Answer Key

Geometry Regents Exam Questions

2020-21 Edition Donny Brusca

ISBN 978-1-67816-319-8 © 2020 Course Workbooks. All rights reserved. www.CourseWorkbooks.com

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Unit Page Chapter 1. Lines, Angles and Proofs ........................ 4 Chapter 2. Triangles ................................................ 5 Chapter 3. Right Triangles and Trigonometry ....... 12 Chapter 4. Quadrilaterals ...................................... 20 Chapter 5. Perimeter And Area ............................. 26 Chapter 6. Coordinate Geometry .......................... 27 Chapter 7. Polygons In The Coordinate Plane ....... 30 Chapter 8. Rigid Motions ....................................... 37 Chapter 9. Dilations ............................................... 39 Chapter 10. Transformation Proofs ......................... 42 Chapter 11. Circles .................................................. 49 Chapter 12. Solids ................................................... 56 Chapter 13. Constructions ....................................... 65 Notation A code next to each Regents Question number states from which Geometry Common Core Regents exam the question came. For example, AUG ’18 [25] means the question appeared on the August 2018 as question 25.

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Chapter 1. Lines, Angles and Proofs

1.1 Geometric Terms and Symbols JAN '16 [6] Ans: 1

1.2 Deductive Reasoning There are no Regents exam questions on this topic. 1.3 Parallel Lines and Transversals

JUN '15 [17] Ans: 1 AUG ’16 [1] Ans: 2

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Chapter 2. Triangles

2.1 Triangle Inequality Theorem JAN ’19 [19] Ans: 3

2.2 Angles of Triangles AUG ’16 [4] Ans: 2 JUN ’17 [17] Ans: 4 JAN ’18 [9] Ans: 3 JAN ’18 [18] Ans: 2 JUN ’18 [2] Ans: 3 AUG ’18 [1] Ans: 4 JAN ’19 [16] Ans: 4 JAN ’20 [1] Ans: 3 FALL '14 [10] The sum of the measures of the angles of a triangle is 180°, so m∠𝐴𝐵𝐶 m∠𝐵𝐶𝐴 m∠𝐶𝐴𝐵 180°. Each interior angle of the triangle and its exterior angle form a linear pair. Linear pairs are supplementary, so m m 180ABC FBC∠ + ∠ = ° , m m 180BCA DCA∠ + ∠ = ° , and m m 180CAB EAB∠ + ∠ = ° . By addition, the sum of these linear pairs is 540°. When the angle measures of the triangle are subtracted from this sum, the result is 360°, the sum of the exterior angles of the triangle. JAN '16 [33] (2) Parallel Postulate (3) Alternate Interior Angles Theorem (4) Consecutive adjacent angles on a straight line add to 180° (5) Substitution

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2.3 Medians, Altitudes and Bisectors There are no Regents exam questions on this topic. 2.4 Isosceles and Equilateral Triangles

AUG ’17 [11] Ans: 4 AUG ’19 [5] Ans: 3 JAN ’20 [12] Ans: 2 FALL '14 [5] In an isosceles triangle, the bisector of the vertex angle is also a median. Therefore, 𝑀𝑂 𝑂𝑃, so MO = 8. FALL '14 [24] △𝑋𝑌𝑍, XY ZY≅ and YW bisects XYZ∠ (Given).

XYZ△ is isosceles (Definition of isosceles triangle). YW is an altitude of XYZ△ (The angle bisector of the vertex of an isosceles triangle is also an altitude). YW XZ⊥ (Definition of altitude).

YWZ∠ is a right angle (Definition of perpendicular). JUN '15 [32] Since linear pairs are supplementary, m 65GIH∠ = ° . Since GH IH≅ , m 65IGH∠ = ° and m 180 (65 65) 50GHI∠ = − + = ° . Since EGB GHI∠ ≅ ∠ , the corresponding angles formed by the transversal and lines are congruent and AB CD . JAN ’17 [30] m m 25DAC ECA∠ = ∠ = ° m 180 2(25) 130AXC∠ = − = °

2.5 Congruent Triangles AUG ’16 [22] Ans: 3 JAN ’17 [3] Ans: 1

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JUN ’17 [9] Ans: 2 AUG ’18 [10] Ans: 4 JUN ’19 [8] Ans: 4 JUN ’17 [33] (Givens omitted)

TX XV≅ and RX XS≅ (def. of segment bisector); TXR VXS∠ ≅ ∠ (vertical angles); TXR VXS≅△ △ (SAS); T V∠ ≅ ∠ (CPCTC);

TR SV (alternate interior angles converse). AUG ’17 [35] (Givens omitted)

AD BC≅ (congruent legs of isosceles trapezoid). DEA∠ and CEB∠ are right angles (def. of perpendicular). DEA CEB∠ ≅ ∠ (all right angles are congruent). CDA DCB∠ ≅ ∠ (base angles of an isosceles trapezoid are congruent). CDA CDE DCB DCE∠ −∠ ≅∠ −∠ (subtraction). ADE BCE∠ ≅∠ (substitution). ADE BCE≅△ △ (AAS).

EA EB≅ (CPCTC). AEB△ is an isosceles triangle (def. of isosceles).

AUG ’18 [32] (2) Reflexive; (4) ∠𝐵𝐷𝐴 ≅ ∠𝐵𝐷𝐶; (6) CPCTC; (7) If points B and D are equidistant from the endpoints of 𝐴𝐶, then B and D are on the perpendicular bisector of 𝐴𝐶.

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AUG ’19 [33] △𝐴𝐵𝐸 ≅△ 𝐶𝐵𝐷 (given); ∠𝐴 ≅ ∠𝐶 (CPCTC); ∠𝐴𝐹𝐷 ≅ ∠𝐶𝐹𝐸 (vertical angles are congruent); 𝐴𝐵 ≅ 𝐶𝐵,𝐷𝐵 ≅ 𝐸𝐵 (CPCTC); 𝐴𝐷 ≅ 𝐶𝐸 (subtraction); △𝐴𝐹𝐷 ≅△ 𝐶𝐹𝐸 (AAS) JAN ’20 [35] (Givens omitted); 𝐵𝐷 ≅ 𝐵𝐷 (reflexive prop); △𝐴𝐵𝐷 ≅△ 𝐶𝐷𝐵 (SAS); ∠𝐶𝐵𝐷 ≅ ∠𝐴𝐷𝐵 (CPCTC); 𝐵𝐶 ≅ 𝐷𝐴 (CPCTC); 𝐵𝐶 𝐶𝐸 𝐷𝐴 𝐴𝐹, so 𝐵𝐸 ≅ 𝐷𝐹 (subtraction prop); ∠𝐵𝐺𝐸 ≅ ∠𝐷𝐺𝐹 (vertical angles are congruent); △𝐸𝐵𝐺 ≅△ 𝐹𝐷𝐺 (AAS); 𝐹𝐺 ≅ 𝐸𝐺 (CPCTC)

2.6 Similar Triangles JUN '15 [15] Ans: 3 JUN '15 [21] Ans: 4 AUG '15 [19] Ans: 2 JAN '16 [13] Ans: 1 JAN '16 [20] Ans: 4 JAN '16 [24] Ans: 3 JUN '16 [5] Ans: 3 JUN '16 [17] Ans: 1 AUG ’16 [12] Ans: 3 AUG ’17 [5] Ans: 4 AUG ’17 [9] Ans: 4 JAN ’18 [13] Ans: 3

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JAN ’18 [17] Ans: 4 JUN ’18 [4] Ans: 3 JUN ’18 [9] Ans: 4 JAN ’19 [8] Ans: 1 JUN ’19 [15] Ans: 2 AUG ’19 [18] Ans: 3 JAN ’20 [3] Ans: 2 JAN ’20 [6] Ans: 3 JAN ’20 [24] Ans: 4 AUG '15 [29] 6 914 21= ; Yes (SAS~) JAN ’17 [29] (Givens omitted)

I N∠ ≅ ∠ , G T∠ ≅ ∠ (alternate interior angles theorem); GIA TNA△ △ (AA~).

JUN ’18 [30] Yes, both triangles are 5-12-13 triangles by the Pythagorean Theorem, so they are congruent by SSS. All congruent triangles are also similar. AUG ’18 [29] Triangles are similar by AA. The triangles share the same angle at the stake, so these angles are congruent; the angles at the bases of the two poles are corresponding angles formed by parallel lines, so they are congruent. (The angles formed by the poles with the support wire are also congruent since they are corresponding angles formed by parallel lines.)

2.7 Triangle Angle Bisector Theorem There are no Regents exam questions on this topic.

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2.8 Triangle Midsegment Theorem JAN ’17 [4] Ans: 4 AUG ’17 [16] Ans: 4 JUN ’19 [23] Ans: 3 JAN ’20 [9] Ans: 3

2.9 Side Splitter Theorem JUN '15 [11] Ans: 3 AUG '15 [17] Ans: 4 JUN '16 [21] Ans: 2 JUN ’17 [5] Ans: 4 JUN ’17 [10] Ans: 2 AUG ’17 [7] Ans: 4 JUN ’18 [11] Ans: 2 JUN ’18 [21] Ans: 4 AUG ’18 [12] Ans: 2 AUG ’18 [16] Ans: 3 JAN ’19 [6] Ans: 2 JUN ’19 [11] Ans: 1 JUN '15 [31]

1.654.15 16.6x= ; 𝑥 6.6

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AUG '15 [27] 120230 315BC= ; 𝐵𝐶 164 JUN '16 [27] 3.75 4.55 6= ; 22.5 22.5

AB CD because AB divides the sides proportionately. 2.10 Incenter and Circumcenter There are no Regents exam questions on this topic. 2.11 Orthocenter and Centroid

JUN ’18 [18] Ans: 1 AUG ’19 [4] Ans: 1 JAN ’20 [30] 𝐶𝑋 2 𝐶𝐸 10; 𝐶𝐹 𝑌𝐹 7; 𝑋𝐹 𝑋𝑍 7.5; 𝑃 10 7 7.5 24.5

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Chapter 3. Right Triangles and Trigonometry

3.1 Pythagorean Theorem AUG '15 [11] Ans: 2 AUG ’16 [8] Ans: 3 JAN '16 [32] 169 20.6x= ; 𝑥 36.6 2 2 2236.6 20.61763.9242 c

cc

+ ==≈

AUG ’17 [34] 2 2 22 582 33641682

x xx

x

+ ===

2( 1682 8) 2402.2A = + ≈

3.2 Congruent Right Triangles JUN '16 [7] Ans: 3

3.3 Equidistance Theorems JUN '16 [19] Ans: 2 AUG ’16 [11] Ans: 4 AUG ’18 [22] Ans: 4

3.4 Geometric Mean Theorems JAN '16 [22] Ans: 2 JUN '16 [13] Ans: 2 AUG ’16 [10] Ans: 2

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AUG ’17 [18] Ans: 2 JAN ’18 [23] Ans: 2 JUN ’18 [23] Ans: 1 AUG ’18 [7] Ans: 3 AUG ’18 [20] Ans: 2 JAN ’19 [10] Ans: 3 AUG ’19 [16] Ans: 1 AUG ’19 [20] Ans: 2 JAN ’20 [16] Ans: 4 JUN '15 [34] √0.55 0.25 0.49 No, 20.49 0.250.9604 y

y==

0.9604 + 0.25 < 1.5 JUN ’17 [29] If an altitude is drawn to the hypotenuse of a triangle, it divides the triangle into two right triangles that are similar to each other and to the original triangle. JUN ’19 [30] 1515 17x = ; 17𝑥 225; 𝑥 13.2

3.5 Trigonometric Ratios JUN '16 [15] Ans: 4 JAN ’17 [14] Ans: 3 JAN ’19 [17] Ans: 4

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3.6 Use Trigonometry to Find a Side JUN '15 [5] Ans: 3 JUN '16 [11] Ans: 4 JAN ’17 [7] Ans: 2 JAN ’17 [12] Ans: 3 JUN ’17 [21] Ans: 4 AUG ’17 [19] Ans: 1 JAN ’18 [4] Ans: 1 AUG ’18 [6] Ans: 4 JAN ’19 [13] Ans: 2 AUG ’19 [15] Ans: 2 AUG ’19 [24] Ans: 1 FALL '14 [13]

tan3.47 6336M= ; 𝑀 384; 4960 + 384 = 5344

tan0.64 20493A= ; 𝐴 229; 5344 – 229 = 5115

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FALL '14 [22] x represents the distance between the lighthouse and the canoe at 5:00; y represents the distance between the lighthouse and the canoe at 5:05. 112 1.5tan6

x−= ; 𝑥 1051.3 112 1.5tan55

y−= ; 𝑦 77.4 1051.3 77.4 1955− ≈

AUG '15 [32] tan 7 ; 𝐴𝐶 1018.0 tan 16 ; 𝐷𝐶 435.9 𝐴𝐷 𝐴𝐶 𝐷𝐶 1018.0 435.9 582 JAN '16 [29] 30sin70

x= ; 𝑥 32

JAN '16 [36] tan 52.8 ; ℎ 𝑥 tan 52.8 1.32𝑥 tan 34.9 ; ℎ 𝑥 8 tan 34.9 0.70 𝑥 8 ; 1.32𝑥 0.7 𝑥 8 ; 𝑥 9.0 tan 52.8 ; ℎ 9 tan 52.8 11.86 11.86 1.7 13.6 AUG ’16 [31] 15sin75

x= ; 𝑥 15.5

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JUN ’17 [36] 6250tan15x

= ; 𝑥 23,325.3 6250tan52y

= ; 𝑦 4,883.0 23,325.3 – 4,883.0 = 18,442.3 Plane traveled 18,442 ft. in 1 min. 18442ft 1mile 60min 210mph1min 5280ft 1hr⋅ ⋅ ≈ JAN ’18 [34] 4.5cos54

HI= ; 𝐻𝐼 7.7 mi tan54 4.5IM= ; 𝐼𝑀 6.2 mi

JUN ’18 [33] tan72 400ST° = ; 𝑆𝑇 1231.07 1231.07sin55 STCT CT

° = ≈ ; 𝐶𝑇 1503 AUG ’18 [33] m∠𝐴𝐺𝐻 36 tan 36 ; 𝐻𝐴 7.2654 𝐻𝐴 𝐹𝐺 𝐷𝐸 7.3 𝐴𝐺 𝐻𝐴 𝐻𝐺 √7.2654 10 12.3607 𝐴𝐶 3 12.3607 37 JAN ’19 [34] sin 4.76° ⋅ ; 𝑥 216.914 in. 18.1 ft. tan 4.76° ; 𝑦 216.166 in. 𝑑 216.166 16 200.166 in. 16.7 ft. JUN ’19 [27] cos 68 ; 𝑥 27

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JUN ’19 [34] tan 30 ; 𝑦 254 tan 38.8 ; ℎ 353.8 353.8 254 100 JAN ’20 [26] sin 38° . ; 𝑥 40 inches JAN ’20 [33] tan 56° . ; 𝑥 1.927; 𝑥 1.5 3.427; 3.427 1.3 𝑐 ; 𝑐 3.7 m

3.7 Use Trigonometry to Find an Angle FALL '14 [1] Ans: 1 JAN '16 [16] Ans: 3 JUN ’17 [13] Ans: 1 AUG ’17 [15] Ans: 1 JUN ’18 [6] Ans: 2 AUG ’18 [9] Ans: 1 JUN ’19 [22] Ans: 4 JAN ’20 [7] Ans: 1 JUN '15 [28] 4.511.75sin x = ; 𝑥 23° JUN '16 [30] 104tan x = ; 𝑥 68° AUG ’16 [34] 1275tan x = ; 𝑥 9.09; 7275tan y = ; 𝑦 43.83; 𝑦 𝑥 34.7°

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JAN ’18 [31] 618cos x = ; 𝑥 71° AUG ’19 [26] 525sin x = ; 𝑥 11.5°

3.8 Special Triangles JAN ’17 [9] Ans: 2

3.9 Cofunctions JUN '15 [12] Ans: 4 AUG '15 [4] Ans: 1 JAN '16 [9] Ans: 4 AUG ’16 [6] Ans: 1 JUN ’17 [3] Ans: 3 AUG ’17 [21] Ans: 4 JUN ’18 [8] Ans: 1 AUG ’18 [24] Ans: 2 JAN ’19 [22] Ans: 1 JUN ’19 [9] Ans: 2 AUG ’19 [19] Ans: 1 JAN ’20 [21] Ans: 3 JAN ’20 [23] Ans: 2 FALL '14 [7] 2𝑥 0.1 4𝑥 0.7; 𝑥 0.4

A and B are complementary angles, and cofunctions of complementary angles are equal.

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FALL '14 [20] The acute angles in a right triangle are always complementary. The sine of any acute angle is equal to the cosine of its complement. JUN '16 [28] 𝑅 90 73 17° Cofunctions of complementary angles are equal. JAN ’17 [27] Yes, because 28º and 62º angles are complementary. The sine of an angle equals the cosine of its complement. JAN ’18 [27] Since angles A and B are complentary and sine and cosine are cofunctions, sin cosA B= . Therefore, when sin A increases, cos B increases.

3.10 SAS Sine Formula for Area of a Triangle There are no Regents exam questions on this topic.

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Chapter 4. Quadrilaterals

4.1 Properties of Quadrilaterals AUG '15 [8] Ans: 3 JAN '16 [3] Ans: 3 AUG ’16 [24] Ans: 1 AUG ’17 [8] Ans: 4 JAN ’18 [2] Ans: 2 AUG ’18 [13] Ans: 4 JAN ’19 [7] Ans: 2 JAN ’19 [12] Ans: 2 JUN ’19 [17] Ans: 2 JUN ’19 [21] Ans: 2 AUG ’19 [7] Ans: 2 JUN '15 [26] Opposite angles in a parallelogram are congruent, so m∠𝑂 118°. The interior angles of a triangle equal 180°. 180 – (118 + 22) = 40. AUG ’17 [26] The four small triangles are 8-15-17 triangles. 4 17 68 AUG ’18 [26] Ans: 90°

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JAN ’19 [26] m∠𝐴𝐷𝐶 m∠𝐵 118° (opp. ∠’s of ▱ are ≅) m∠𝐷𝐺𝐹 m∠𝐴𝐷𝐶 118° (alt. int. ∠’s of ∥ lines 𝐹𝐺 and 𝐸𝐷𝐶) m∠𝐺𝐹𝐻 m∠𝐴𝐻𝐶 m∠𝐷𝐺𝐹 138° 118° 20° (∠𝐴𝐻𝐶 is an ext. angle of △𝐹𝐺𝐻) AUG ’19 [25] m∠𝐷 46° because the angles of a triangle add to 180°; m∠𝐵 46° because opposite angles of a parallelogram are congruent

4.2 Trapezoids There are no Regents exam questions on this topic. 4.3 Use Quadrilateral Properties in Proofs

AUG '15 [28] (Givens omitted) 𝐷𝐶 ∥ 𝐴𝐵; 𝐷𝐴 ∥ 𝐶𝐵 (opposite sides of a parallelogram are parallel). ∠𝐴𝐶𝐷 ≅ ∠𝐶𝐴𝐵 (alternate interior angles theorem). JUN '16 [33] (Givens omitted) ∠𝐷𝐹𝐸 ≅ ∠𝐵𝐹𝐺 (vertical angles are congruent); 𝐴𝐷 ∥ 𝐶𝐵 (opposite sides of a parallelogram are parallel); ∠𝐸𝐷𝐹 ≅ ∠𝐺𝐵𝐹 (alternate interior angles theorem); △𝐷𝐸𝐹 ∼△ 𝐵𝐺𝐹 (AA~) JAN ’18 [25] (Givens omitted)

AB CD≅ , AD BC≅ (opposite sides of a parallelogram are congruent); AC AC≅ (Reflexive property);

ABC CDA≅△ △ (SSS)

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4.4 Prove Types of Quadrilaterals JUN '15 [13] Ans: 4 AUG '15 [1] Ans: 2 JUN '16 [9] Ans: 1 AUG ’16 [7] Ans: 3 JAN ’17 [5] Ans: 4 JAN ’17 [16] Ans: 1 JUN ’17 [11] Ans: 4 JUN ’17 [20] Ans: 2 AUG ’17 [14] Ans: 3 JAN ’18 [19] Ans: 4 JUN ’18 [13] Ans: 4 JUN ’19 [12] Ans: 3 JUN ’19 [24] Ans: 3 AUG ’19 [13] Ans: 3 JAN ’20 [4] Ans: 1 AUG '15 [35] (Givens omitted) ∠𝐵𝐸𝐶 and ∠𝐷𝐹𝐶 are right angles (def. of perpendicular) ∠𝐵𝐸𝐶 ≅ ∠𝐷𝐹𝐶 (all right angles are congruent). ∠𝐹𝐶𝐷 ≅ ∠𝐵𝐶𝐸 (reflexive property). △𝐵𝐸𝐶 ≅△ 𝐷𝐹𝐶 (ASA). 𝐵𝐶 ≅ 𝐶𝐷 (CPCTC).

ABCD is a rhombus (a parallelogram with consecutive congruent sides is a rhombus).

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JAN '16 [35] (Givens omitted) 𝐴𝑁 ≅ 𝑅𝐷, 𝐴𝑅 ≅ 𝐷𝑁 (Opposite sides of a parallelogram are congruent). 𝐴𝑅 𝐷𝑁 (division prop) 𝑅𝐸 𝐴𝐸 𝐴𝑅, 𝑁𝑊 𝑊𝐷 𝐷𝑁 (def. of bisector) 𝑅𝐸 ≅ 𝑁𝑊 and 𝐴𝐸 ≅ 𝑊𝐷 (substitution). ∠𝑅 ≅ ∠𝑁 (Opposite angles of a parallelogram are congruent). △𝐴𝑁𝑊 ≅△𝐷𝑅𝐸 (SAS). 𝐴𝐸𝑅 ∥ 𝑁𝑊𝐷 (Opposite sides of a parallelogram are parallel). AWDE is a parallelogram (A quadrilateral with a pair of opposite sides that are congruent and parallel is a parallelogram).

JUN '16 [35] (Givens omitted) quadrilateral ABCD is a parallelogram (if the diagonals of a quadrilateral bisect each other, it is a parallelogram); quadrilateral ABCD is a rhombus (if a diagonal of a parallelogram bisects its angle, it is a rhombus); 𝐴𝐷 ≅ 𝐷𝐶 (the sides of a rhombus are congruent); △𝐴𝐶𝐷 is an isosceles triangle (def. of isosceles); 𝐴𝐸 ⊥ 𝐵𝐸 (diagonals of a rhombus are perpendicular); ∠𝐵𝐸𝐴 is a right angle (def. of perpendicular); △𝐴𝐸𝐵 is a right triangle (definition of right triangle).

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JAN ’17 [35] (Givens omitted) ∠𝐴𝐸𝐷 and ∠𝐶𝐹𝐵 are right angles (def. of perpendicular). ∠𝐴𝐸𝐷 ≅ ∠𝐶𝐹𝐵 (All right angles are congruent). ABCD is a parallelogram (A quadrilateral with one pair of sides congruent and parallel is a parallelogram). 𝐴𝐷 ∥ 𝐵𝐶 (Opposite sides of a parallelogram are parallel). ∠𝐷𝐴𝐸 ≅ ∠𝐵𝐶𝐹 (alternate interior angles theorem). 𝐷𝐴 ≅ 𝐵𝐶 (Opposite sides of a parallelogram are congruent). △𝐴𝐷𝐸 ≅△ 𝐶𝐵𝐹 (AAS). 𝐴𝐸 ≅ 𝐶𝐹 (CPCTC).

JUN ’18 [35] (Givens omitted) BC AD (opp sides of a parallelogram are parallel); BE FD (parts of || lines are ||); BF DE (two lines ⊥ to same line are ||); BEDF is a parallelogram (quadrilateral with both pairs of opposite sides || is a parallelogram);

DEB∠ is a right angle (def. of ⊥); BEDF is a rectangle (parallelogram with a right angle is a rectangle)

JAN ’19 [35] (Givens omitted); 𝐻𝐹 ≅ 𝐻𝐹 (Reflexive property); 𝐻𝐹 𝐶𝐹 ≅ 𝐻𝐹 𝐴𝐻, so 𝐴𝐹 ≅ 𝐶𝐻 (Addition property); ABCD is a ▱ (Both pairs of opp. sides are ≅); 𝐴𝐵 ∥ 𝐶𝐷 (Opp. sides of a ▱ are parallel); ∠𝐵𝐴𝐶 ≅ ∠𝐷𝐶𝐴 (alternate interior angles theorem); △𝐸𝐴𝐹 ≅△ 𝐺𝐶𝐻 (SAS); 𝐸𝐹 ≅ 𝐺𝐻 (CPCTC)

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JUN ’19 [35] (Givens omitted); ∠𝐻𝐸𝐴 and ∠𝑇𝐴𝐻 are right angles (definition of ⊥); ∠𝐻𝐸𝐴 ≅ ∠𝑇𝐴𝐻 (all right angles are congruent); MATH is a parallelogram (a quadrilateral with two pairs of congruent opposite sides is a parallelogram); 𝑀𝐴 ∥ 𝑇𝐻 (opposite sides of parallelogram are parallel); ∠𝑇𝐻𝐴 ≅ ∠𝐸𝐴𝐻 (alternate interior angles theorem); △𝐻𝐸𝐴~ △ 𝑇𝐴𝐻 (AA~); (CSSTP); 𝑇𝐴 ∙ 𝐻𝐴 𝐻𝐸 ∙ 𝑇𝐻 (cross products are equal)

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Chapter 5. Perimeter And Area

5.1 Perimeter and Circumference There are no Regents exam questions on this topic. 5.2 Area

JAN ’17 [8] Ans: 1 AUG ’17 [20] Ans: 1 JAN ’19 [18] Ans: 1 JUN ’19 [2] Ans: 3 AUG ’19 [17] Ans: 4 JAN '16 [30] Dish A

240,000 19.6(25.5)Adπ

= ≈ 272,000 16.3(37.5)Bdπ

= ≈ JAN ’19 [31] Area of outer circle – Area of inner circle 𝜋30 𝜋20 500𝜋 Area of each rectangle = 90 10 900 Total area = 500𝜋 2 900 3,371 sq. ft

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Chapter 6. Coordinate Geometry

6.1 Forms of Linear Equations There are no Regents exam questions on this topic. 6.2 Parallel and Perpendicular Lines

JUN '15 [9] Ans: 1 AUG '15 [10] Ans: 1 JAN '16 [2] Ans: 4 JUN '16 [14] Ans: 4 JAN ’17 [1] Ans: 3 JUN ’17 [19] Ans: 2 JAN ’18 [20] Ans: 1 JUN ’18 [12] Ans: 2 AUG ’18 [11] Ans: 1 JUN ’19 [16] Ans: 2 AUG ’19 [8] Ans: 1 JAN ’19 [25] 𝑦 𝑥 , 𝑚 𝑦 6 𝑥 2 JAN ’20 [31] 𝑦 𝑥 , 𝑚 𝑦 12 𝑥 5

6.3 Distance Formula JUN '15 [3] Ans: 3 JAN '16 [15] Ans: 2

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6.4 Midpoint Formula JAN ’19 [30] No. (a) If 𝐸𝐺 is a median, then G is the midpoint of 𝐷𝐹. But midpoint of 𝐷𝐹 is 1 4 1 2, (2.5,0.5)2 2+ − + =

, or (b) If 𝐸𝐺 is a median, then DG = GF. But, 𝐷𝐺 √2 2√8 and 𝐺𝐹 √1 1 √2.

6.5 Perpendicular Bisectors JUN '16 [12] Ans: 1 AUG ’17 [24] Ans: 4

6.6 Directed Line Segments FALL '14 [14] Ans: 4 AUG ’16 [18] Ans: 4 JAN ’17 [20] Ans: 1 JUN ’17 [15] Ans: 2 AUG ’17 [17] Ans: 1 JAN ’18 [6] Ans: 1 JUN ’18 [14] Ans: 2 AUG ’18 [15] Ans: 1 JAN ’19 [15] Ans: 1 JUN ’19 [19] Ans: 4 AUG ’19 [3] Ans: 3 JAN ’20 [5] Ans: 4 JUN '15 [27] 256 (4 6) 2xP = − + + = − ; 255 (0 5) 3yP = − + + = − 𝑃 – 2, – 3

29

AUG '15 [31] 251 (16 1) 7xE = + − = ; 254 (14 4) 8yE = + − = 𝐸 7,8 JAN '16 [27] 232 (4 2) 2xJ = − + + = ; 231 (7 1) 5yJ = + − = 𝐽 2,5 JUN '16 [26] 494 (22 4) 12xP = + − = ; 492 (2 2) 2yP = + − = 𝑃 12,2

30

Chapter 7. Polygons In The Coordinate Plane

7.1 Triangles in the Coordinate Plane JAN '16 [18] Ans: 1 AUG ’16 [15] Ans: 3 AUG '15 [33] 32BCm = − 23m⊥ = The right angle may be at B or C. Right angle at B means: 231 ( 3)1 b

b− = − +

= 233 13 x

x= +=

Right angle at C means: 23 1034 ( 1) b

b

− = − +

= − 1023 3192

3 9.5x

x

= −

= = Either answer, 3x = or 9.5x = , is correct.

31

AUG ’17 [32]

A(–2,3) and B(1,5) 5 3 21 ( 2) 3ABm −= =

− − 3 ( 1) 4 23 ( 3) 6 3PRm − −= = =

− −

AB PR because same slopes. 2 22 23 2 136 4 52 2 13AB

PR

= + =

= + = =

Therefore, 12AB PR= . JUN ’18 [32] 2 22 22 2

5 1 264 4 321 5 26AB

BC

AC

= + =

= + =

= + =

ABC△ has two congruent sides, not three, so it is isosceles and not equilateral.

32

JAN ’19 [32] 𝐴𝐶 3 5 √34 𝐵𝐶 5 3 √34 Isosceles because AC = BC 𝑚 𝑚 𝐴𝐶 ⊥ 𝐵𝐶 because their slopes are opposite reciprocals, so ∠𝐶 is a right angle 7.2 Quadrilaterals in the Coordinate Plane

AUG '15 [22] Ans: 4 AUG ’16 [14] Ans: 1 JAN ’17 [19] Ans: 3 AUG ’19 [2] Ans: 3 FALL '14 [11] Midpoint of MT is ( )520 4 1 6, 2,2 2+ − + =

.

746 14 0MTm += =−

. In a rhombus, diagonals are perpendicular, so 47AHm = − . 5 42 7 ( 2)y x− = − − The diagonals, MT and AH , of rhombus MATH are perpendicular bisectors of each other.

33

JUN '15 [36] 53TSm = − 35SRm = Since the slopes of TS and SR are opposite reciprocals, they are perpendicular and form a right angle. RST△ is a right triangle because S∠ is a right angle. P(0,9) 53RPm = − 35PTm = Since the slopes of all four adjacent sides are opposite reciprocals, they are perpendicular and form right angles. Quadrilateral RSTP is a rectangle because it has four right angles.

JAN ’17 [31] (1,3) and (–3,–3) JUN ’17 [35] 2 22 22 22 2

5 5 50( 7) 1 50( 5) ( 5) 50( 7) 1 50PQ

QR

RS

PS

= + =

= − + =

= − + − =

= − + =

PQRS is a rhombus because all sides are congruent. The slope of PQ is 1 and the slope of QR is –7. Because the slopes of adjacent sides are not opposite reciprocals, they are not perpendicular and do not form a right angle. Therefore PQRS is not a square.

34

JAN ’18 [35] 2 22 2( 4 1) (5 6) 130(5 4) ( 2 5) 130PA

AT

= − + + + =

= + + − − =

PA AT= , so PAT△ is isosceles. R(2,9). 113PAm = − and 113RTm = − , so PA RT . 2 2(5 2) ( 2 9) 130RT = − + − − = , so PA RT≅ . PART is a parallelogram because it has a pair of opposite sides that are both congruent and parallel.

AUG ’18 [35] 6 310 5MHm = = and 6 310 5ATm = = 53MAm = − and 53HTm = − 𝑀𝐻 ∥ 𝐴𝑇 and 𝑀𝐴 ∥ 𝐻𝑇, so MATH is a parallelogram since both sides of opposite sides are parallel. Since their slopes are negative reciprocals, 𝑀𝐴 ⊥ 𝐴𝑇 and ∠𝐴 is a right angle. MATH is a rectangle because it is a parallelogram with a right angle. JUN ’19 [32]

( )0 6 31 1ADm −= = −− −

and 1 8 36 3BCm − −= = −−

𝐴𝐷 ∥ 𝐵𝐶 because their slopes are equal, so ABCD is a trapezoid. 𝐴𝐶 1 6 6 1 √98 and 𝐵𝐷 8 0 3 1 √68. 𝐴𝐵𝐶𝐷 is not an isosceles trapezoid because its diagonals are not congruent.

35

AUG ’19 [35] 𝐴𝐵 5 1 3 2 √37 and 𝐵𝐶 5 6 3 3 √37; △𝐴𝐵𝐶 is isosceles because 𝐴𝐵 𝐵𝐶. 𝐷 0, 4 𝐴𝐷 1 0 2 4 √37 and 𝐶𝐷 6 0 3 4 √37; 𝑚 and 𝑚 6; 𝐴𝐵 ⊥ 𝐵𝐶 (slopes are opposite reciprocals), so ∠𝐵 is a right angle; ABCD is a square because all four sides are congruent and it has a right angle. JAN ’20 [32] 𝑁𝐴 1 4 2 3 √50, 𝐴𝑇 8 1 1 2 √50, 𝑇𝑆 3 8 4 1 √50, and 𝑆𝑁 4 3 3 4 √50. All four sides are congruent, so NATS is a rhombus.

7.3 Perimeter and Area using Coordinates JUN '16 [22] Ans: 3 JUN ’17 [2] Ans: 3 AUG ’17 [3] Ans: 3 JUN ’18 [15] Ans: 1 AUG ’18 [8] Ans: 4 JAN ’19 [21] Ans: 4 JAN ’20 [18] Ans: 2

36

JUN ’19 [26]

𝐴 5 10 25 AUG ’19 [28]

𝐴 5 12 30

37

Chapter 8. Rigid Motions

8.1 Translations There are no Regents exam questions on this topic. 8.2 Line Reflections

JAN '16 [25]

8.3 Rotations

FALL '14 [2] Ans: 4 JAN '16 [11] Ans: 4 AUG ’16 [5] Ans: 1 JAN ’20 [15] Ans: 4 AUG ’16 [29] m m 47P L∠ = ∠ = ° because rotations preserve angles. m 180 (47 57) 76M∠ = − + = because the angles of a triangle add to 180°.

8.4 Point Reflections There are no Regents exam questions on this topic.

38

8.5 Map a Polygon onto Itself FALL '14 [15] Ans: 2 JUN '15 [10] Ans: 1 AUG '15 [5] Ans: 1 JAN ’17 [17] Ans: 4 JUN ’17 [7] Ans: 1 AUG ’17 [6] Ans: 3 AUG ’17 [22] Ans: 4 JAN ’18 [15] Ans: 3 JUN ’18 [19] Ans: 3 AUG ’18 [17] Ans: 3 JAN ’19 [4] Ans: 3 JUN ’19 [4] Ans: 4 AUG ’19 [23] Ans: 4 JAN ’20 [11] Ans: 3 AUG ’16 [27] 360 606 =

39

Chapter 9. Dilations

9.1 Dilations of Line Segments JUN '15 [18] Ans: 1 JAN '16 [10] Ans: 2 JUN '16 [2] Ans: 4 AUG ’16 [21] Ans: 4 JAN ’17 [13] Ans: 1 JUN ’17 [6] Ans: 3 AUG ’17 [10] Ans: 1 JAN ’19 [1] Ans: 4 JUN ’19 [5] Ans: 1 AUG ’19 [1] Ans: 2 AUG ’17 [29] 12 2 21212

(5 2) ( 1 3)25 2.5A B AB′ ′ =

= − + − −

= = 9.2 Dilations of Polygons

JUN '15 [16] Ans: 2 AUG '15 [6] Ans: 4 AUG '15 [20] Ans: 1 AUG '15 [23] Ans: 1 JAN ’18 [11] Ans: 1 JUN ’18 [5] Ans: 4 AUG ’18 [23] Ans: 3

40

JAN ’17 [32]

A dilation preserves slope, so the slopes of QR and Q R′ ′ are equal. Therefore, Q R QR′ ′ . JUN ’18 [26]

9.3 Dilations of Lines

FALL '14 [3] Ans: 2 FALL '14 [16] Ans: 2 JUN '15 [22] Ans: 1 AUG '15 [24] Ans: 4 JAN ’18 [14] Ans: 1 JUN ’18 [24] Ans: 2

41

JAN ’19 [24] Ans: 4 JUN ’19 [7] Ans: 2 AUG ’19 [10] Ans: 1 JAN ’20 [8] Ans: 1 JAN '16 [31] : 3 4y x= −ℓ ; : 3 8m y x= − JUN ’17 [31] The center of dilation is on the original line, so the line does not change. Line p is 3 4 20x y+ = . AUG ’18 [30] No, the line passes through the center of dilation, so the dilated line is not distinct. To show the given lines are distinct, 43

4 3 243 4 248x yy xy x

+ == − += − +

This has a different y-intercept than 43 16y x= − + .

42

Chapter 10. Transformation Proofs

10.1 Properties of Transformations AUG '15 [13] Ans: 2 JAN ’16 [5] Ans: 3 JUN '16 [4] Ans: 1 AUG ’16 [2] Ans: 2 JAN ’18 [4] Ans: 4 JAN ’18 [8] Ans: 4 JUN ’18 [1] Ans: 1 JAN ’19 [2] Ans: 4

10.2 Sequences of Transformations JUN '15 [4] Ans: 4 AUG '15 [7] Ans: 1 JAN '16 [8] Ans: 1 JUN '16 [8] Ans: 4 JAN ’17 [10] Ans: 3 JUN ’17 [1] Ans: 2 JUN ’18 [3] Ans: 4 AUG ’18 [4] Ans: 1 JAN ’19 [3] Ans: 3 JUN ’19 [1] Ans: 4 AUG ’19 [9] Ans: 2 JAN ’20 [17] Ans: 2 JAN ’20 [22] Ans: 1

43

JUN '16 [25] translation 6 units right and reflection over x-axis AUG ’16 [26]

JAN ’17 [26] 0, 2T − and y axisr − JUN ’17 [30] Rotate ABC△ about point C until DF AC . Translate

ABC△ along CF so that C maps onto F. AUG ’17 [27] 180R ° about ( )1 12 2,− JUN ’18 [27] Reflection over the y-axis followed by a translation up 5. AUG ’18 [28] rotation 180º about (0,–1); or rotation 180° about the origin, translation 2 units down; or reflection over x-axis, translation 2 units down, reflection over y-axis JAN ’19 [28] reflection over line: 𝑟 ; or rotation 𝑅 ° around 5, 1 followed by reflection 𝑟 ; or rotation 𝑅 ° around 5,2 followed by 𝑇 , and 𝑟 JUN ’19 [29] 𝑅 ° around origin; or 𝑇 , followed by 𝑅 , .

44

AUG ’19 [27] reflection 𝑟 and reflection 𝑟 10.3 Transformations and Congruence

JUN '15 [24] Ans: 3 JUN '16 [16] Ans: 3 JAN ’17 [6] Ans: 4 AUG ’17 [2] Ans: 4 JAN ’18 [1] Ans: 1 JUN ’19 [14] Ans: 4 FALL '14 [4] Translate ABC△ such that point C maps onto point F, then reflect over DF . or Reflect ABC△ over the perpendicular bisector of EB . FALL '14 [21] a) 𝐿𝐴 ≅ 𝐷𝑁, 𝐶𝐴 ≅ 𝐶𝑁, and 𝐷𝐴𝐶 ⊥ 𝐿𝐶𝑁 (Given).

LCA∠ and DCN∠ are right angles (Definition of perpendicular). LAC△ and DNC△ are right triangles (Definition of a right triangle). LAC DNC≅△ △ (HL). b) Rotate LAC△ counterclockwise 90º about point C such that point L maps onto point D.

JUN '15 [30] Reflections are rigid motions that preserve congruency.

45

JUN '15 [33] Quadrilateral ABCD is a parallelogram with diagonals 𝐴𝐶 and 𝐵𝐷 intersecting at E (Given). AD BC≅ (Opposite sides of a parallelogram are congruent).

AED CEB∠ ≅ ∠ (Vertical angles are congruent). BC DA (Definition of parallelogram).

DBC BDA∠ ≅ ∠ (Alternate interior angles theorem). AED CEB≅△ △ (AAS). 180° rotation of AED△ around point E.

AUG '15 [30] The transformation is a rotation of 180° around the origin. Rotations are rigid motions that preserve congruency. AUG '15 [34] Translations preserve distance, so if point D is mapped onto point A, point F would map onto point C.

DEF ABC≅△ △ since DEF△ can be mapped onto ABC△ by a sequence of rigid motions.

JAN '16 [28] Yes. The sequence of transformations consists of a reflection y axisr − and a translation 0, 3T − , which are rigid motions which preserve congruency.

46

AUG ’16 [33]

( , ) rotationpoint ( , ) rotationpoint(2, 9) (2, 3) (0, 6) (6,0) (2, 3) (8, 3)(6, 8) (2, 3) (4, 5) (5, 4) (2, 3) (7,1)x y y xC CB B

− → − +′− − − = − → + − = −

′− − − = − → − + − =

DEF A B C′ ′ ′≅△ △ because DEF△ is a reflection of A B C′ ′ ′△ over 1x = − and reflections preserve congruency.

JUN ’17 [32]

Reflection 1xr =− . The triangles are congruent because reflections are rigid motions that preserve distance. AUG ’17 [30] Yes, a sequence of rigid motions preserves distance and angle measure, so ABC XYZ≅△ △ by ASA. BC YZ≅ by CPCTC.

47

JAN ’18 [30] 2 23 8 73AB = + = and 2 23 7 58RS = + = , so AB is not congruent to RS . Therefore, ABC△ is not congruent to RST△ . Since they are not congruent, there is no sequence of rigid motions that would map ABC△ onto RST△ .

JUN ’18 [25] Yes, translations are rigid motions that preserve distance and angles. JAN ’20 [25] Various answers, such as: ∠𝑄 ≅ ∠𝑀, ∠𝑃 ≅ ∠𝑁, and 𝑄𝑃 ≅ 𝑀𝑁.

10.4 Transformations and Similarity JUN '15 [2] Ans: 4 AUG '15 [2] Ans: 3 AUG '15 [14] Ans: 4 AUG ’16 [9] Ans: 4 JAN ’17 [2] Ans: 2 JUN ’17 [14] Ans: 1 AUG ’18 [2] Ans: 1 FALL '14 [17] Circle A can be mapped onto circle B by first translating circle A such that A maps onto B, and then dilating circle

A, centered at A, by a scale factor of 53 . Since there exists a sequence of transformations that maps circle A onto circle B, circle A is similar to circle B.

48

FALL '14 [19] Let X Y Z′ ′ ′△ be the image of XYZ△ after a rotation about point Z such that ZX ′ coincides with ZU . Since rotations preserve angle measure, ZY ′ coincides with ZV . Then, dilate X Y Z′ ′ ′△ by a scale factor of ZU

ZX ′ with its center at point Z. Since dilations preserve angles,

X Y′ ′ maps onto UV . Therefore, XYZ UVZ△ △ . JUN '16 [34] A dilation of 52 about the origin. Dilations preserve similarity. JAN ’18 [32] Dilation by a scale factor of 3 with its center at point A. Dilations preserves similarity. JUN ’19 [25] No, dilations do not preserve distance, and therefore do not preserve congruence.

49

Chapter 11. Circles

11.1 Circumference and Rotation JAN '16 [23] Ans: 1

11.2 Arcs and Chords JUN '15 [8] Ans: 1 AUG '15 [15] Ans: 3 JUN '16 [10] Ans: 2 AUG ’16 [23] Ans: 1 JUN ’17 [4] Ans: 4 JUN ’17 [8] Ans: 2 AUG ’17 [4] Ans: 1 JAN ’18 [16] Ans: 4 JAN ’18 [21] Ans: 4 JUN ’18 [17] Ans: 3 JAN ’19 [5] Ans: 4 JUN ’19 [13] Ans: 3 AUG ’19 [22] Ans: 4

50

JAN '16 [26] Ans: 120° B C∠ ≅ ∠ (alternate interior angles) A B∠ ≅ ∠ (base angles of isosceles triangle formed by two radii)

AUG ’18 [27] Ans: 118°

134 102 1182+ = 11.3 Tangents

JUN '15 [20] Ans: 1 AUG '15 [12] Ans: 3 JAN '16 [21] Ans: 3 AUG ’18 [14] Ans: 2 AUG ’16 [25] 38 56 21⋅ =

51

11.4 Secants JAN ’17 [15] Ans: 2 AUG ’17 [12] Ans: 2 JUN ’19 [18] Ans: 1 JAN ’17 [28] 152 56 482− = ° JUN ’18 [28] 10 6 154 x

x⋅ =

= JAN ’19 [27] m∠𝑅𝑃𝑆 35 𝑥 70 121 𝑥 51° AUG ’19 [30] 34 JAN ’20 [28] 𝐷𝐴 𝐴𝐵 𝐴𝐶 ; 𝐷𝐴 8 12.5 ; 𝐷𝐴 100; 𝐷𝐴 10

52

11.5 Circle Proofs FALL '14 [26] (Givens omitted) Chords BC and BD are drawn (auxiliary lines).

A A∠ ≅ ∠ , BC BC≅ (Reflexive property). 12m mBDC BC∠ = (an inscribed angle is half the measure of its intercepted arc). 12m mCBA BC∠ = (an angle formed by a tangent and chord measures half the intercepted arc). m mBDC CBA∠ = ∠ (substitution).

BDC CBA∠ ≅ ∠ (definition of congruence). ABC ADB△ △ (AA~).

AB ADAC AB

= (Side Proportionality). 2AC AD AB⋅ = (cross products are equal). AUG ’16 [35] (Givens omitted) Chords CB and AD are drawn (auxiliary lines);

CEB AED∠ ≅ ∠ (vertical angles); C A∠ ≅ ∠ (inscribed angles that intercept the same arc are congruent); BCE DAE△ △ (AA~);

AE EDCE EB

= (Side Proportionality); (cross products are equal). AE EB CE ED⋅ = ⋅

53

AUG ’17 [33] (Givens omitted) B∠ is a right angle (angle inscribed in semi-circle is a right angle).

EC OC⊥ (radius drawn to a point of tangency is ⊥ to the tangent).

ECA∠ is a right angle (def. of perpendicular). B ECA∠ ≅ ∠ (all right angles are congruent). BCA CAE∠ ≅ ∠ (the transversal of parallel lines creates congruent alternate interior angles). ABC ECA△ △ (AA~).

BC ABCA EC

= (CSSTP). 11.6 Arc Lengths and Sectors

JUN '15 [23] Ans: 2 AUG '15 [18] Ans: 3 JAN '16 [12] Ans: 3 JUN '16 [24] Ans: 3 AUG ’16 [19] Ans: 2 JAN ’17 [21] Ans: 4 JAN ’18 [24] Ans: 3 JUN ’18 [22] Ans: 4 AUG ’18 [18] Ans: 2 JAN ’19 [14] Ans: 2 AUG ’19 [12] Ans: 4 JAN ’20 [13] Ans: 3

54

FALL '14 [23] 180 20m 802BOD −∠ = = ° 2360 Sr

θπ

=°

, 80360 36Sπ= , 8S π=

JUN '15 [29] 2360 S

rθ

π=

°, 12360 36θ π

π= , 120θ = °

JUN ’17 [26] 2360 S

rθ

π=

°, 40360 20.25S

π= , 2.25S π=

JAN ’18 [28] 625 500 125S π π π= − = 2360 S

rθ

π=

°, 125360 625θ π

π= , 15 360 72θ = ⋅ = °

JUN ’19 [28] 2360 S

rθ

π=

° , 72360 100S π= , 𝑆 20𝜋

11.7 Radians (–) FALL '14 [4] Ans: 3 AUG ’17 [23] Ans: 2 JUN '16 [29] 44AA

ππ

= ⋅

= 13 6.58 4B

B

π

π

= ⋅

=

Yes, both angles are equal.

55

11.8 Circles in the Coordinate Plane JUN '15 [14] Ans: 2 AUG '15 [9] Ans: 3 JAN '16 [17] Ans: 4 JUN '16 [3] Ans: 2 JUN '16 [23] Ans: 1 AUG ’16 [16] Ans: 1 JAN ’17 [18] Ans: 1 JAN ’17 [22] Ans: 3 JUN ’17 [12] Ans: 1 JAN ’18 [12] Ans: 2 JUN ’18 [20] Ans: 2 AUG ’18 [21] Ans: 4 JAN ’19 [20] Ans: 1 JUN ’19 [20] Ans: 4 AUG ’19 [6] Ans: 4 JAN ’20 [20] Ans: 2 AUG ’16 [30] Yes. 2 2 22 2 2( 1) ( 2) 4(3.4 1) (1.2 2) 45.76 10.24 16x y− + + =

− + + =+ =

AUG ’17 [31] 2 22 22 2

6 8 566 9 8 16 56 9 16( 3) ( 4) 81x x y y

x x y yx y

− + + =− + + + + = + +

− + + = (3,–4); r = 9

56

Chapter 12. Solids

12.1 Volume of a Sphere JAN '16 [14] Ans: 3 JUN ’19 [10] Ans: 1 JUN ’17 [28]

343V rπ= , so 3 34Vrπ

= . 3 33(294) 3(180) 0.64 4π π

− ≈ JUN ’18 [31] 2 29.529.52r

r

π

π

=

= 334 43 3 29.5 4342V rπ π

π = = ≈

in3 12.2 Volume of a Prism or Cylinder

JAN '16 [4] Ans: 2 AUG ’16 [20] Ans: 4 JAN ’17 [11] Ans: 2 JUN ’17 [23] Ans: 3 JUN ’18 [7] Ans: 1 JUN '16 [32] 2(11.25) (33.5) 57.7231π ≈

57

JUN ’17 [34] 3 31 ft20,000 2673.8 ft7.48V gg

= ⋅ ≈ 𝑉 𝜋𝑟 ℎ 22673.8 (34.5)4.967rr

π=≈

2(4.967) 110.9pp

≈ +≈

AUG ’17 [36] tan16.5 413.5x x= ≈ 4.5 4 8.5 ft. 12 13 24 31 2 3 4

9 16 4.5 64813.5 16 4.5 97213.5 16 4 43212.5 16 8.5 17003752 ftVVVVV V V V

= × × == × × == × × × == × × =+ + + =

3752 (35 16 0.5) 34723472 7.48 25,97125,971 2473.410.52473.4 41hrs60

− × × =× ≈

≈

≈

JAN ’18 [33] 31 42 3 323 4 134.04hemisphereV rπ

π

= ⋅

= ⋅ ≈

2 24 (13 4) 452.39cylinderV r hπ

π

=

= ⋅ ⋅ − ≈ 3134.04 452.39 586mV ≈ + ≈

58

AUG ’18 [31] 2 $3.25 $19.50 JAN ’19 [33] 𝑉 30 ⋅ 15 ⋅ 3.5 1,575 cu. ft. 1575 ⋅ 7.48 11,781 gallons $3.95 ⋅ 117.81 $465.35 𝑉 𝜋 ⋅ 12 ⋅ 3.5 504𝜋 cu. ft. 504𝜋 ⋅ 7.48 11,843.553 gallons $200 per 6000 = $1 per 30 gallons 11,843.553 30 $394.79 Theresa paid more. JUN ’19 [33] 𝑟 . 3.25; 𝑉 𝜋 3.25 1 22; 22 7.48 165 AUG ’19 [31] 𝑟 . 4.125; 𝑉 𝜋 4.125 2.5 134 AUG ’19 [34] Altitude of triangular base = √4 3 √7; Area of triangular base = 6 √7 3√7; Area of rectangular base = 10 6 60; Volume = 60 3√7 6.5 442 JAN ’20 [34] 𝑉 𝜋 7 18 2770.8847; 𝑉 16𝑥 ; 16𝑥 2770.8847; 𝑥 173.1803; 𝑥 13.2; . 6.1 and . 4.5, so 6 4 24 containers fit

12.3 Volume of a Pyramid or Cone AUG '15 [21] Ans: 4 JAN '16 [7] Ans: 2 JUN '16 [6] Ans: 4

59

JAN ’17 [24] Ans: 1 JUN ’17 [16] Ans: 1 JAN ’18 [7] Ans: 3 JAN ’18 [22] Ans: 2 JUN ’18 [10] Ans: 1 AUG ’18 [19] Ans: 2 JAN ’19 [9] Ans: 2 JAN ’19 [23] Ans: 1 JUN ’19 [6] Ans: 2 AUG ’19 [21] Ans: 3 JAN ’20 [2] Ans: 2 JAN ’20 [10] Ans: 1 JUN '16 [36] Similar triangles are required to model and solve a proportion. Let x = height of the rest of the cone

height of cone height below glassradius of top radius of bottom

= 51.5 1x x+ = x = 10 h = 10 + 5 = 15

glass cone below glassV V V= − 2 21 13 3(1.5) (15) (1) (10)24.9V π π= −≈

JAN ’17 [34]

21 13 3231.416 231.416 52 (25)(13) 340

C rr

r

V r h

ππ

ππ π

==

= ≈

= = ≈

60

12.4 Density JUN '15 [7] Ans: 3 AUG '15 [16] Ans: 1 JAN '16 [19] Ans: 2 JUN '16 [18] Ans: 2 JUN '16 [20] Ans: 1 AUG ’16 [17] Ans: 2 AUG ’19 [14] Ans: 2 JAN ’20 [14] Ans: 1 FALL '14 [6] 5.1 10.2 20.3 1,056.006× × = cm3 500 1,056.006 528,003× = cm3 33 331m528,003cm 0.528003m100cm× =

331920kg 0.528003m 1013kgm × ≈ No, the weight of the bricks is greater than 900 kg FALL '14 [25] 1m25cm 0.25m100cmr = × = 2(0.25) (10) 0.625V π π= = m3 3 3380kg0.625 m 746.1kgmW π= × ≈ $4.75 746.1 $3,544× ≈ per tree $50,000 14.1$3,544 ≈ Need to sell 15 trees

61

JUN '15 [35] tan 47 8.5x° = 9.115x ≈ 213 (8.5) (9.115) 689.6coneV π= ≈ 2(8.5) (25) 5674.5cylinderV π= ≈ ( )31 42 3 (8.5) 1286.3hemisphereV π= ≈ 3689.6 5674.5 1286.3 7650fttowerV ≈ + + ≈ 3 362.4lb85% 7650ft 405,756lbft× × = No, the weight exceeds 400,000 lbs

AUG '15 [25] 33137.8 0.638g/cm6 ≈ ; Ash AUG '15 [36]

( )2 3313 2 (8) 18.85inconeV π= ≈ Total 3100 18.85 1885in≈ × ≈ (0.52 1885) $0.10 $98.02× × = (100 $1.95) ($98.02 $37.83) $59.15× − + = AUG ’16 [36]

( ) ( )2 38.3 8.31 1 43 2 2 3 2 3(10.2)183.961 149.693 333.65 cmV π π= + ⋅

≈ + ≈

3333.65 50 16,682.7cm× = 16,682.7 0.697 11,627.8g× = 11,627.8 11.6278g kg= 11.6278 3.83 $44.53× =

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JAN ’17 [36] 2 2(26.7) (750) (24.2) (750) 95,437.5cylinderV π π π= − = 3 32.7 1 $0.3895,437.5 $307.621000 11 g kgcmg kgcm

× × × =2 2(40 )(750) (35 )(750) 281,250prismV = − =3 32.7 1 $0.38281,250 $288.561000 11 g kgcmg kgcm

× × × = Prisms cost less. 307.62 288.56 $19.06− = JAN ’18 [29]

WDV

= , so 7.95 1015W= (7.95)(1015) 8,069.25W = = g 500 8069.25 4,034,625 g 4034.625 kg× = =4034.625 0.29 $1,170× ≈ JUN ’18 [34] 2 2(10 )(18) 1800V r hπ π π= = = 33 3 31ft1800 in 1.041712 inπ π⋅ ≈ 1.0417 (95.46)(0.85) 266π ≈ lbs 266 + 270 = 536 lbs AUG ’18 [34] Diameter of hollow is 4 2 0.5 3. Radius of outer sphere is 2 and radius of inner sphere is 1.5. 𝑉 43𝜋 2 1.5 19.4 19.4 1.308 8 203 JAN ’20 [27] 𝑉 8 3 2 43 2 86 lbs

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12.5 Lateral Area and Surface Area JUN '15 [19] Ans: 2

12.6 Rotations of Two-Dimensional Objects JUN '15 [1] Ans: 4 AUG '15 [3] Ans: 4 JUN '16 [1] Ans: 3 AUG ’16 [3] Ans: 1 JUN ’17 [18] Ans: 1 AUG ’17 [13] Ans: 3 JAN ’18 [10] Ans: 4 JUN ’18 [16] Ans: 3 AUG ’18 [3] Ans: 4 JAN ’19 [11] Ans: 3 JUN ’19 [3] Ans: 2 AUG ’19 [11] Ans: 4

12.7 Cross Sections JUN '15 [6] Ans: 2 JAN '16 [1] Ans: 1 AUG ’16 [13] Ans: 3 JAN ’17 [23] Ans: 4 AUG ’17 [1] Ans: 2 JAN ’18 [5] Ans: 2 AUG ’18 [5] Ans: 3 JAN ’20 [19] Ans: 4

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12.8 Cavalieri’s Principle (–) FALL '14 [18] Each quarter in both stacks has the same base area. Therefore, each corresponding cross-section of the stacks will have the same area. Since the two stacks of quarters have the same height of 23 quarters, the two volumes must be the same. JUN ’17 [27] Each triangular prism has the same base area. Therefore, each corresponding cross-section of the prisms will have the same area. Since the two prisms have the same height of 14, the two volumes must be the same. AUG ’17 [25] Yes. The bases of the cylinders have the same area and the cylinders have the same height.

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Chapter 13. Constructions

13.1 Copy Segments, Angles, and Triangles JAN '16 [34] solutions vary, such as

SAS AUG ’16 [32]

A C′ ′ is twice AC

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JAN ’19 [29] copy length of 𝐴𝐵 from C and length of 𝐵𝐶 from A for point D at intersecting arcs

AUG ’19 [32]

Yes, because a dilation preserves angle measure. 13.2 Construct an Equilateral Triangle There are no Regents exam questions on this topic. 13.3 Construct an Angle Bisector

AUG ’19 [29] △𝐴𝐵𝐶 is anequilateral triangle, so m∠𝐶𝐴𝐵 60°; 𝐴�⃗� is an angle bisector, so m∠𝐶𝐴𝐷 30°.

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13.4 Construct a Perpendicular Bisector AUG ’16 [28]

JAN ’17 [25]

JUN ’18 [29]

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AUG ’18 [25]

JAN ’20 [29]

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13.5 Construct Lines Through a Point FALL '14 [9] The altitude is perpendicular to the extended side.

JUN '16 [31] The tangent is OA⊥ .

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JUN ’17 [25]

13.6 Construct Inscribed Regular Polygons

FALL '14 [12]

The sides of the square are four congruent chords in the circle, so they intercept four congruent arcs. Each arc therefore measures one-fourth of 360°, or 90°. Therefore, an arc intercepted by two adjacent sides measures 2 90 180× ° = ° .

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JUN '15 [25]

AUG '15 [26]

JAN ’17 [33]

COF is a diameter, so FBC∠ is an inscribed angle of a semicircle, and is therefore a right angle. This means

FBC△ is a right triangle.

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AUG ’17 [28]

JAN ’18 [26]

JUN ’19 [31]

13.7 Construct Points of Concurrency There are no Regents exam questions on this topic. 13.8 Construct Circles of Triangles There are no Regents exam questions on this topic.