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Name: ________________________ Class: ___________________ Date: __________
ID: A
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 J K, KL, and J L.


→
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)°
1
Name: ________________________
ID: A
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).
2
Name: ________________________
ID: A
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°.
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Name: ________________________
ID: A
21. Given: RT⊥SU , ∠SRT ≅ ∠URT , RS ≅ RU . T is the midpoint of SU .
Prove: ∆RTS ≅ ∆RTU
Complete the proof.
Proof:
Statements
1. RT⊥SU
2. ∠RTS and ∠RTU are right angles.
3. ∠RTS ≅ ∠RTU
4. ∠SRT ≅ ∠URT
5. ∠S ≅ ∠U
6. RS ≅ RU
7. T is the midpoint of SU .
8. ST ≅ UT
9. RT ≅ RT
10. ∆RTS ≅ ∆RTU
Reasons
1. Given
2. [1]
3. Right Angle Congruence Theorem
4. Given
5. [2]
6. Given
7. Given
8. Definition of midpoint
9. [3]
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 =
∆ABC ≅ [3] by [4], and ∠ABC ≅ ∠RPQ by [5].
26 . So AB ≅ [2], BC ≅ PQ, and CA ≅ QR. Therefore
23. Write an equation in point-slope form for the perpendicular bisector of the segment with endpoints A(−2,2)
and B(5,4).
4
Name: ________________________
ID: A
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.
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Name: ________________________
ID: A
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
1. ABDF and FBCD are parallelograms.
2. ∠BCD ≅ ∠DFB
3. DF Ä AB
4. ∠DFB ≅ ∠ABF
5. ∠BCD ≅ ∠ABF
Reasons
1. Given
2. [1]
3. Opposite sides in a parallelogram are
parallel.
4. [2]
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.
6
Name: ________________________
ID: A
33. Show that all four sides of square ABCD are congruent and that AB ⊥ BC .
7
Name: ________________________
ID: A
34. Given: ABCD is a rectangle. W, X, Y, and Z are midpoints.
Prove: WXYZ is a rhombus.
Complete the proof.
Proof:
Statements
1. ABCD is a rectangle.
W, X, Y, and Z are midpoints.
2. ∠A, ∠B, ∠C, and ∠D are right angles
3. ∠A ≅ ∠B ≅ ∠C ≅ ∠D
4. AB ≅ CD, BC ≅ AD
1
2
5. AW =
1
AD, WD = 2 AD,
1
1
1
1
Reasons
1. Given
2. Definition of a rectangle
3. Right Angle Theorem
4. Theorem: Both pairs of opposite sides are
congruent.
5. [1]
AX = 2 AB, XB = 2 AB,
BY = 2 BC, YC = 2 BC,
1
1
CZ = 2 CD, ZD = 2 CD
6.
1
2
1
AB = 2 CD,
1
2
1
BC = 2 AD
6. Division Property of Equality
7. AB = AB, BC = BC
7. Reflexive Property of Equality
CD = CD, AD = AD
8. AW = WD = BY = YC,
8. Substitution
AX = XB = CZ = ZD
9. AW ≅ WD ≅ BY ≅ YC,
9. Definition of Congruent Segments
AX ≅ XB ≅ CZ ≅ ZD
10. ∆AXW ≅ ∆BXY ≅ ∆DZW ≅ ∆CZY
11. WX ≅ XY ≅ YZ ≅ ZW
12. WXYZ is a rhombus
10. [2]
11. CPCTC
12. Definition of a rhombus
8
Name: ________________________
ID: A
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.
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Name: ________________________
ID: A
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 (2x 2 + 3x − 5)° and (x 2 + 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.
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Name: ________________________
ID: A
44. ∆ABF ≅ ∆EDG. ∆ABF and ∆GCF are equilateral. AG = 21 and CG =
1
4
AB. 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.
11
ID: A
Geometry Midterm 2012
Answer Section
NUMERIC RESPONSE
1. ANS: 74
PTS: 1
DIF:
TOP: 1-4 Pairs of Angles
2. ANS: 20
PTS:
TOP:
KEY:
3. ANS:
Average
NAT: 12.2.1.f
STA: 6MG2.2
KEY: supplementary angles | complementary angles
1
DIF: Advanced
NAT: 12.2.1.f
1-7 Transformations in the Coordinate Plane
transformations | application | rotations
3.35
PTS: 1
DIF: Advanced
TOP: 3-6 Lines in the Coordinate Plane
4. ANS: 11
NAT: 12.3.3.d
STA: GE15.0
PTS: 1
DIF: Average
NAT: 12.3.3.f
TOP: 5-4 The Triangle Midsegment Theorem
5. ANS: 49.8
PTS: 1
DIF: Average
TOP: 6-2 Properties of Parallelograms
NAT: 12.2.1.h
STA: GE7.0
SHORT ANSWER
6. ANS:
CE = 105
CE = CD + DE
6x = (4x + 8) + 27
6x = 4x + 35
2x = 35
2x 35
=
2
2
35
x=
or 17.5
2
Segment Addition Postulate
Substitute 6x for CE and 4x + 8 for CD.
Simplify.
Subtract 4x from both sides.
Divide both sides by 2.
Simplify.
CE = 6x = 6 (17.5) = 105
PTS:
OBJ:
STA:
KEY:
1
DIF: Average
REF: Page 15
1-2.3 Using the Segment Addition Postulate
NAT: 12.3.5.a
GE1.0
TOP: 1-2 Measuring and Constructing Segments
segment addition postulate
1
ID: A
7. ANS:
J K = 6, KL = 6, J L = 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 J K, KL, and J L.
JK = 6x = 6 (1) = 6
KL = 3x + 3 = 3(1) + 3 = 6
JL = JK + KL = 6 + 6 = 12
PTS: 1
DIF: Average
REF: Page 16
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
3x = 9
x=3
Add 1 to both sides.
Subtract 4x from both sides.
Divide both sides by 3.
Step 2 Find m∠ABD.
m∠ABD = 7x − 1 = 7(3) − 1 = 20°
PTS: 1
DIF: Average
REF: Page 23
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)°
PTS:
OBJ:
NAT:
KEY:
1
DIF: Average
REF: Page 29
1-4.2 Finding the Measures of Complements and Supplements
12.3.3.g
STA: 6MG2.2
TOP: 1-4 Pairs of Angles
complementary angles | supplementary angles
2
ID: A
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.
3
Then the width of the mirror is 4 l.
A = lw
3
192 = l( 4 l)
3
192 = 4 l 2
256 = l 2
16 = l
3
The length of the mirror is 16 inches. The width of the mirror is 4 (16) = 12 inches.
PTS: 1
DIF: Advanced
TOP: 1-5 Using Formulas in Geometry
NAT: 12.2.1.h
STA: GE8.0
KEY: area | rectangles | application
3
ID: A
12. ANS:
7.8 units
Method 1 Substitute the values for the
coordinates of T and U into the Distance
Formula.
TU =
Method 2 Use the Pythagorean Theorem.
Plot the points on a coordinate plane. Then
draw a right triangle.
2
2
ÁÊ x 2 − x 1 ˜ˆ + ÁÊ y 2 − y 1 ˜ˆ
Ë
¯
Ë
¯
2
=
(−2 − 4) + (3 − −2)
=
(−6) + (5)
=
61
2
2
2
≈ 7.8 units
Count the units for sides a and b. a = 6 and
b = 5. Then apply the Pythagorean Theorem.
c 2 = a 2 + b 2 = 62 + 5 2 = 36 + 25 = 61
c ≈ 7.8 units
PTS: 1
DIF: Average
REF: Page 45
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 (x 1 , y 1 ) and (9, –9) for (x 2 , y 2 ) in the slope formula.
y 2 − y 1 −9 + 7 −2
m=
=
=
9−6
3
x2 − x1
PTS: 1
NAT: 12.3.5.a
DIF: Average
STA: 7AF3.3
REF: Page 182
OBJ: 3-5.1 Finding the Slope of a Line
TOP: 3-5 Slopes of Lines
4
ID: A
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 x 1 , and 7 for y 1 .
y − 7 = 2(x − 4)
PTS: 1
NAT: 12.3.5.a
16. ANS:
DIF: Average
STA: 1A7.0
REF: Page 191
OBJ: 3-6.1 Writing Equations of Lines
TOP: 3-6 Lines in the Coordinate Plane
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
NAT: 12.3.5.a
DIF: Average
STA: 1A6.0
REF: Page 191
OBJ: 3-6.2 Graphing Lines
TOP: 3-6 Lines in the Coordinate Plane
5
ID: A
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
NAT: 12.3.5.a
STA: 1A7.0
18. ANS:
AB = 24
Step 1 Find the value of x.
BC
=
CA
REF: Page 192
OBJ: 3-6.3 Classifying Pairs of Lines
TOP: 3-6 Lines in the Coordinate Plane
Step 2 Find AB.
AB = 4x + 4
8x + 3
=
7x + 8
=
4(5) + 4
x
=
5
=
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
TOP: 4-2 Angle Relationships in Triangles
6
STA: GE12.0
ID: A
21. ANS:
[1] Definition of perpendicular lines
[2] Third Angles Theorem
[3] Reflexive Property of Congruence
Proof:
Statements
1. RT⊥SU
2. ∠RTS and ∠RTU are right angles.
3. ∠RTS ≅ ∠RTU
4. ∠SRT ≅ ∠URT
5. ∠S ≅ ∠U
6. RS ≅ RU
7. T is the midpoint of SU .
8. ST ≅ UT
9. RT ≅ RT
10. ∆RTS ≅ ∆RTU
PTS: 1
NAT: 12.3.5.a
DIF: Average
STA: GE5.0
Reasons
1. Given
2. Definition of perpendicular lines
3. Right Angle Congruence Theorem
4. Given
5. Third Angles Theorem
6. Given
7. Given
8. Definition of midpoint
9. Reflexive Property of Congruence
10. Definition of congruent triangles
REF: Page 232
OBJ: 4-3.3 Proving Triangles Congruent
TOP: 4-3 Congruent Triangles
7
ID: A
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=
AB =
=
BC =
=
CA =
=
ÊÁ x − x ˆ˜ 2 + ÊÁ y − y ˆ˜ 2
1¯
1¯
Ë 2
Ë 2
2
2
(5 − 3) + ÊÁË 2 − (−1) ˆ˜¯
4+9 =
RP =
13
2
(−2 − 5) + (0 − 2)
49 + 4 =
=
2
53
ÊÁ 3 − (−2) ˆ˜ 2 + (−1 − 0) 2
Ë
¯
25 + 1 =
PQ =
=
QR =
26
=
ÊÁ −3 − (−6) ˆ˜ 2 + (4 − 2) 2
Ë
¯
9+4 =
13
ÊÁ −5 − (−3) ˆ˜ 2 + (−3 − 4) 2
Ë
¯
4 + 49 =
53
ÊÁ −6 − (−5) ˆ˜ 2 + ÊÁ 2 − (−3) ˆ˜ 2
Ë
¯
Ë
¯
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.
PTS: 1
DIF: Average
REF: Page 261
OBJ: 4-6.4 Using CPCTC in the Coordinate Plane
STA: GE5.0
TOP: 4-6 Triangle Congruence: CPCTC
8
NAT: 12.2.1.e
ID: A
23. ANS:
7
y − 3 = − 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.
ÊÁ −2 + 5 2 + 4 ˆ˜
˜˜ = (1.5,3)
,
Midpoint of AB = ÁÁÁÁ
˜˜
2
2
Ë
¯
Step 3 Find the slope of the perpendicular bisector.
(4) − (2) 2
=
Slope of AB =
(5) − (−2) 7
7
Since the slopes of perpendicular lines are opposite reciprocals, the slope of the perpendicular bisector is − 2 .
Step 4 Use point-slope form to write the equation.
y − y 1 = m(x − x 1 )
7
y − 3 = − 2 (x − 1.5)
PTS: 1
DIF: Average
REF: Page 303
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
2
BO = 3 BY
Centroid Theorem
2
BO = 3 (26.4) = 17.6
Substitute 3.3 for BY and simplify.
PTS: 1
DIF: Average
REF: Page 315
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
9
ID: A
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.
Pythagorean Theorem
a2 + b2 = c2
2
2
2
Substitute 4x for a, 3x for b, and 27 for c.
(4x) + (3x) = 27
2
Multiply and combine like terms.
25x = 729
729
x2 =
Divide both sides by 25.
25
x=
729
= 5.4 in.
25
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.
Substitute the longest side length for c.
14 2 ? 62 + 13 2
196 ? 36 + 169
Multiply.
196 < 205
Add and compare.
Since 14 2 < 6 2 + 13 2 , 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.
Hypotenuse = leg 2
17 = l 2
17
l=
≈ 12 m
Divide both sides by 2 and round.
2
PTS: 1
NAT: 12.3.3.d
DIF: Average
STA: GE20.0
REF: Page 357
OBJ: 5-8.2 Application
TOP: 5-8 Applying Special Right Triangles
10
ID: A
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.
Simplify.
= 7740
Step 2 Find the measure of one interior angle.
7740
= 172
The interior angles are ≅ , so divide by 45.
45
PTS: 1
DIF: Average
REF: Page 384
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
A regular decagon has 10 congruent exterior
360
= 36°
measure of one exterior angle =
angles, so divide the sum by 10.
10
The measure of each exterior angle of a regular decagon is 36°.
PTS: 1
DIF: Average
REF: Page 384
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
360 °
there are 6 congruent exterior angles, so m∠YXZ = 6 = 60°
PTS: 1
NAT: 12.3.3.f
DIF: Average
STA: GE12.0
REF: Page 385
OBJ: 6-1.5 Application
TOP: 6-1 Properties and Attributes of Polygons
11
ID: A
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. Opposite sides in a parallelogram are
3. DF Ä AB
parallel.
4. ∠DFB ≅ ∠ABF
4. Alternate Interior Angles Theorem
5. ∠BCD ≅ ∠ABF
5. Substitution
PTS: 1
DIF: Average
REF: Page 394
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.
X = midpoint of AC
Ê 2+x 3+y ˆ
X = (7,6) = ÁÁÁÁ 2 , 2 ˜˜˜˜
Ë
¯
7=
2 +x
2
,6=
3 +y
2
x = 12, y = 9
C(12, 9)
PTS: 1
DIF: Advanced
TOP: 6-3 Conditions for Parallelograms
Step 2 Solve for point D.
X = midpoint of BD
Ê 8 + x 11 + y ˆ
X = (7,6) = ÁÁÁÁ 2 , 2 ˜˜˜˜
Ë
¯
7=
8 +x
2
,6=
11 + y
2
x = 6, y = 1
D(6, 1)
NAT: 12.3.3.f
12
STA: GE7.0
ID: A
33. ANS:
1
AB = 3 5 , BC = 3 5 , CD = 3 5 , DA = 3 5 , slope of AB = 2, slope of BC = − 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)
0−3
=2
1
slope of BC = 5 − (−1) = − 2
Ê 1ˆ
2 ⋅ ÁÁÁ − 2 ˜˜˜ = −1
Ë ¯
Since the product of the slopes is −1, AB ⊥ BC .
PTS: 1
DIF: Average
REF: Page 410
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.
PTS: 1
DIF: Average
REF: Page 411
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
13
ID: A
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)
−1 − 0
=3
1
slope of BD = 1 − (−2) = − 3
Ê 1ˆ
Since (3) ÁÁÁ − 3 ˜˜˜ = −1, AC ⊥ BD and ABCD is a rhombus.
Ë ¯
Since ABCD is both a rhombus and a rectangle, it is also a square.
PTS: 1
DIF: Average
REF: Page 420
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
14
ID: A
36. ANS:
square
Find the slope of each side to determine if the sides are perpendicular.
(5 − 9)
(6 − 2) 4
−4 4
=
=
=
slope of MN =
slope of OP =
(−3 − 0) −3 3
(4 − 1) 3
(6 − 9) −3
(5 − 2)
3
=
=
slope of NO =
slope of PM =
(4 − 0)
(−3 − 1) −4
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.
PTS: 1
DIF: Advanced
NAT: 12.3.3.f
STA: GE7.0
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°
m∠QOP = 90°
Diagonals of a kite are
perpendicular.
m∠QRO + m∠RQO = 90°
m∠QPO + m∠PQO = 90°
The acute angles of a right
triangle are complementary.
70° + m∠RQO = 90°
50° + m∠PQO = 90°
Substitute the given values.
m∠RQO = 20°
m∠PQO = 40°
Subtract.
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
NAT: 12.3.3.f
DIF: Basic
STA: GE12.0
REF: Page 429
OBJ: 6-6.2 Using Properties of Kites
TOP: 6-6 Properties of Kites and Trapezoids
15
ID: A
38. ANS:
YD = 6.9
BD ≅ AC
BD = AC
BD = 17.2
BY + YD = BD
10.3 + YD = 17.2
YD = 6.9
Diagonals of an isosceles trapezoid are congruent.
Definition of congruent segments
Substitute 17.2 for AC.
Segment Addition Postulate
Substitute the given values.
Subtract 10.3 from both sides.
PTS: 1
DIF: Average
REF: Page 430
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
Theorem: A trapezoid is isosceles if and only if its diagonals are
QS ≅ RT
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.
PTS: 1
DIF: Average
REF: Page 430
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).
1
The area of a triangle is given by A = 2 bh .
From the diagram, the base of the triangle is b = RT = 4.
From the diagram, the height of the triangle is h = 4.
1
Therefore the area is A = 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 + (y 2 − y 1 ) 2 =
(−3 − 1) 2 + (−3 − 5) 2 =
PTS: 1
DIF: Advanced
NAT: 12.2.1.e
TOP: 1-6 Midpoint and Distance in the Coordinate Plane
16
16 + 64 =
80 ≈ 8.9
STA: GE17.0
KEY: area | distance formula | triangles
ID: A
41. ANS:
x = 6; 85°; 95°
Step 1 Create an equation
The angles are supplements and their sum equals 180°.
(2x 2 + 3x − 5) + (x 2 + 11x − 7) = 180
Step 2 Solve the equation
3x 2 + 14x − 12 = 180
3x 2 + 14x − 192 = 0
(3x + 32)(x − 6) = 0
32
x = − 3 or 6.
32
When x = − 3 , the measurement of the second angle is
x 2 + 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 2x 2 + 3x − 5 = 2(6) 2 + 3(6) − 5 = 85°.
The measurement of the second angle is x 2 + 11x − 7 = (6) 2 + 11(6) − 7= 95°.
PTS: 1
DIF: Advanced
TOP: 2-6 Geometric Proof
NAT: 12.2.1.f
17
STA: 6MG2.2
ID: A
42. ANS:
x = 7; y = 9; 147°; 147°; 33°; 33°
Step 1 Create a system of equations.
m∠1 = m∠3
20x + 7 = 5x + 7y + 49
15x − 7y = 42
The sum of the measures of supplementary angles equals 180°.
m∠1 + ∠2 = 180
20x + 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
Multiply the first equation by −2.
Multiply the second equation by 7.
161x − 14y = 1001
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
18
TOP: 2-7 Flowchart and Paragraph Proofs
ID: A
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 = 100
−(4x + y) = −60
Subtract the first equation from the second.
4x
x
8(10) + y
y
Divide both sides by 4.
Substitute 10 for x.
Simplify.
= 40
= 10
= 100
= 20
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.
1
Since CG = 4 AB, use substitution to get GF =
1
4
AF .
1
4
AF = AG + GF = AG + AF
3
4
AF = AG
3
4
AF = 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
4
AB, 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.
PTS: 1
DIF: Advanced
TOP: 4-3 Congruent Triangles
NAT: 12.3.3.f
KEY: multi-step
19
STA: GE5.0
ID: A
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
360 °
the dodecagon is 360º. So, each exterior angle is 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°.
PTS: 1
DIF: Advanced
NAT: 12.3.3.f
TOP: 6-1 Properties and Attributes of Polygons
20
STA: GE12.0
ID: A
47. ANS:
ZW = 12.9, WX = 8.6, XY = 25.8, and ZY = 8.6
1
(XY + ZW) = 19.35
Trapezoid Midsegment Theorem
2
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
3(ZY) + ZY + WX + [38.7 − 3(ZY)] = 55.9
3(WX) + WX + WX + [38.7 − 3(WX)] = 55.9
2(WX) = 17.2
WX = 8.6
Perimeter of the trapezoid
Substitute for XY and ZW.
Substitute for ZY.
Simplify.
Solve.
ZW = 12.9, WX = 8.6, XY = 25.8, and ZY = 8.6.
PTS: 1
DIF: Advanced
NAT: 12.2.1.h
TOP: 6-6 Properties of Kites and Trapezoids
21
STA: GE7.0