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1. 2. 3. 4. 5. 6. 7. A triangle with three acute angles and no congruent sides is acute scalene triangle. A triangle with side lengths of 5 cm, 3 cm, and 5 cm is isosceles triangle. The triangle shows 3 different tick marks which means 3 different measures and it shows a 90° angle so it is D-right scalene. Two 35° angles are shown so the third angle has to be 110°. Therefore the answer is C-obtuse isosceles. Because the directions say the triangle is isosceles, the two legs are congruent. Therefore 3x+2=5x-14. So solving x = 8. Plugging in 8 for x the legs measure 26 each. So to find the base 26 + 26 + x = 82 and the third side (base) is 30. <BCD is exterior so it equals the sum of the two remote interior angles. So 3x+10 =40+2x+20. X calculates to be 50. Plugging in m<BCD is 160, choice D.e Looking at the triangle I notice it is isosceles. So base angles are equal. So 40 + x + x = 180. X equals B-70°. 8. This is a bowtie so the vertical angles are congruent. In the triangle to the left that has 90° and 60° labeled, the third angle has to be 30°. That means that we have 30° plus 65° given in the triangle on the right. When I subtract 95 from 180 the measure of <1 is E-85°. 9. If ABC congruent to XYZ, order matters on stating congruence. So the answer is C – CBA cannot be congruent to ZXY. 10. Since EFG congruent to HIJ, then m<E=m<H, m<F=m<I so m<F=65 and now we can find m<E =80. And we can use substitution to find m<H is D-80°. 11. Since EFG congruent to HIJ, then EG = HJ. 12. Since X congruent to N, m<N is 20°. So 20+75+5x-10=180. Solving we find x=19. 13. Since m<Z = m<0, m<Z is 5x – 10. Substituting 10 for x, m<Z is 85°. 14. Since K congruent to C, M<K is 3x – 5. So 3x-5+40+2x = 180. X calculates to be 29°. 15. Using Choice B we could get SSS, Using choice C we could get SAS. So answer is E- B or D. 16. Using Choice C we would have SAS. 17. For the triangles to be congruent we could use choice A or Choice C. That means our answer is Choice D – A and C. 18. Knowing that side AB is congruent by reflexive property, we would have C- ASA. 19. Knowing that the vertical angles are congruent, we would have D-AAS. 20. We have one marked pair of congruent angles and we can mark the congruent side BD because of reflexive. Therefore we need another angle to use AAS. So answer is B. 21. We have congruent vertical angles and one pair of congruent sides. So we need another angle to use ASA. 22. Since M is midpoint of KQ, then we know KM and MQ are congruent. Adding the congruent vertical angles, we would have SSA which is not valid. Looking closer I see parallel lines KL and QP. That means that <K and <Q are alternate interior angles as well as <P and <L. This gives me enough information to use ASA, Choice C. 23. Looking at the two congruent angles lets me know that their opposite sides are congruent. So 3x+1 = x+11 and x = 5, Choice B. 24. This is a 45-45-90 triangle, making it isosceles. So the legs are congruent! Therefore 3x+4=7x-8. X = 3. Plugging in 3 for x means my legs are each 13, choice D. 25. The markings on this triangle indicate it is isosceles, so the base angles are congruent. That means y = 35. Then 35 + 35 + x = 180. X = 110. Choice E. 26. When I look at the triangle I see it is equilateral which automatically means it is equiangular. That means x + x + x = 180. X = 60, Choice B. 27. This problem takes a few steps to complete. First of all look at the reflexive side that is shared. It is congruent to itself. Then the unmarked leg on the lower triangle is the same as the left side of the triangle because the base angles are marked as congruent. That gives us Choice A-SSS. 28. The triangle on the left shows all three sides congruent. That means each angle has a measure of 60°. X is the angle that is supplementary to one of those 60° angles so x has to be 120. Looking at the right side angle, it is isosceles. That means since one base angle is y, so is the other. So 120+y+y=180. Y = 30. Choice D 29. Looking at the lower triangle, I see it is isosceles. So both base angles are 35°. 35 + 35 + y = 180. This means y = 110. X is supplementary to 110 so x = 70. Choice C. 30. Looking at the entire big triangle (ignore internal segment) you wee that z+35 + 90 = 180. Z = 55. And w + 35 = 180. This means w = 145. Choice D. 31. Since we are told that YW is perpendicular bisector then we know that YB has same measure as YA. That means 7x-4=5x+6. X-5. Choice E. 32. Only III has enough information to say that AP and PB have the same measure. So choice C. 33. PQ is marked congruent to PO so PQ is 10, choice B. 34. Because M is on the perpendicular bisector the measure of LM is the same as the measure of MN which is 8. 35. From choices shown, the only one that is correct is Choice B. 36. This distance from P to B is the same as the distance from P to C so 15 = 3x+6. That means x is 3, choice B. 37. This problem uses Pythagorean theorem and answer is 3 Choice C. 38. With QK = 20, we can use Pythagorean theorem to find NQ = 15. That is the same for NP. Choice E. 39. Remember that I told you with centroids the distance from the median to the centroid is half the distance from the centroid to the vertex. This means that since CW is 4, then CX is 8, choice D. 40. VZ = 4 + 8 so Choice E – 12. 41. UY goes from vertex to opposite side at a right angle, making it an altitude, a perpendicular bisector, a median and an angle bisector. So Choice E. 42. UV is midsegment so XY is 10. W is midpoint of XY so WX is half of 10 or 5, choice A. 43. WU is half of UZ so it is half of 8 – which is 4. 44. KL is twice BP so it is 50. KM is 30. Using Pythagoren Theorem LM is 40. So 50 + 30 + 40 = 120, choice B. 45. FG is half of KM. so (5x+7) +5x+7) = 12x + 8 and x comes out to be 3. Plugging in 5(3)+7 gives us 22. 46. Since H is smallest angle, I is middle sized angle, and J is largest angle, then the opposite sides are proportionally sized. So JI < HJ<HI, choice C. 47. Since the third angles measures 70, the smallest angle is B and the shortest side is AC, choice A. 48. Since <G is 80, the largest side down to smallest are EF, GF, and EF, choice D. 49. The shortest measurement would be 14-8 or 6, while the largest side could be 8+14 which is 22. So all sides possible are between 6 and 22. That omits the side that is exactly 22, Choice D. 50. 3x+2x+2 = 42. X = 8 Then plugging in 8 for x, LM is 3(8) or 24. 51. DE is half of AC so 2(y+6) = 3y+4 and y =8. Sub in 4 for y and AC is 3(8) + 4 or 28. 52. Since the triangle is isosceles, the base angles are congruent. So x + x + 50 = 180, or x = 65. Choice J. 53. Each time we divide by 2 so next number is 4. 54. H, Median 55. Choice G – because MP is shown as a segment that connects midpoints of two sides of the triangle it is a midsegment. 56. Triangle is isosceles,so base angles congruent. So x+x+110 = 180. X = 35 Choice D. 57. Using congruent vertical angles we have SAS, choice G. 58. Choice D – point B. 59. Because the base angles are congruent, ABD is isosceles triangle – Choice H. 60. Lines are parallel so we have SSI interior angles that are supplementary. So 3x – 45 + 90 = 180. X = 45, Choice A. 61. If <2 congruent to <6 the lines would be parallel because <2 and <6 are corresponding. Choice J 62. X + 42 + 36 = 180. So x = 102. And y + 36 + 42 = 180. So y =102 also! 63. OMIT THIS PROBLEM. 64. <A, <C, <B, Choice B 65. Inverse negates the hypothesis and the conclusion so : If the sky is NOT blue, then it IS raining. Choice A. 66. The two angles together form a right angle so they add up to 90. So 16y+2+13y+1=90 and y = 3 Choice H. 67. 22 = ½ (4x) or 22=2x, = 11, Choice G. 68. Using the congruency statement TV is congruent to LP, choice G. 69. This is an isosceles triangle so base angles are congruent. Therefore x + x + 48 = 180 and x = 66, choice A. 70. All right triangles have two acute angles – Choice G. 71. 70 + x = 115 so x = 45, choice C.