ssrs barcode font Parallel Lines, Distances, and Angle Sums in Objective-C

Generating QR Code 2d barcode in Objective-C Parallel Lines, Distances, and Angle Sums

CHAPTER 4 Parallel Lines, Distances, and Angle Sums
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(a) Since P is on the bisector 4 j A, it is equidistant from AB and AC. Since Q is on the bisectors 4 j4 it is of of B, 4 equidistant from AB and BC. Since R is on the bisectors of j A and jB, it is equidistant from AB, BC, and 4 AC. R is the incenter of ^ ABC, that is, the center of its inscribed circle. (b) Since P is on the ' bisector of AB, it is equidistant from A and B. Since Q is on the ' bisector of AC, it is equidistant from A and C. Since R is on the ' bisectors of AB and AC, it is equidistant from A, B, and C. R is the circumcenter of ^ ABC, that is, the center of its circumscribed circle.
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4.10 Applying principles 1, 3, 6, and 7 In each part of Fig. 4-34, find two points equidistant from the ends of a line segment, and find the perpendicular bisector determined by the two points.
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Fig. 4-34
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(a) B and D are equidistant from A and C; hence, BE is the ' bisector of AC. (b) A and D are equidistant from B and C; hence, AD is the ' bisector of BC. (c) B and D are equidistant from A and C; hence, BD is the ' bisector of AC. A and C are equidistant from B and D; hence, AC is the ' bisector of BD.
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4.3 Sum of the Measures of the Angles of a Triangle
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The angles of any triangle may be torn off, as in Fig. 4-35(a), and then fitted together as shown in (b). The three angles will form a straight angle. We can prove that the sum of the measures of the angles of a triangle equals 180 by drawing a line through 4 one vertex of the triangle parallel to the side opposite the vertex. In Fig. 4-35(c), MN is drawn through B parallel to AC. Note that the measure of the straight angle at B equals the sum of the measures of the angles of ^ ABC; that is, a b c 180 . Each pair of congruent angles is a pair of alternate interior angles of parallel lines.
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Fig. 4-35
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CHAPTER 4 Parallel Lines, Distances, and Angle Sums
4.3A Interior and Exterior Angles of a Polygon
An exterior angle of a polygon is formed whenever one of its sides is extended through a vertex. If each of the sides of a polygon is extended, as shown in Fig. 4-36, an exterior angle will be formed at each vertex. Each of these exterior angles is the supplement of its adjacent interior angle.
Fig. 4-36
Thus, in the case of pentagon ABCDE, there will be five exterior angles, one at each vertex. Note that each exterior angle is the supplement of an adjacent interior angle. For example, mja mja 180 .
4.3B Angle-Measure-Sum Principles
PRINCIPLE
The sum of the measures of the angles of a triangle equals the measure of a straight angle or 180 .
mj B mjC 180 .
Thus in ^ ABC of Fig. 4-37, mj A
Fig. 4-37
PRINCIPLE 2:
If two angles of one triangle are congruent respectively to two angles of another triangle, the remaining angles are congruent.
Thus in ^ ABC and ^ A B C in Fig. 4-38, if jA > j A and jB > jB , then jC > jC .
Fig. 4-38
PRINCIPLE 3:
The sum of the measures of the angles of a quadrilateral equals 360 .
mjB mjC mjD 360 .
Thus in quadrilateral ABCD (Fig. 4-39), mjA
Fig. 4-39
CHAPTER 4 Parallel Lines, Distances, and Angle Sums
PRINCIPLE 4:
The measure of each exterior angle of a triangle equals the sum of the measures of its two nonadjacent interior angles.
mjA mjB.
Thus in ^ ABC in Fig. 4-40, mjECB
Fig. 4-40
PRINCIPLE 5:
The sum of the measures of the exterior angles of a triangle equals 360 .
mjb mjc 360 .
Thus in ^ ABC in Fig. 4-41, mj a
Fig. 4-41
PRINCIPLE 6:
The measure of each angle of an equilateral triangle equals 60 .
60 , mj B 60 , and mj C 60 .
Thus if ^ ABC in Fig. 4-42 is equilateral, then mj A
Fig. 4-42
PRINCIPLE 7:
The acute angles of a right triangle are complementary.
90 , then mj A mj B 90 .
Thus in rt. ^ ABC in Fig. 4-43, if mjC
Fig. 4-43
PRINCIPLE 8:
The measure of each acute angle of an isosceles right triangle equals 45 .
90 , then mj A 45 and mj B 45 .
Thus in isos. rt. ^ ABC in Fig. 4-44, if mjC
Fig. 4-44
PRINCIPLE 9:
A triangle can have no more than one right angle.
s 90 , then j A and j B cannot be rt. j
Thus in rt. ^ ABC in Fig. 4-43, if mjC
PRINCIPLE 10:
A triangle can have no more than one obtuse angle.
Thus in obtuse ^ABC in Fig. 4-45, if jC is obtuse, then j A and jB cannot be obtuse angles.
Fig. 4-45
CHAPTER 4 Parallel Lines, Distances, and Angle Sums
Two angles are supplementary if their sides are respectively perpendicular to each other.
PRINCIPLE 11:
Thus if l1 ' l3 and l2 ' l4 in Fig. 4-46, then ja > jb, and a and jc are supplementary.
Fig. 4-46
SOLVED PROBLEMS
4.11 Numerical applications of angle-measure-sum principles In each part of Fig. 4-47, find x and y.
Fig. 4-47
Solutions
(a) x y 35 110 70 x 25 y 180 75 180 45 (Pr. 1) (Pr. 1) (c) In ^ABC, x In ^ I, x 25
y y y
65 x 90 90 65
90 25
(Pr. 7) (Pr. 7)
Check: The sum of the measures of the angles of quad. ABCD should equal 360 . 70 120 110 60 360
360 360 (Pr. 4)
(b) x is ext. j of ^ I. x 30 40 x y y 70 mjB 85 40 40 125 y is an ext. j of ^ABC.
(d) Since DC ' EB, x 90 x y 120 360 90 y 120 360 y 150 (e) Since AB iDE, x y x 50
4 4 4
(Pr. 7)
50 (Pr. 4)
45 45
(Pr. 4)
95 80 2x 180 100 (Pr. 4) 130
(f) Since AB iCD, 2x x y y 50 x 80 50 80
CHAPTER 4 Parallel Lines, Distances, and Angle Sums
4.12 Applying angle-measure-sum principles to isosceles and equilateral triangles Find x and y in each part of Fig. 4-48.
Fig. 4-48
Solutions
(a) Since AB > AC, we have j1 > jx x 180 125 55 2x y 180 (Pr. 1) 110 y 180 y 70 (c) Since AB > AC, jABC > jACB 2x 80 180 (Pr. 1) x 50 1 1 In ^ I, 2x 2x y 180 (Pr. 1) x 50 (Pr. 6) (Pr. 8) y y y 180 180 130
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