Math 531, Exam 1 Information.

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1 Math 531, Exam 1 Information. 9/21/11, LC 310, 9:05-9:55. Exam 1 will be based on: Sections 1A - 1F. The corresponding assigned homework problems (see boylan/sccourses/531fa11/531.html) At minimum, you need to understand how to do the homework problems. Topic List (not necessarily comprehensive): ou will need to know: theorems, results, and definitions from class. 1B: Congruent triangles. Definition: ABC = RST if and only if 1. 3 pairs of corresponding sides are equal: (a) AB = RS, (b) BC = ST, (c) CA = T R and 2. 3 pairs of corresponding angles are equal: (a) A = R, (b) B = S, (c) C = T. Note: ABC = RST does not imply that ABC = T SR. Notation: S = side; A = angle. Example: For triangles T 1 and T 2, SAS ( side-angle-side ) means that the triangles have two pairs of equal sides with the angle between the sides equal.

2 Table 1: Which conditions guarantee a congruence? condition SSS SSA SAS ASS SAA ASA AAS AAA congruence? (/N) N N N Definition: Let ABC be a right triangle. Its largest side is the hypotenuse. The other two sides are its arms. Fact: Suppose that ABC and DEF are right triangles with equal hypotenuses and one pair of equal arms. Then the triangles are congruent. I.e., we have HA = congruence. (Thm. 1.3). Definitions: Consider ABC. 1. median. Let X = midpt(bc). Then AX = med(a), the median from A to side BC. 2. altitude. Let be the point on side BC for which A is perpendicular to BC. Then A = alt(a), the altitude from A to BC. 3. angle bisector. Let Z be the point on side BC such that BAZ = CAZ. Then AZ = bis( A), the bisector of A. Facts on isosceles triangles: (1) A triangle ABC is isosceles with base BC its base angles are equal: B = C. Note: = is Theorem 1.2 (pons asinorum); the converse is homework problem 1B.1. (2a) Suppose: ABC is isosceles with base BC. Then we have med(a) = alt(a) = bis( A). Note: This conclusion only applies to the angle opposite the base. (Theorem 1.2) (2b) Consider ABC. Suppose that any two of the following are equal: {med(a), alt(a), bis( A)}. Then the triangle is isosceles. (Homework problems 1B.2, 1B.3, and Problem 1.13.) 2

3 1C: Angles and parallel lines. Vocabulary: 1. Vertical angles. 2. Exterior angles. 3. Remote interior angles with respect to an exterior angle. 4. Alternate interior angles. 5. Corresponding angles. Facts: 1. Vertical angles are equal. 2. Corresponding angles are equal 3. Alternating interior angles are equal. 4. An exterior angle is larger than either of its two corresponding remote interior angles. Moreover, an exterior angle is the sum of its two corresponding remote interior angles. (Thm. 1.4). Definition: An interior angle of an n-gon is a reflex angle it is > 180. Theorem: The sum of the interior angles of an n-gon is 180(n 2) degrees. 1D: Parallelograms. Definition: A quadrilateral is a 1. parallelogram its opposite sides are parallel. 2. rhombus all 4 sides are equal. 3. rectangle all 4 angles are equal (hence, 90 ). 4. square it is a rhombus and a rectangle. Fact: A quadrilateral ABCD is a parallelogram if and only if any one of the following is true: 1. Its opposite sides are equal: AB = CD, AD = BC. (Theorems 1.6 and 1.7) 2. Its opposite angles are equal: A = C, B = D. (Homework 1D.7, 1D.9) 3. It has one pair of opposite sides which are parallel and have equal length. (Example: AB = CD and AB CD.) (Theorem 1.8, homework problem 1D.2) 4. Its diagonals, AC and BD, bisect each other. (Theorem 1.9) 3

4 Further results: 1. Adjacent interior angles of a parallelogram are supplements (add to 180 ). 2. Suppose that parallelogram ABCD has an interior right angle. Then it is a rectangle. 3. Let BC be a given line segment. Then the set of all points equidistant from B and C, L = {P : BP = CP }, is the perpendicular bisector of BC. (Theorem 1.10) 4. Parallelogram ABCD is a rhombus its diagonals are perpendicular. (Corollary 1.11 and homework 1D.4). 1E: Area. Basic facts and definitions. 1. Parallelogram: Area = base height. 2. Triangle: Area = (1/2) Base height. 3. sine: In a right triangle, the sine of an angle θ is sin θ = opposite. It is well - defined hypotenuse (only depends on θ, not on side lengths of the triangle). Law of Sines: Consider ABC. We have a sin A = b sin B = c sin C. Further triangle area formulas. Consider ABC. Its area is expressible as: 1. (SAS): Area = 1ab sin C = 1ac sin B = 1 bc sin C (SSS): Heron s Formula: Let s = a+b+c be the semiperimeter of ABC. Then we 2 have Area =» s(s a)(s b)(s c). 3. There are other formulas corresponding to the fact that AAS, SAA, and ASA are valid triangle congruence criteria. Homework 1E.3 addresses ASA. Auxiliary result: Consider ABC. Suppose that bis( A) = AX. Then we have BX XC = AB AC. 4

5 1F: Circles and arcs. Vocabulary: 1. Arc. 2. Central angle corresponding to an arc. 3. Chord. 4. Secant. 5. Tangent. 6. Circumcircle of ABC. 7. Angle inscribed in a circle. 8. Polygon inscribed in a circle. Notation: The symbol means equal in degrees or radians. Therefore, ĀB θ means that the arc ĀB has corresponding central angle θ. Note that we can have ĀB CD (corresponding central angles equal in degrees) but ĀB CD (arcs are not the same length). Theorem: Let A, B, and C be non-collinear points. There exists a unique circle passing through these points. (3 non-collinear points uniquely determine a circle.) (Theorem 1.15) Facts: The following are basic relations between arcs and angles formed different ways. 1. Suppose that inscribed angle P subtends arc ĀB. Then we have P (1/2)ĀB. (Theorem 1.16) 2. Suppose that two secants meet at a point X outside of the circle, and suppose that X subtends large arc ĀB and small arc RP. Then we have X (1/2)(ĀB RP ). (Theorem 1.18) 3. Suppose that two chords meet at a point X inside of the circle, and suppose that X subtends the two arcs ĀB and RP. Then we have X (1/2)(ĀB + RP ). (Theorem 1.19) 4. Suppose that T is the line tangent to the circle at point P, and suppose that QP is a chord. Then we have QP T (1/2) QP. (Theorem 1.23) Facts on tangents: 1. Let P be a point on circle C. There exists a unique line tangent to C at P. 2. The line tangent to C at point P is perpendicular to the radius through P. 5

6 Further results. 1. A quadrilateral inscribed in a circle has opposite angles supplementary. (Theorem 1.17) 2. Suppose that C is the circumcircle of ABC. Then we have C = 90 AB is the diameter of C. (Theorem 1.22) 3. Suppose that AB and CD are chords on two circles with equal radii. Then the chords are equal (AB = CD) if and only if ĀB CD. (Homework 1F.1) 6