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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°. 3 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. 5 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. 9 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. 10 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