Chapter 3 Cumulative Review Answers


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1 Chapter 3 Cumulative Review Answers 1a. The triangle inequality is violated. 1b. The sum of the angles is not 180º. 1c. Two angles are equal, but the sides opposite those angles are not equal. 1d. The exterior angle theorem is violated. / The sum of the angles is not equal to 180º. 1e. The angles shown should add up to 180º. 1f. The angles should add up to 360º. 1g. The vertical angles are equal, so the triangles should be congruent by SAS. By corresponding angles, the remaining angles should equal 60º, but that would make the triangles equilateral. However the sides are not equal. 1h. The 85º angle should equal 80º.. The sum of the exterior angles is 360º, so n x=360. We also know that 9 x=180º so x=0º and n= A polygon with n sides, where all the sides and angles are equal. 4. Construct B*, the reflection of B, across the line and connect AB*. 5a. 60º 5b. 30º 5c. 48.5º (isosceles) 5d. 3 (corresponding sides) 5e. 83º (sum is 360º) 5f. 6º ( 6 x= x+130º ) 5g. 0º 5h. 35º 6. Outline of proof: Vertical angles are equal. Triangles are congruent by SAS. Corresponding angles of the triangles are Alternate Interior angles for the two lines. Therefore they are parallel. 7a. 90º 7b. right triangle 7c. converses of each other 8a. 8b. The angle is not the included angle. 9. x=7.
2 Chapter 7 Cumulative Review Answers 1a. Pythagorean Theorem 1b. Pythagorean Theorem 1c. The base angles are equal so they must be 60º, so the triangle has three equal angles, but is not equilateral. 1d. Triangle Inequality 1e. Converse of Pythagorean Theorem 1f. The sum of the angles must be 180º. 1g. The sum of the angles must be 360º. 1h. Two equal angles but not isosceles. 1i. Violates the midsegment theorem. Violates the triangle inequality. 1j. Top and bottom sides are equal and parallel, therefore this should be a parallelogram, but opposite sides not equal. 1k. Exterior Angle Theorem. 1l. Triangle Inequality (area of ADC  area of ADB) a b. 5 (Triangle ABC is equilateral) 6a. Construct equilateral triangle. 6b. 90º minus 60º (or bisect 60º)
3 6d. Construct perpendicular lines then circle. 6c/6e. Bisect 30º to get 15º. 60º + 15º = 75º. 7. A B = = (half the length of the hypotenuse) 9a. Copy the segments end to end. 9b. Make a right triangle with sides a and b 9c. Construct a square of side a then construct its diagonal. 9d. Construct an equilateral triangle of side a then construct its altitude. 10. Extend AM an equal distance to D. ABDC is a quadrilateral whose diagonals bisect each other and are equal. Therefore it is a rectangle, so triangle ABC is a right triangle.
4 Chapter 9 Cumulative Review Answers (8 0) = 41 a. 90º 5 b. 3 c d. e. 6 f , 4 3, 6 5. The triangles are similar. The areas are in the ratio 4:1, therefore the scale factor is. Therefore DE = 4. 6., 3, 4=, 5 7a. Yes; SSS 7b. Yes; SAS 7c. Yes; SSS 7d. Yes; ASA 7e. Not necessarily congruent 7f. Yes; ASA 8. congruent 9a. Yes; SAS; k= 3 9b. Not similar 9c. Yes; AAA; k= 1 9d. Yes; AAA; k = 3 4 9e. Yes; SSS; k =1 9f. Yes; AAA; k= 3 4
5 9 10a. 10b. Arrange PQ along the river and QB the width of the river. AQ and QP can be measured. The proportion ft AQ QP = QP QB can be solved to give the width of the river. 1. Let h be the height. h =a c =b d. The result follows from difference of two squares times the area inches 15a b. 9 15c. 1 15d inches 17. 3a b 18a. 3 18b. 4 18c. 6 18d Δ AEC Δ ADB by SAS 0a. Construct a perpendicular bisector to bisect the segment.
6 0b. Trisect the segment. 0c. Construct an equilateral triangle on /3 of the segment. 0d. Construct the perpendicular bisector then construct a square on half of the segment. 0e. Trisect the segment, then construct a square on /3 of the segment and duplicate it so its dimensions are Trisect a side and draw a parallel line.. All except c.
7 3. The four triangles are congruent by SAS. All sides of PQRS are equal by the Pythagorean Theorem, so it is a rhombus. Each angle of PQRS is supplementary to a pair of complementary angles so they are all right angles. Therefore PQRS is a square.
8 Chapter 1 Cumulative Review Answers cents. 10 ft a b. 10 4c d. 84 4e a. 6b. 6c Δ PQS has the same base and height as rectangle ABQS, so it has half the area. Δ QRS has the same base and height as rectangle QCDS, so it has half the area. Therefore PQRS has half the area of ABCD. 8a. 68º 8b. 15º 8c. 0º 8d. 107º 8e. 115º 8f. 135º 8g. 3 8h. 4 8i. 58º 8j. 15 8k. 10º 8l m. 4
9 9. 5 π sq. units 10. Area = 1 (AE )(BC) = 1 (BD)( AC ) therefore (AE )(BC )=( BD)( AC) π 1. x 6 = 6 x+5... x = Congruent by AAS 15. Δ ABC Δ DEC by SAS. Δ DEC Δ FEG by AAS. Therefore Δ ABC Δ FEG ft º, 73º, 107º 18. The triangles are congruent by SSS, so the corresponding angles are equal. 19a. If two triangles have the same area, then they are congruent. False. 19b. If two squares have the same area, then they are congruent. True. 19c. If the diagonals of a quadrilateral bisect each other, then it is a parallelogram. True. 19d. If two angles of a triangle are equal, then it is isosceles. True. 0. Thales Theorem. 1. Circle I has half the diameter of the large circle, so it has 1/4 of the total area. Circle II also has 1/4 of the total area. Therefore the remaining part of the circle has half the total area. Therefore regions III and IV each have 1/4 of the total area. a π yd. b π sq. yd. 3a. 36 3b. 54 4a b. 8(1+ 3) 5a. 5 5b. 6 5c. 5d. 65
10 6. The sides have length 5, 45, and 50. By the converse Pythagorean Theorem the triangle is a right triangle. 7a. 7b. 8. The side lengths are 5, 5, and 18 therefore it is isosceles. 9a. 6 9b. 17 9c. 6 9d inches 31a. 31b. 3a. Duplicate the triangle with a compass to mark the sides. 3b. Trisect ED 33. According to the intersecting chords theorem x(x+)=( x+3)( x+4) which leads to x= 5. But in this context x must be positive. 34a. 4 cubic units 34b. 0 square units 35. Major axis = 14; Minor axis = 4 6
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