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Chapter 9 Summary Project

Period 2 12/17. Chapter 9 Summary Project. Starring:. Parallelism. &. Triangles. Quadrilaterals. By Matthew Donoghue. P a r a e i s m. Corresponding Angles:. Definitions.

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Chapter 9 Summary Project

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  1. Period 2 12/17 Chapter 9 Summary Project Starring: Parallelism & Triangles Quadrilaterals By Matthew Donoghue

  2. P a r a e i s m Corresponding Angles: Definitions Two angles that are formed by two lines cut by a transversal. Each angle is located on the same side of the transversal and one is interior, and the other is exterior. Alternate Interior Angles: Two angles that are formed by two lines cut by a transversal. Each angle is located on an opposite side of the transversal and neither share a common ray. Skew Lines: Transversal: Parallel Lines: A line that intersects two lines at different points. Two lines that are non-coplanar and never intersect. Two lines that are coplanar and never intersect.

  3. Theorem 9-5 T The AIP Theorem A L2 C If two lines cut by a transversal form two congruent alternate interior angles, then the lines are parallel. a b L1 B D Restatement. Given: Lines L1 and L2 cut by a transversal T. a  b, then L1 L2 < < = Statement Reason 1. Given 1. AC = BD, AB = CD Parallelism Problem #1. 2. RPE 2. CB = CB 3. ∆ABC  ∆DCB 3. SSS Use Diagram above right. Given: AC = BD, AB = CD. Lines L1 and L2 are cut by a transversal T. 4. ACB  CBD 4. CPCTC < < = 5. AIP 5. L1 L2 = Prove L1 L2.

  4. Theorem 9-30 Say bye-bye to parallelism. It has no name, so don’t ask for one. Just 9-30. T1 T2 L1 A E L2 B F If two congruent segments are cut by three parallel lines, then any other transversals along the lines are also cut in to equal segments. L3 C G T1 Statement Reason = = 1. L1 L2 L3, 1. Given = = L4 L5 L6, Restatement Given: L1 L2 L3. All three lines are intersected by transversals T1 and T2; AB =BC. Then EF = FG. AB = BC. T2 = = 2. 9-30 2. GH = HI. D E F 3. 9-30 3. FE = DE. L1 A I Parallelism Problem #2 L2 B H Use figure below right. Given: L1 L2 L3, L4 L5 L6. L1, L2, & L3 are cut by transversals T1 & T3. L4, L5, & L6 are cut by transversals T1& T2. AB = BC. L3 C G = = = = L6 T3 L5 Prove: EF = DE. L4

  5. Trangle Theorem 9-27 Definitions Theorems The 30-60-90 Triangle Theorem Statement Reason B Isosceles Triangle 1. MAT and ROX are right triangles; MT = 20; R & T = 30˚ 1. Given In a right triangle, if the smallest angle measures 30˚, then the shortest side, which is opposite the 30˚ angle, is 1/2 the length of the hypotenuse. D 30˚ A C < < 2. MA = 10 2. 30-60-90 3. MA = RX 3. Given Restatement Given: ABC is a right triangle; A has a measure of 30˚; D is the midpoint of AB. Then BC = 1/2 AB. 4. RX = 10 4. Substitution < Right Triangle 5. OX = 5 5. 30-60-90 Triangles Problem #1 Use the. figure on the left. Given: MAT and ROX are right triangles; MA = RX; R & T = 30˚; MT = 20. 5 An isosceles triangle has a pair of congruent angles and sides. The two congruent sides will always be opposite the two congruent angles, and vice versa. O R M < < 30˚ 10 10 A right triangle has one right and two acute angles. P.S. Right triangles can be isosceles as well. X 20 T 30˚ A

  6. Corollary 9-13.3 Statement Reason Un-named. Y 1. RSU  ( R + RSQ). 1. Given < < B The exterior angle of any triangle, has the same degrees as the two remote interior angles added together. < A C E 2. Corollary 9-13.3 2. TQR  ( R + RSQ). < < X Z < Restatement Given: XYZ with angles A, B, C, and E. E is an exterior angle adjacent to angle C. Then the sum of A & B = E. 3. TQR  RSU. 3. TPE < < 4. TQR is supp. to SQR. 4. Supp. Pos. < < 5. USR is supp. to QSR. 5. Supp. Pos. < < Triangles Problem #2 6. Supp. Theorem. 6. QSR  SQR. < < 7. Def of Isosc.  7. RSQ is isosc. Use figure to below. Given: RSU  ( R + RSQ). < < < R Prove: RSQ is isosceles. T U Q S

  7. Quadrilateral Square Definitions Rectangle Quadrilateral A parallelogram with 4 right angles and 4 congruent sides. Any parallelogram with 4 right angles. Any 2 dimensional figure with exactly 4 sides. Rhombus Parallelogram Trapezoid Any quadrilateral with only one pair of opposite sides being parallel. Any parallelogram with 4 congruent sides. Any quadrilateral with every pair of opposite sides being parallel.

  8. Theorem 9-16 Un-named B C In a parallelogram, the opposite angles are congruent. Restatement Given: Parallelogram ABCD, then A  C & B  D. A < < < < D B F C Quadrilaterals Problem #1 Use the figure to the right. Given: Parallelogram ABCD with FB = HD & BE = GD. G Statement Reason E 1. ABCD is a parallelogram; BE = GD; FB = HD 1. Given A H D Prove: EF = GH. 2. B  D. 2. 9-16 < < 3. FBE  HDG 3. SAS 4. EF = GH 4. CPCTC

  9. Theorem 9-25 B Un-named If a quadrilaterals diagonals are perpendicular to each other and bisect each other, then the quadrilateral is a rhombus. A C Restatement Given: ABCD, with AC  BD, and AC and BD bisecting each other, then it is a rhombus. D B A Quadrilateral Problem #2 Use the figure on the right. Given: ABCD , with AC  BD, and AC and BD bisecting each other. C Statement Reason D 1. Given 1. ABCD , with AC  BD, and AC and BD bisecting each other. Prove: All sides are equal. 2. ABCD is a rhombus. 2. 9-25 3. All sides are equal. 3. Def of rhombus

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