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Tampa Bay Tech Invitational Geometry Individual For all questions, E. NOTA means “none of the above answers is correct”. Diagrams are NOT drawn to scale. 1. Points A, B, and C are collinear, but they do not necessarily lie on a line in the order named. If AB 5, BC 3, what is the length of segment AC ? A. either 2 or 8 C. 2 January 11, 2014 4. Which proportion is not equivalent to a c ? b d (no denominator is equal to 0) a b c d a c D. d b A. B. b d ab cd C. b d a c E. NOTA B. either 2 or 4 D. 8 E. NOTA 5. In ABC with right angle at B, D is a point on AC such that BD AC , AD 6 , 2. Which statement would not guarantee that A and B are congruent? A. A and B are adjacent angles formed by perpendicular lines. BD 4 3 . Find the length of BC . A. 4 7 B. 6 3 D. 2 21 E. NOTA C. 6 7 B. A and B are vertical angles. C. A and B are both complements of C. D. A and B are same side interior angles formed by two lines and a transversal. 6. The median to the hypotenuse of a right triangle divides the triangle into two triangles that are both E. NOTA A. similar D. isosceles 3. In ABC , point D is on AB , point E is on AC B. right E. NOTA C. scalene AB 2 x 1, AE 2 x 2, EC x 1 , find the 7. Given rectangle ABCE and right isosceles triangle DEC with common side EC. DE DC EA BC x . If the perimeter of length of AC . ABCDE = 16 4 2 , find the value of x. such that DE || BC . If AD x 2 A. 7 D. 15 B. 8 E. NOTA C. 12 A. 1 B. 2 D. 4 E. NOTA C. 3 Tampa Bay Tech Invitational Geometry Individual 8. Given: m1 x 2 3 y , m2 20 y 3 , m3 3 y 4 x , x 0. Find the m1 in degrees. 1 12. The perimeter of ABCD is 85. If AB 3x 4, BC 4 x, CD 6 x 11, and AD 5x 7, find the length of AB . 2 3 A. 5.5 D. 42.5 A. 37 D. 67 B. 40 E. NOTA January 11, 2014 B. 20.5 E. NOTA C. 21.5 C. 53 13. Given parallelogram ABCD with mA x, mD 3x 4 . Find mB mD. 9. AB CD at point E. mAEC ( x 2 y), mAED x 2 y , mDEB 50 . By how much does x exceed y in degrees? A. 70 B. 90 D. cannot be determined C. 110 E. NOTA A. 46 D. 268 B. 134 E. NOTA C. 180 14. The perimeter of equilateral triangle ABC is 36 3 . The bisectors of A and B meet at D. Find the length of AD . 10. Which group of numbers can be the lengths of the sides of an obtuse triangle? A. 3,5,7 3 5 D. ,1, 4 4 B. 2,4,6 A. 6 B. 6 3 D. 12 3 E. NOTA C. 12 C. 0.5,0.6,0.7 E. NOTA 15. Point X is equidistant from vertices T and N of TEN . Point X must lie on which of the following? A. bisector of E 11. Which information does not prove that quadrilateral ABCD is a parallelogram? A. AC and BD bisect each other B. median to TN C. perpendicular bisector of TN D. altitude to TN E. NOTA B. AD || BC ; AD BC C. AB || CD; AD BC D. A C; B D E. NOTA 16. One of two complementary angles is twice the other, Find the supplement of the smaller angle. A. 30 D. 150 B. 60 E. NOTA C. 120 Tampa Bay Tech Invitational Geometry Individual January 11, 2014 17. ABCD is an isosceles trapezoid with upper base 22. In ZYW , M is a point on WZ and N is a point AD and diagonals intersecting at point E. Given BE x 7, CE y 3, AE x 5, BD y 4. Find on YZ such that NM || YW . If the numerical length of AC . A. 10 D. 16 B. 12 E. NOTA C. 14 18. If the sum of the measures of the interior angles of a polygon is increased by 900, how many sides will have been added to the polygon? A. 4 D. 10 B. 5 E. NOTA C. 9 19. The lengths of the diagonals of a rhombus are in the ratio 3:4 and the perimeter is 40. Find the sum of the lengths of the diagonals. A. 14 D. 28 B. 16 E. NOTA C. 24 ZY 2 x 9, ZM 10, ZN x 3, and MW x, then find the value of x. A. 12 D. 12 B. 5 E. NOTA C. 2 34 23. The inverse of the contrapositive of the converse of a conditional statement is equivalent to the __?__ A. inverse B. converse D. conditional C. contrapositive E. NOTA 24. The average lengths of the sides of ABC is 14. If AB x 5, BC 2 x 1, and AC 4 x 6, how much longer than the average is the longest side? A. 2 D. 24 B. 4 E. NOTA C. 22 25. HGF is equilateral. If the perimeter is 6 y 24 20. In PRT , S is a point on PT such that RS is an angle bisector. If PS 6, ST 8, and the perimeter of PTR is 42, find the length of PR . A. 12 D. 29 B. 12.75 E. NOTA perimeter is 4 18, B. 3 12 E. NOTA HGF. A. 38 D. 120 B. 72 E. NOTA C. 114 C. 18 21. Find the length of the diagonal of a square if the A. 2 18 D. 72 and HG 3 y 7, find the numerical perimeter of C. 36 26. One side of a triangle is 4 cm shorter than a second side. The ray bisecting the angle formed by these sides divides the opposite side into 4 cm and 6 cm segments. Find the perimeter of the triangle. A. 30 B. 32 D. cannot be determined C. 44 E. NOTA Tampa Bay Tech Invitational Geometry Individual 27. Each interior angle of a regular polygon measures 160 and each side has length 5. Find the perimeter of the polygon. A. 20 D. 90 B. 55 E. NOTA C. 85 28. In a trapezoid, the length of the bases are 4 and 16, and the lower base angles have measures 30 and 60 . Find the distance between the midpoints of the bases. A. 2 3 B. 3 3 D. 12 3 E. NOTA C. 6 29. Regular hexagon ABCDEF has side AF in common with regular hexagon AFGHIJ, and side BC in common with regular hexagon BCKLMN. All three hexagons are coplanar and nonoverlapping. If AB = 64, find the length of JN . A. 64 D. 128 B. 80 E. NOTA C. 96 30. In right triangle ABC, the altitude to the hypotenuse AB is drawn intersecting the hypotenuse at point E. If BC 3 5, BE 5 , find the length of AC . A. 3 D. 9 B. 4 E. NOTA C. 6 January 11, 2014 Tampa Bay Tech Invitational Geometry Individual SOLUTIONS 1. A 2. D 3. E 4. D 5. A 6. D 7. D Let the legs of the triangle be x which makes the width of the rectangle x. EC has 7. D 8. A 9. C 10. A 11. C 12. B 13. D 14. C 15. C 16. D 17. D 18. B 19. D 20. A 21. E 22. D 23. D 24. B 25. C 26. A 27. D 28. C 29. D 30. C 1. A There are two possibilities. With AB-C, B is between A and C and the length of AC would be 8. With A-C-B, C is between A and B and the length of AC is 2. 2. 3. D January 11, 2014 A. Both would be right angles and B. Vertical, congruent. C. Both complements of 90 are congruent. D. Same side interior angles are supplementary. E The triangles are similar by AA so 2x 2 x 2 , x 7,0 . the sides are proportional. x 1 x 1 Reject the 0. This makes AC equal to 20. 4. D Check all the rules of proportions or put numbers in and see if the cross products are the same. 5. A The triangles are similar by AA so the geometric mean formulas can be used. DC 6 4 3, DC 8, BC 8 8 6 4 7 . 6. D The median to the hypotenuse is ½ the length of the hypotenuse so the triangles both have 2 sides congruent making the triangles isosceles. length x 2 . So the perimeter is 4 x x 2. Which now gives 4 x x 2 16 4 2 , x 4 2 16 4 2, solving this gives x 4. 8. A Use systems: x 2 3 y 20 y 3 180, x 2 23 y 177. 20 y 3 3 y 4 x 180, 4 x 23 y 177. x 2 23 y 4 x 23 y, x 2 4 x 0, x 4,0. Reject the 0 since the question says x 0. Substituting x 4 in x 2 23 y 177 gives y 7. m1 x 2 3 y 16 21 37. mAEC mAED 180, so we 9. C have x 2 y x 2 y 180, x 90. And angles AEC and DEB are vertical so x 2 y 50, substituting for x, 90 2 y 50, y 20. x exceeds y by 110, 90 20 110. 10. A Use c 2 a 2 b 2 which makes the triangle isosceles. A. 49 9 25 , obtuse B. This is not a triangle. 2 4 6 . C. 0.49 0.25 0.36 , acute 25 16 9 , right D. 16 16 16 11. C Follow the rules for proving a quad is a parallelogram. A. Yes, diagonals bisect each other. B. Yes, one pair of sides is both parallel and congruent. C. No. D. Yes, opposite angles are congruent. 12. B Find the value of x by adding the sides and setting equal to the perimeter 85. 41 11 18x 11 85, x 5.5. AB 3 4 20.5 . 2 2 Tampa Bay Tech Invitational Geometry Individual 13. D Consecutive angles in a parallelogram are supplementary. 4 x 4 180, x 46.mD 134, mD mB 268. 14. C After drawing the diagram, ABD is an isosceles triangle with base angles of 30 .Draw the altitude from D to AB forming 2 30 60 90 triangles. The length of a side of the original January 11, 2014 20. A Since the perimeter is 42 and the sum of the lengths of PS and ST is 14, the sum of the other two sides is 28. Use the Triangle Angle Bisector Theorem to get the proportion: x 28 x , x 12 . 6 8 21. E Simplifying the perimeter is 12 2 . One side of the square is 3 2 . The diagonal would be 2 times the length of the side 3 2 which is 6. triangle is 12 3 . That makes the base of each of the 30-60-90 triangles have a length of 6 3 . Since it is opposite a 60 angle, divide by 3 to get the length of the altitude DE which is 6. Double that to get the length of AD – 12. 22. D Triangles are similar by AA. Sides are proportional. x3 10 , x 12, 5 . 2 x 9 x 10 15. 23. C By definition, this point would be on D q p,~ p ~ q, p q, conditional the perpendicular bisector of TN . 16. Let the angles be x and 90 x . D x 2 90 x , x 60 and the complement will be 30. The supplement of the smaller angle is 150. 17. D The diagonals are congruent. BD AE CE ; y 4 x 5 y 3 . Solving this 24. B The average length of the sides is 14 so the perimeter is 42. x 5 2 x 1 4 x 6 42 , x 6. That makes AB 11, BC 13, AC 18. So 18 – 14 is 4. 25. C 3 3 y 7 6 y 24, y 15. The perimeter would be 6 15 24 114. gives x 2. Now we have x 7 y 3, substituting 2 for x, we have y 12. This makes AC 16. 18. (n 2)180 900 n 2 x 180, B 180n 540 180n 180 x 360, x 5. 19. D Since the diagonals of a rhombus are perpendicular, the half diagonals would be 3x and 4x. Using Pythagorean theorem, 3x 2 4 x 100, x 2. This makes the 2 3x 6,4 x 8 and the sum of the diagonals would be twice (6+8)=28. 26. A Two possibilities for the 4 and 6 on the side of the triangle. Use the triangle angle x4 x , x 12 or bisector theorem to get 4 6 x4 x , x 8 . Reject this solution as the side 6 4 of the triangle can’t be negative. The lengths of the sides of the triangle are 12+8+10 = 30. 27. D One exterior angle would be 20 . The sum of the exterior angles of a polygon, one at each vertex is 360, so divide 360 by 20 to get 18 angles/sides. The perimeter is 18 5 90. Tampa Bay Tech Invitational Geometry Individual 28. C Draw the altitudes to get a rectangle and two 30-60-90 triangles. In the triangle with 30 base angle, let the side opposite the 60 angle x be x, the side opposite the 30 angle will be . 3 In the triangle with the 60 base angle, the side opposite the 30 angle will be 12 x (the length of the base minus [length of the upper base 4 and x]), the side opposite the 60 angle will be 12 x 3 . x and 12 x 3 are both altitudes of the 3 trapezoid so they are equal. Set them equal and solve for x gives 9. So the segments of the base are 3 – 4 – 9. We want the distance between the midpoints of the bases. ½ the longer base is 8. So the midpoint of the base is 8 units from either vertex of the lower base. Using the triangle with the 30 base angle, go in 8 units – this is one vertex of the right triangle we need to find the distance between the midpoints. The other vertex is the midpoint of the upper base. The third vertex is the perpendicular from the midpoint of the upper The base. The altitude of the trapezoid is 3 3 , the length of the side on the lower base is 3. Using Pythag, the distance we want is 6. Whew, that was a long solution. 29. D Draw the diagram and you will see that JNBA is a trapezoid. AB is the shorter base and JN is the longer base. The measure of angle AJN and JNB is 60 and the upper base angles are 120. Draw an altitude and you have a 30-60-90 triangle. The legs of the trapezoid are 64 making the part of the lower base opposite the 30 degree angle having a length of 32. So the 3 lengths of the base JN are 32 + 64 + 32 = 128. 30. C The 3 triangles are similar. Using the geometric mean to find the length of AB : 3 5 5 AB , AB 9 . Then to find AC: AC 4 9, AC 6 . January 11, 2014