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4.4C Polygon-Angle Principles
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If S is the sum of the measures of the interior angles of a polygon of n sides, then S n 2 straight angles (n 2)180
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The sum of the measures of the interior angles of a polygon of 10 sides (decagon) equals 1440 , since 8(180) 1440.
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The sum of the measures of the exterior angles of any polygon equals 360 .
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Thus, the sum of the measures of the exterior angles of a polygon of 23 sides equals 360 .
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If a regular polygon of n sides (Fig. 4-53) has an interior angle of measure i and an exterior angle of measure e (in degrees), then 180(n n 2) 360 n , and i
Fig. 4-53
CHAPTER 4 Parallel Lines, Distances, and Angle Sums
Thus for a regular polygon of 20 sides, 180(20 20 2) 360 20
SOLVED PROBLEMS
4.15 Applying angle-measure formulas to a polygon (a) Find the sum of the measures of the interior angles of a polygon of 9 sides (express your answer in straight angles and in degrees). (b) Find the number of sides a polygon has if the sum of the measures of the interior angles is 3600 . (c) Is it possible to have a polygon the sum of whose angle measures is 1890
Solutions
(a) S (in straight angles) (b) S (in degrees) (c) Since 1890 n 2 9 2 7 straight angles; mjS (n 2)180 22. sides. 7(180) 1260 .
(n 2)180. Then 3600 (n 2)180, then n 121. 2
(n 2)180, from which n A polygon cannot have
1 122
4.16 Applying angle-measure formulas to a regular polygon (a) Find each exterior angle measure of a regular polygon having 9 sides. (b) Find each interior angle measure of a regular polygon having 9 sides. (c) Find the number of sides a regular polygon has if each exterior angle measure is 5 . (d) Find the number of sides a regular polygon has if each interior angle measure is 165 .
Solutions
(a) Since n 9, m/e 360 n (n 360 9 2)180 n e 40. Ans. 40 (9 2)180 9 180 e
(b) Since n
9, m/ i
Ans. 140 140. 360, so n 15. Ans. 24 sides 72. Ans. 72 sides
Another method: Since i (c) Substituting e (d) Substituting i Then, using e 5 in e 165 in i
180, i
180 40
360 n , we have 5 e
360 n . Then 5n e
180, we have 165 15, we have 15
180 or e 24.
360 n with e
360 n , or n
4.17 Applying algebra to the angle-measure sums of a polygon Find each interior angle measure of a quadrilateral (a) if its interior angles are represented by x 2x 20, 3x 50, and 2x 20; (b) if its exterior angles are in the ratio 2:3:4:6.
CHAPTER 4 Parallel Lines, Distances, and Angle Sums
Solutions
(a) Since the sum of the measures of the interior is 360 , we add (x 10) (2x 20) (3x 50) (2x 20) 8x 40 x Then x 10 60; 2x 20 120; 3x 50 100; 2x 20 80. 360 360 50 Ans. 60 , 120 , 100 , 80
(b) Let the exterior angles be represented respectively by 2x, 3x, 4x, and 6x. Then 2x 3x 4x 6x 360. Solving gives us 15x 360 and x 24. Hence, the exterior angles measure 48 , 72 , 96 , and 144 . The interior angles are their supplements. Ans. 132 , 108 , 84 , 36
4.5 Two New Congruency Theorems
Three methods of proving triangles congruent have already been introduced here. These are 1. Side-Angle-Side (SAS) 2. Angle-Side-Angle (ASA) 3. Side-Side-Side (SSS) Two additional methods of proving that triangles are congruent are 4. Side-Angle-Angle (SAA) 5. Hypotenuse-Leg (hy-leg)
4.5A Two New Congruency Principles
PRINCIPLE 1:
(SAA) If two angles and a side opposite one of them of one triangle are congruent to the corresponding parts of another, the triangles are congruent.
Thus if jA > jA , jB > jB , and BC > B C in Fig. 4-54, then ^ ABC > ^ A B C .
Fig. 4-54
Fig. 4-55
PRINCIPLE 2:
(hy-leg) If the hypotenuse and a leg of one right triangle are congruent to the corresponding parts of another right triangle, the triangles are congruent.
Thus if hyAB > hy A B and leg BC > leg B C in Fig. 4-55, then rt. ^ABC > rt. ^A B C .
A proof of this principle is given in 16.
CHAPTER 4 Parallel Lines, Distances, and Angle Sums
SOLVED PROBLEMS
4.18 Selecting congruent triangles using hy-leg or SAA In (a) Fig. 4-56 and (b) Fig. 4-57, select congruent triangles and state the reason for the congruency.
Fig. 4-56
Fig. 4-57
Solutions
(a) ^ I > ^ II by hy-leg. In ^ III, 4 is not a hypotenuse. (b) ^ I > ^ III by SAA. In ^ II, 7 is opposite 60 instead of 35 . In ^ IV, 7 is included between 60 and 35 .
4.19 Determining the reason for congruency of triangles In each part of Fig. 4-58, ^ I can be proved congruent to ^ II. Make a diagram showing the congruent parts of both triangles and state the reason for the congruency.
Fig. 4-58
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