
C. In convex pentagon ABCDEm∠ A = 6x,m∠ B = (4x + 13),m∠ C = (x + 9),m∠ D = (2x - 8), andm∠ E = (4x - 1). What arethe measures of all theangles?
Geometry Indicators p. 9
A
C
E
G
J
F. ACEGJ is an equilateralpolygon and ∠ E and ∠ G areright angles. What is theequation of the line ofsymmetry of ACEGJ?
D. Point B is a mutual vertex ofa regular hexagon, a square,and a third regular polygonas shown below. If two ofthe sides of this thirdpolygon are AB and BC,what is this polygon?
A
B
C
E. What geometric figure doesa regular polygonincreasingly resemble as thenumber of sides increases?
Geometry Indicators p. 10
2.10 Recognize, identify, and model regular and non-regular polyhedra.
B. What geometric figure doesa regular pyramidincreasingly resemble as thenumber of sides in its baseincreases?
C. What geometric figure doesa regular prism increasinglyresemble as the number ofsides in its base increases?
A. The figure shown was builtwith cubes. The bottomhorizontal edge of the figureis 14 cm long. What is thevolume of the figure? Whatis its surface area?
D. A semiregular polyhedron isa solid that has faces in theshape of more than one kindof regular polygon, eachvertex is surrounded by thesame kinds of polygons inthe same order, and eachedge is congruent.Construct a truncated cube(a cube with its corners cutoff) that is semiregular froma 4 by 4 by 4 cube.What polyhedra is removedfrom each corner of thecube to form the truncatedcube?How long is each edge ofthe truncated cube?What polygons make thefaces of the truncated cube?What is the area of eachface of the truncated cube?What is the surface area ofthe truncated cube?What is the volume of thetruncated cube?
Geometry Indicators p. 11
z
x
y
A B
CD
G
H
E. GABCDH is a regular octahedron with vertices G(a, b, c),A(0, 0, 5), B(5, 0, 5), C(5, 0, 0), and D(0, 0, 0).What is the ordered triple for H?What is the volume of GABCDH?What is the surface area of GABCDH?
Geometry Indicators p. 12
2.12 Develop and use properties of quadrilaterals (parallelograms, rectangles, rhombi, squares,trapezoids, kites) to solve problems and write proofs.
A. In the diagram of isoscelestrapezoid ABCD,m∠ A = 53, DE = 6,and DC = 10.Find the perimeter of ABCDto the nearest tenth.
A B
CD
E
S
C
O
T
B. Which quadrilateral isTOCS? Justify.
C. The measures of twoadjacent angles of aparallelogram are (x + 55)o
and (7x - 71)o. What are theangle measures?
D. A parallelogram has vertices(-4, 5), (-1, -4), and (6, 4).What are the ordered pairsthat can be the fourthvertex? Justify each.
E. Given: quadrilateral PQRT,
QSV, RST, PTV, QV
bisects RT, and QR isparallel to PV.Prove QS = VS.
P
Q R
S
TV
Geometry Indicators p. 13
F. The bisectors of the lower base angles (C and D) meet atpoint H. Find the relationship between the measures of theupper base angles (A and B) and ∠ DHC.
A B
CD
H
Geometry Indicators p. 14
2.13 Develop and use properties of triangles to solve problems and write proofs related to:a) The relationships of the lengths of the sides and measures of the angles.b) Similar triangles and the relationship of their corresponding parts.c) Congruent triangles and their corresponding parts.d) Isosceles and equilateral triangles.e) Altitudes, perpendicular bisectors, angle bisectors, and medians.
A. If ∆ADE and ∆ABC aresimilar, what is theapproximate value of x?
C15.6
B
Ex
D
3.3 2.8
A
4.1
B. ∆ABC is isosceles; the basehas endpoints A(4, 4) andB(10, 6). Give the equationfor the line that includes thealtitude.
C. ∆ABC is equilateral withvertices A (4, 4) andB (10, 8). Locate vertex C.
D. Using a map of NorthCarolina, identify by countythe area in the stateequidistant from Charlotte,Greensboro, and Raleigh.
E. ∆ABC is similar to ∆DEF.If AC = 10.5, AB = 6.5, andDE = 8, find DF.
F. In ∆ABC, AB = 10 andBC = 5. What is true about
the length of AC?
Geometry Indicators p. 15
A
B
C
G. List the angles in ∆ABCfrom greatest to least. Findthe measure of each angle.
H. If the interior anglemeasures of a triangle arexo, (3x + 20)o, and (6x)o,what kind of triangle must itbe?
A
B
C
D
I. CD is the bisector of∠ ACB, m∠ A = 46 andm∠ B = 82. Find m∠ ACD.
Geometry Indicators p. 16
J. In parallelogram ABCD, thebisectors of two consecutiveangles (A and D) meet at apoint P on a non-adjacentside. Describe trianglesABP, PCD, and APD.
A
B C
D
P
A
B
K. A triangle can be formedusing points A and B andanother point not shown. If∠ ABC is a right angle,identify the coordinate pairsfor C such that ∆ABC is aright isosceles triangle?Justify.
A
B
C
D
L. ∆ABC and ABD are shown.Where should point E belocated so that ∆ABC and∆ADE are similar? Justify.
Geometry Indicators p. 17
2.14 Investigate and use properties of triangles to solve problems and write proofs related to:a) The interior and exterior angles of a triangle.b) The segment joining the midpoints of two sides of a triangle.c) Segments divided proportionally.
A. The sides of ∆ABC are16.8, 11.3, and 7.7 meters.Find the perimeter of thetriangle that is formed byjoining the midpoints of thesides of ∆ABC.
B. BC and DE are parallel.Find the perimeter ofBCED.
A
BC
D
E
A
B
C
D
E
C. In isosceles ∆ADE,
AB, BC, CD, and DE arecongruent. Find m∠ A.
Geometry Indicators p. 18
2.15 Apply properties of right triangles to solve problems using:a) The geometric mean.b) The Pythagorean Theorem and its converse.c) The relationships in special right triangles.d) The definitions of sine, cosine, and tangent.
A B
C
D
B. Five congruent squares arearranged in a T-shape asshown. If x = 15 cm, thenfind the area of the T-shapedfigure.
12.8 m
5.1 mA
X
C. Find sin A.A. ∆ABC is equilateral.
If AB = 6.8 cm, find CD.
D. In order to avoid a large oilspill, a fishing vessel wasdirected by the US CoastGuard to proceed 12 milessouth and then travel east7 miles before resumingtheir previous course. Howmany miles out of their waydid the vessel travel?
Geometry Indicators p. 19
H. Each edge of a cube is10 cm. From the midpointof a diagonal of a face, findthe distance to each vertexnot in the same face.A
B
C
D
E
F. In right ∆ABC, CD isdrawn perpendicular to
hypotenuse AB. If AB = 40and DB = 7, find BC.
G. A 6-by-9 inch sheet of paperis folded so that oppositevertices touch. Find thelength of the fold.
E. Find the measures ofinterior ∠ CDE andexterior ∠ ABC.
Geometry Indicators p. 20
2.16 Develop and use properties of circles to solve problems and write proofs related to:a) The definition of a circle and sets of points related to the circle.b) The equation of a circle, its center and radius length.c) Congruent and concentric circles.d) Circles and their common tangents.e) Circumscribed and inscribed figures.
A. In the diagram, ABCD is arectangle inscribed in thecircle O. The ratio AB toBC is 4:3. The area of therectangle is 65 cm2. Findthe area of the shadedportion.
A B
CD
O
B. What is the equation of acircle with center (-2, 7) anddiameter 18?
C. What is the circumferenceof the circle with equationx2 + (y + 4.2)2 = 15?
D. What is the equation of thecircle if the endpoints of adiameter are (-6, 2) and(4, 6)?
E. If a point is selected atrandom in the interior of acircle, find the probabilitythat the point is closer to thecircle than the center.
Geometry Indicators p. 21
H. Sides AD and BC ofrectangle ABCD passthrough the centers ofcircles R and T. If thecircumference of each circleis 8π, what is the area of theshaded region?
A B
CD
R S T
F. An equilateral triangle isinscribed in a circle andanother circle incribedwithin the triangle. Find theratio of the areas of thecircles.
x
y
p
t
O
G. In the circle with center O,the two triangles have legsof length x, y, p, and t.If x2 + y2 + p2 + t2 = 288,what is the circumference ofthe circle?
Geometry Indicators p. 22
2.17 Apply properties of circles to solve problems involving:a) Arcs and angles of circles.b) The chords, tangents, secants, and radii of a circle.
A. PQ is a tangent to circle Rat point Q. The circle has aradius of 6. If m∠ R = 35,find RP.
P
QR
B. If AB and CD are chords
of circle O and ON = OM,
then find AB.
A
B
C
D
N
M
O
5
C. AB, BC, CD, and AD aretangent to circle O,
AB = 19, BC = 6, and
CD = 14. Find AD.
A
B
C
D
D. Given circle C, find thelength of KL and m∠ KJL.
J
K
LC
Geometry Indicators p. 23
2.18 Use spheres to solve problems related to the definition of a sphere and sets of points relatedto the sphere.
A. A baseball has a surfacearea of 620 cm2. What is itsdiameter?
B. The volume of a sphere is1000 in3. Find the surfacearea of the sphere.
C. Find the surface area of thehemisphere shown.
12.5 cm
D. Catawba Bay (1 unit =1 km) covers an area ofabout 35 km2 and has anaverage depth of 120 m.Microphones are placed onthe floor of the bay at thelocations indicated (X, Y,and Z) to detect dolphinswhich come to feed. Thelocations have the followingdepths: X, 90 m, Y, 105 m,and Z, 350 m.The microphones have arange of 800 m. What arethe chances that themicrophones will detect adolphin at any time while itis in the bay?
X
Y
Z
Geometry Indicators p. 24
2.19 Use formulas to solve problems related to:a) The perimeter of a geometric figure and circumference of a circle.b) The area of a triangle, parallelogram, rhombus, trapezoid, square, rectangle, regularpolygons, and circles.c) Arc lengths and the area of sectors of a circle.d) The ratio of the perimeters, areas, and volumes of similar geometric figures.e) The lateral area, surface area, and volume of a right prism, pyramid, right circularcylinder, cone, and sphere.
A. The plastic cube shownoriginally had a volume of1500 cm3. The front face isdrawn to proportion. Squareholes were cut through tothe opposite face. Howmuch surface is exposed?
B. A cylindrical pipe has anoutside radius of 5.6 inchesand an inside radius of5.3 inches. The pipe is sixfeet long. To the nearesttenth, how much totalsurface is exposed?
14 cm
21 cm
C. If the diameter of the coneshown is increased by2.5 cm, the volume of thenew cone is what percent ofthe original?
D. There is a figure similar tothe one shown that has abase of 23. What is the areaof the other figure?
8
12
Geometry Indicators p. 25
F. Find the exact area andperimeter of ABCDE. Findthe measures of the interiorangles to the nearest tenth.
A
B
C
D
E
G. Find the perimeter.
15.7
25o
A
B
CD
H. In ∆ABC, m∠ A = m∠ C andaltitude BD is three morethan AD. The area of∆ABC is 54 cm2. Find theperimeter of ∆ABC.
E. The circle has area 376 in2
and is divided into sixcongruent sectors. Find the
length of AC.
A
C
Geometry Indicators p. 26
J. A cylinder is cut as shown.The height goes from 19 to23 cm and the radius is5 cm. Find the volumeand total surface area. Thearea of the ellipse is πRrwhere R is the length ofthe semi-major axis or, inthis case, half the length ofthe diagonal cut.
19 23
5
I. The top and side views ofthe new museum are shownbelow. On the grid,1unit = 3 m. How muchspace will be heated orcooled during the year?
TOP
SIDE
Geometry Indicators p. 27
K. The area of a parallelogramis 80 cm2. A segment isdrawn from one vertex tothe midpoint of the oppo-site side. The diagonal isdrawn between the othervertices as shown. Find thearea of the four regions.
W
X
Y
Z
7
12
24
37
L. Find the area of thetrapezoid whose bases are12 and 37 and legs are 7 and24.
M. Describe a strategy to findthe volume of solid FGHJwhere the vertices areF(1, 6, 3), G(4, 4, 8),H(3, 7, 0), and J(2, 5, 10)
F
H
GJ
z
x
y
A B
CD
O
N. Side AD of square ABCDpasses through the center ofcircle O. If the length of theside of square ABCD is 3 m,what is the area of theshaded region exactly?
Geometry Indicators p. 28
Patterns, Relationships, and Functions
3.01 Use slopes to determine if two lines are parallel or perpendicular.
M
A
T
H
B. Is GUB a right triangle? Explain.
G
U
B
A. Prove MATH is a trapezoid.
Geometry Indicators p. 29
3.02 Write the equation of a line parallel or perpendicular to a given line through a given point.
A. ∆ABC is a right trianglewith vertices A (-3, 3) andB (2, 2) and m∠ A = 90.What is the equation forAC?
B. A, B, and C are threevertices of a parallelogram.What is an equation of aline, through A, B, or C, thatincludes the fourth vertex ofthe parallelogram?
A
B
C
C. Give the equation of the linethat includes an altitude ofparallelogram MATH.
M
A
T
H
Geometry Indicators p. 30
D. Give an equation of a line on which is located the pointequidistant from E, F, and G. Find the point. Give an equationof a line that includes an altitude of ∆EFG.
E
F
G
Geometry Indicators p. 31
3.03 Transform (translate, reflect, rotate, dilate) polygons in the coordinate plane; describe thetransformation in simple algebraic terms.
A. ABCD is rotated 90o
clockwise about the origin.Describe the transformationalgebraically.
A
BC
D
B. ABCD is reflected acrossthe y-axis and translated3 units to the right. Describethe transformationalgebraically.
A
B
C
D
C. ∆ABC, with verticesA(2, 8), B(5, 3), and C(6, 8)is transformed according to(x, y) → (-2x + 3, y - 4).Graph ∆ABC and ∆A´B´C´;describe the transformation.
D. Algebraically describe thetransformation of ABCD toA′B′C′D′.
A
B C
D
A'
B'C'
D'
E. Transform ∆RST, withvertices R(2, 2), S(3, 6), andT(8, 3), so that its lineardimensions double, butvertex R´ is located at (2, 2).Describe the transformationalgebraically.
Geometry Indicators p. 32
Data, Probability, and Statistics
4.01 Use length, area, and volume to solve problems involving probability.
A. In the area shown, thestripes are equal in width,and the figure is drawn toscale. What is the chance apoint selected in the areawill be in the shadedregion?
B. In the areashown, what isthe chance apoint selected inthe area will bein the shadedregion?
Geometry Indicators p. 33
C. On a circle, mAB = 40. LetD be any point randomlyselected on the circle so thatD ≠ A and D ≠ B. Find theprobablility thatm∠ ADB = 20.
D. A bag contains sticks oflength 5, 6, 7, 8, 10, 12, and13. If three sticks arerandomly selected, what isthe probability that thesticks will form a triangle?What is the probability thatthe sticks will form a righttriangle?
A B
CD
10
15
E. In rectangle ABCD,AD = 10 m and CD = 15 m.If F is randomly selectedfrom the interior of ABCD,what is the probability thatthe area of ∆DFC is lessthan 30 m2? What is theprobability that the area of∆DFC is between 18 and51 m2?
Geometry Indicators p. 34 Commentaries
NOTES2.03A (also 2.01, 2.02, 2.05, 2.11, 2.12, 2.19)Since the diagonals of a parallelogram bisect
one another, the midpoint of AC (or BD ) is
the point of intersection. Midpoint of AC
(and BD ) is (5, 7).
P = 2 53 2 37+
2.03B (also 1.01, 2.01, 2.02, 2.05, 2.11, 2.12,
3.01)
E (-7, -1) is the midpoint of AB, F (-1, 7) is
the midpoint of AD, G (10, 3) is the midpoint
of DC , and H (4, -5) is the midpoint of CB.
EF and HG are parallel (slopes of
4
3) and FG
and EH are parallel (slopes of − 4
11). By
definition EFGH is a parallelogram. Since
EF ≠ FG (10 ≠ 137 ), EFGH is not a
rhombus.
2.03C (also 1.01, 2.01, 2.05, 2.11, 2.12, 3.01)
Since MN = PO (lengths of 194 ) and MN is
parallel to PO (slopes of
5
13), MNOP is a
parallelogram. Since MN ≠ NO, MNOP is not
a rhombus. Since NO (slope of -2) is not
perpendicular to PO, MNOP is not a
rectangle.
2.03D (also 1.01, 2.01, 2.02, 2.04, 2.10)
The center is the intersection of the diagonals
AH, CF , BE, and DG, (5.5, 8, 5.5), which isalso the midpoint of each of the diagonals.
Geometry Indicators p. 35 Commentaries
NOTES2.06A (d)
2.06BI do not buy new shoes implies I do not earnmoney.I do not earn money implies I do not get asummer job.I do not get a summer job, therefore I go tocamp.
2.06C (c)
2.06DThe car breaks down therefore we will not goon the trip.We will not go on the trip implies I will notreceive a check for $500.I will not receive a check for $500 thereforewe will not buy souvenirs.
2.06E (b)
2.06FSue does not fail mathematics therefore shestudies.Sue studies implies Sue goes out on saturdaynight.
2.06G (c)
2.08A (also 1.01, 2.05)
Since BD bisects ∠ ABC, then m∠ ABD =m∠ DBC. Since m∠ ABD = m∠ DBC, then4x + 19 = x + 58. Solving algebraically usingproperties of equality, x = 142.
Geometry Indicators p. 36 Commentaries
NOTES2.08B (also 1.01, 2.05, 2.09, 2.14)
Since CE bisects ∠ ACB, then m∠ ACE =m∠ ECB. Let a = m∠ ACE = m∠ ECB. Since
BE bisects ∠ ABD, then m∠ ABE = m∠ EBD.Let b = m∠ ABE = m∠ EBD. By exteriorangle theorem, m∠ ABD = m∠ A + m∠ ACB.By substitution, 2b = 80 + 2a. By divisionproperty of equality, b = 40 + a. In ∆BCE,180 = m∠ E + m∠ ABE + m∠ ABC + m∠ ECB.By substitution, 180 = m∠ E + (40 + a) +(100 - 2a) + a. Simplifying, 180 = m∠ E +140. By subtraction property of equality,40 = m∠ E.
2.08C (also 1.01, 2.05, 2.14)
Since AC is parallel to DB, ∠ D and ∠ C arealternate interior angles. Since ∠ D and ∠ Care alternate interior angles, m∠ C = 56. In∆ACE, 180 = m∠ C + m∠ A + m∠ AEC. Bysubstitution, 180 = 56 + 41 + m∠ AEC. Bysubtraction property of equality,83 = m∠ AEC.
2.08D (also 1.01, 2.05)
Since BE bisects DBC, m∠ DBE = m∠ EBC.By angle addition, 180 = m∠ ABD + m∠ DBE+ m∠ EBC. By substitution, 180 = 8x +(2x + 15) + (2x + 15). By simplifying andusing properties of equality, x = 12.5. Bysubstitution, m∠ EBC = 40.
2.08E (also 1.01, 2.05)Since ∠ RAN and ∠ RBQ are correspondingangles, m∠ RAN = m∠ RBQ. By substitution,3x + 24 = 7x - 16. By using properties ofequality, x = 10. By substitution, m∠ RBQ =54. Since ∠ ABP and ∠ RBQ are a linear pairand therefore supplementary, 180 = m∠ ABP +m∠ RBQ. By substitution and subtractionproperty of equality, m∠ ABP = 126.
Geometry Indicators p. 37 Commentaries
NOTES2.08F (also 1.01, 2.02, 2.05, 2.11, 2.14, 2.15,3.01)Using the grid and right triangle relationships,
m∠ ACD =
tan tan− −+
1 12
3
1
2. Since
∠ ADC and ∠ ADE are a linear pair andsupplementary, then
m∠ ADC =
180 31
2
1 1− ( ) +
− −tan tan . For
∆ACD, 180 = m∠ ADC + m∠ ADC +m∠ CAD. By substitution and properties of
equality, m∠ CAD =
tan tan− −( ) −
1 132
3;
m∠ CAD ≈ 37.87.
CD is parallel to GH (slopes of -0.5). Since∠ ACD and ∠ CGH are corresponding angles,m∠ CGH = m∠ ACD. Since ∠ CGH and∠ FGJ are a vertical pair, m∠ CGH = m∠ FGJ.By transitive property, m∠ FGJ = m∠ ACD;m∠ FGJ ≈ 60.26. Since ∠ ADC and ∠ DHGare corresponding angles, m∠ DHG =m∠ ADC. Since ∠ DHG and ∠ KHI are avertical pair, m∠ DHG = m∠ KHI. By thetransitive property, m∠ KHI = m∠ ADC;m∠ KHI ≈ 81.87.
2.08G (also 1.01, 2.05)Since line p is perpendicular to line l andline l is parallel to line m, then, byperpendicular transversal theorem, line p isperpendicular to line m. Since the angle withrespect to line m, which corresponds to theangle at line l, measures x and the angle whichmeasures y are a complementary pair andx = y, then x = 45.
2.08H (also 1.01, 2.05, 2.14)For the triangle shown, 180 = 90 + x + 55,where the third angle is the correspondingvertical angle to the one shown measuring 55.By subtraction property of equality, x = 35.
Geometry Indicators p. 38 Commentaries
NOTES2.08I (also 1.01, 2.05, 2.14)Since ∠ b and ∠ a are a linear pair andsupplementary, m∠ b = 180 - m∠ a. Since ∠ dand ∠ e are a linear pair and supplementary,m∠ d = 180 - m∠ e. In the triangle shown, thecorresponding vertical angle to ∠ c measures180 - m∠ b - m∠ d; therefore m∠ c = 180 -m∠ b - m∠ d. By substitution, m∠ c = 180 -(180 - m∠ a) - (180 - m∠ e). After simplifying,m∠ c = m∠ a + m∠ e - 180.
2.09A (also 2.01, 2.04, 2.09)Construct arcs through B with vertices at Aand C. The second intersection (not theoriginal B) of the arcs would be the vertex B.
2.09B (also 2.02, 2.04, 2.05, 2.08, 2.11, 2.15)Use the rectangular grid and coordinate systemto find the interior angle measures.m∠ A = 90 + tan-1(1) = 135
m B∠ = +≈−90
3
2146 31tan .
m C∠ =+
≈− −tan tan .1 12
3
5
485 0
m D∠ = +≈−90
4
5128 71tan .
m∠ E = 90m∠ F = 90 + tan-1(1) = 135Since the exterior angles are supplements ofthe corresponding interior angles, then theexterior angle measures are:
m A∠ = − + ( )[ ] =−180 90 1 451tan
m B∠ = − +
≈
−180 903
233 71tan .
m C∠ = −+
≈
− −1802
3
5
495 01 1tan tan .
m D∠ = − +
≈
−180 904
551 31tan .
m∠ E = 90
m F∠ = − + ( )[ ] =−180 90 1 451tan
Geometry Indicators p. 39 Commentaries
NOTES2.09C (also 2.05)Since the sum of the interior angles of aconvex pentagon is 3 • 180 = 540, then540 = 6x + (4x + 13) + (x + 9) + (2x - 8) + (4x- 1)540 = 17x + 13 (combine like terms)31 = x (properties of equality)By substitution, m∠ A = 186, m∠ B = 137,m∠ C = 40, m∠ D = 54, and m∠ E = 123.
2.09D (also 2.05)Since the sum of the interior angle measuresof a regular hexagon is 720, then m∠ B = 120.By definition, the m∠ B in the square is 90.m∠ CBA = 360 - 120 - 90 = 150.Since the sum of the interior angle measuresof a convex polygon is 180(n - 2), where n isthe number of sides, and 150n is the sum ofthe interior angle measures in the unknownregular polygon (by definition), then180(n- 2) = 150n180n - 360 = 150n (distribution property)n = 12 (properties of equality)
2.09E (also 2.04, 2.16)Circle; one definition of a circle is aninfinitely-sided regular polygon.
2.09F (also 2.01, 2.02, 2.03, 2.04, 2.05, 2.08,2.11, 2.13, 3.02, 3.03)The best candidate looks like the
perpendicular bisector of EG. F (7.5, 2) is the
midpoint of EG. Since the slope of EG is0.4, then the perpendicular bisector isy = -2.5x + b. Since F is on the bisector, then2 = (-2.5)(7.5) + b by substitution.b = 2 + 18.75 = 20.75. The line of symmetryis y = -2.5x + 20.75. Check to see if thisequation also represents the perpendicular
bisector of CJ and the altitude of theequilateral triangle ACJ.
Geometry Indicators p. 40 Commentaries
NOTES2.10A (also 2.19)The edge of each cube is 14 ÷ 4 = 3.5 cm.Volume = 21 • 3.53 = 900.4 cm3.Surface area = 54 • 3.52 = 661.5 cm2.
2.10 BCone
2.10CCylinder
2.10D (also 2.04, 2.19)A pyramid is removed from each corner.
x
x
4 - 2x
x 2
Looking at any face of the solid, let x be thelength removed from the original cube’sedges. The length of the solid’s edges areexpressed two ways as shown. Solve for x:
4 2 2− =x x
4 2 2= +x x
4 2 2= +( )x
4
2 2+= x
1.17 ≈ x.
Length of edge =
4 24
2 21 7−
+
≈ . units
Geometry Indicators p. 41 Commentaries
NOTESRegular triangles and octagons are the faces ofthe solid.area of octagon = 16 - 2x2
∴ =QS VS; corresponding lengths ofcorresponding parts of congruent triangles
2.12F (also 2.05, 2.08, 2.13)Since DH bisects ∠ D, let x = m∠ ADH =
= m∠ HDC.Since CH bisects ∠ C, let y = m∠ BCH =
= m∠ HCD.m∠ A + m∠ B = 360 - 2x - 2ym∠ DHC = 180 -x - y
m∠ DHC =
1
2(360 - 2x - 2y)
=
1
2(m∠ A + m∠ B)
Geometry Indicators p. 44 Commentaries
NOTES2.13A
xx
15 6
2 8
6 96 3
.
.
..= ⇒ ≈
2.13B (also 2.08, 3.02)y = -3x + 26
2.13C (also 2.01, 2.02, 2.03, 2.04, 2.16)
Draw AB on the coordinate plane. Drawcircles of length AB with centers at A and B.The intersections will be the locations forvertex C.To find the exact locations of C, solve asystem of equations with one of the circlescentered at A or B and the perpendicular
bisector of AB. Locations of C are:
7 12 6 3 3+ −( ), and
7 12 6 3 3− +( ),
2.13D (also 2.01, 2.02, 2.04, 2.08)Scotland County
AC = 104 , then the measures of the anglesopposite those sides are unequal in the sameorder. Therefore m∠ C > m∠ A > m∠ B.
m∠ C =
tan tan− −−
1 111
2
2
10 ≈ 68.4
m∠ B =
tan tan− −+
1 18
9
2
11 ≈ 51.9
m∠ A =
tan tan− −+
1 19
8
2
10 ≈ 59.7
Geometry Indicators p. 45 Commentaries
NOTES2.13HSince the sum of the angles of a triangle is180, then 180 = x + (3x + 20) + 6x. Since x =16, the angle measures are 16, 68, and 96.Since one of the angles is greater than 90, thetriangle is obtuse.
2.13J (also 2.05, 2.08, 2.12)Let m∠ A = 2x and m∠ D = 2y. Since ∠ A and∠ D are supplementary, 2x + 2y = 180 andx + y = 90. Therefore ∆APD is a righttriangle. Since m∠ PAD = m∠ BPA, ∆ABP isan isosceles triangle. Since m∠ PDA =mCPD, ∆PCD is an isosceles triangle.
2.13K (also 1.01, 2.01, 2.02, 2.03, 2.05, 3.02)(-12, -2) and (16, -10) are both on the line
through B, y = − −2
7
38
7x , that is
perpendicular to AB. Both points are 53
units from B, the same length as AB.
2.13L (also 1.01, 2.01, 2.02, 2.03, 2.05, 3.02)
y = − −4
3
22
3x is the equation of the line
which contains AC. Since ADAB
= 2, then
AEAC
= 2; therefore AE = 2AC = 20.
The coordinates for E (x, y) satisfies
20 =
− −( ) + + +
10 64
3
22
3
22
x x
(2, -10) is the location for E.
Geometry Indicators p. 46 Commentaries
NOTES2.14A (also 2.05, 2.13, 2.19)Since in each case the segment joining mid-points is half the measure of the third side,then P = 0.5(16.8 + 11.3 + 7.7) = 17.9 m.
2.14B (also 1.01, 2.03, 2.05, 2.13, 2.19)Perimeter = ED + DB + BC + CE
ED = 82 ; DB = 20
BC = ED • ABAD
=
82245
605
EC = DBAD
• AE =
20
605
2
25
2
Perimeter ≈ 23.8 units
2.14C (also 2.05, 2.08, 2.13)Let m∠ A = x. Then m∠ BCA = x andm∠ ABC = 180 - 2x. Since ∠ DBC is thesupplement of ∠ ABC, then m∠ DBC = 2x.Then m∠ BDC = 2x and m∠ DCB = 180 - 4x.m∠ DCE = 180 - (180 - 4x) -x = 3x. Thenm∠ DEC = 3x and m∠ CDE = 180 - 6x. Since∆ADE is isosceles, m∠ D = m∠ E andtherefore 180 - 4x = 3x. Solving for x,
x =
180
7 which is the measure of ∠ A.
2.15A (also 1.01)
CD = 3 4 3. ≈ 5.9 cm.
2.15B (also 1.01)Area = 5 squaresUsing the right triangle formed by thediagonal of two squares shown (X), side ofone square, and adjacent sides of two squares.Using the Pythagorean Theorem, the length of
the side of a square is 45 .
Area = 5 45
2( ) = 225 cm2.
Geometry Indicators p. 47 Commentaries
NOTES2.15C (also 1.01)
sin A =
12 8 5 1
12 8
2 2. .
.
− ≈ 0.9172
2.15D (also 1.01)
19 193− ≈ 5.1 miles
2.15E
m CDE∠ = +≈− −tan tan .1 13
2
885 6
m ABC∠ =+
≈− −tan tan .1 15
3
5
2127 2
2.15FBy AAA, ∆ABC is similar to ∆CBD.
Then
BDBC
BCAB
BD AB BC= ⇒ ( )( ) = ( )2
(BC)2 = 280
BC = 280 ≈ 16.7 units
Geometry Indicators p. 48 Commentaries
NOTES
6
9-x
x
6
AB
CD
E
F
D
6
6
AB
CD
E
F
D
6
2.5
2.5
2.15G (also 1.01, 2.01, 2.04, 2.05, 2.08, 2.13)
For the rectangle ABCD, fold C to A. Let EFrepresent the fold. Let DE = x and AE = 9 - x.
Then x x x2 2
36 9 2 5+ = −( ) ⇒ = .
Since DE is parallel to AF, then∠ ≅ ∠DEA FAE (alternate interior angles).∠ ≅ ∠DAE BAF (complements to congruentangles). ∆ ∆ADE ABF≅ (ASA); thereforeDE = BF.
Geometry Indicators p. 49 Commentaries
NOTES
6
6
AB
CD
E
F
D
6
2.5
2.5 2.5
2.5
4
G
6
Construct EG parallel to CD. Since ABCD is
a rectangle, EG = 6 and EG FG⊥ .
EG EG( ) = ⇒ =
2
52 52 ≈ 7.2 in.
2.15H (also 1.01)
150
Geometry Indicators p. 50 Commentaries
NOTES2.16A (also 1.01, 2.19)
12 65
65
12
2x x= ⇒ =
Area =
π 5
2
65
1265
2
− ≈ 41.4 cm2
2.16B (b)
2.16C (also 1.01, 2.19)
Circumference = 2 15π ≈ 24.3 units
2.16D(x + 1)2 + (y - 4)2 = 29
2.16E (also 1.01, 2.19, 4.01)The region closer to the center is a concentriccircle with a radius half that of the originalcircle.
P =
π π
π
r
r
2
2
2
1
2 3
4
−=
2.16F1:4
2.16G
Since radius = x y p t2 2 2 2+ = + thenx2 + y2 + p2 + t2 = 2r2 = 288. Since radius = 12,then circumference = 24π ≈ 75.4 units.
2.16H (also 1.01, 2.19)Since the circumference is 8π, the radius is 4.The area of the shaded region is128 - 32π ≈ 27.5 units.
Geometry Indicators p. 51 Commentaries
NOTES2.17A (also 2.15)
RP =
6
35cos( ) ≈ 7.3 units
2.17B
Since CD = 10 and AB and CD areequidistant from the center of the circle, thenAB = 10.
2.17C27
2.17D (also 1.01, 2.15)Estimate K at (6, 8.7). Then
2.19H (also 1.01, 2.13, 2.15)Let AD = x and BD = x + 3,then x(x + 3) = 54. Solving, x = 6;therefore AD = 6 and BD = 9.
Perimeter = 12 2 117+ ≈ 33.6 cm.
Geometry Indicators p. 55 Commentaries
NOTES2.19I (also 1.01, 2.09, 2.10, 2.12, 2.13, 2.15,2.17)Partition the hexagon base into rectangles andright triangles to compute the area of the base.Volume of hexagonal prism =594 • 18 = 10692 m3
2.19K (also 1.01, 2.05, 2.08, 2.13)∆X is similar to ∆Z by AAA. Since the lineardimensions of ∆Z are half those of ∆X, thenX = 4Z. ∆(Y+Z) has half the area of ∆(X+Y);2(Y + Z) = X + Y. By applying the distributiveproperty and substitution,2Y + 2Z = 4Z + Y.By properties of equality, Y = 2Z. Since∆(W+Z) is congruent to ∆(X+Y),W + Z = X + Y. By properties of equality andsubstitution, W = 5Z. SinceW + X + Y + Z = 80, by substitution, 12Z = 80.
Solving, Z =
80
12 ≈ 6.7 cm2; by substitution,
X =
480
12
≈ 26.7 cm2,
Y =
280
12
≈ 13.3 cm2, and
W =
580
12
≈ 33.3 cm2.
Geometry Indicators p. 56 Commentaries
NOTES2.19L (also 1.01, 2.15)Partition the trapezoid into two right trianglesand a rectangle. The two triangles make a
7-24-25 right triangle with height
49
25 which is
the height of the trapezoid.
Area =
1
2
49
2512 37
+( ) = 48.02 units2
2.19M (also 2.10)One straegy: Build a rectangular prismenclosing the solid. Delete the volumes ofpyramids and smaller rectangular prisms fromthe volume of the enclosing prism.
2.19N
Area =
91
2
9
49
9
8−= −π π m2
3.01A (also 2.02, 2.05, 2.12)
MA and HT are parallel (slopes of
1
3). HM
and AT are not parallel (slopes of 4 and − 2
5
respectively). Since one one pair of oppositesides are parallel, MATH is a trapezoid.
3.01B (also 2.13)
The slopes of GB, GU, and UB are
9
4, −1
3,
and
8
7. Since none of the sides of the triangle
are perpendicular to one another, ∆GUB is nota right triangle.
3.02Ay - 3 = 5(x + 3)y = 5x + 18
Geometry Indicators p. 57 Commentaries
NOTES3.02B (also 2.05, 2.12)
y = -5x + 14 parallel to BC through A
y =
1
2
5
2x − , parallel to AB through C
y = − −3
5
48
5x , parallel to AC through B
3.02C (also 2.05, 2.12)Any line of the form y = -4x + b where-1 ≤ b ≤ 49 or any line of the form
y = − +3
4x b where
9
4
66
4≤ ≤b .
3.02D (also 2.05, 2.13)The point equidistant is at the intersection of
the perpendicular bisectors of EF, EG, and
FG. Only two of the equations are necessaryto locate the point.
y =
1
5
68
10x − , ⊥ bisector of EF
y =
3
4
6
4x + , ⊥ bisector of EG
y =
5
3
44
6x − , ⊥ bisector of FG
The intersection is at
106
11
96
11,
.
y =
1
5
1
5x − contains the altitude for EF
through G
y =
3
42x + contains the altitude for EG
through F
y =
5
35x + contains the altitude for FG
through E.
Geometry Indicators p. 58 Commentaries
NOTES3.03A
x y y x, ,( )→ −( )
3.03B
x y x y, ,( )→ − +( )3
3.03C
A C
B
C' A'
B'
doubles horizontally, reflects across the y-axis.
3.03D
x y x y, ,( )→ − − +( )2 2 1
3.03E
x y x y, ,( )→ − −( )2 2 2 2
4.01A
7
12
Geometry Indicators p. 59 Commentaries
NOTES4.01B (also 1.01, 2.19)Draw a 9 by 12 rectangle around the shadedregion. From its area delete rectangles andright triangles for the region’s outer boundary.From the interior delete a quarter circle,rectangle, and right triangle to determine thethe shaded region.
