ssrs barcode font Pythagorean Theorem in Objective-C

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7.9 Pythagorean Theorem
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In a right triangle, the square of the length of the hypotenuse equals the sum of the squares of the lengths of the legs.
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Thus in Fig. 7-33, c2 a2 b2.
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A proof of the Pythagorean Theorem is given in 16.
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Fig. 7-33
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7.9A Tests for Right, Acute, and Obtuse Triangles
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If c2 a2 b2 applies to the three sides of a triangle, then the triangle is a right triangle; but if c2 a2 b2, then the triangle is not a right triangle. In nABC, if c2 < a2 b2 where c is the longest side of the triangle, then the triangle is an acute triangle. Thus in Fig. 7-34, 92 < 62 82 (that is, 81 < 100); hence, nABC is an acute triangle. In nABC, if c2 > a2 b2 where c is the longest side of the triangle, then the triangle is an obtuse triangle. Thus in Fig. 7-35, 112 > 62 82 (that is, 121 > 100); hence, nABC is an obtuse triangle.
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Fig. 7-34
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Fig. 7-35
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CHAPTER 7 Similarity
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SOLVED PROBLEMS
7.33 Finding the sides of a right triangle In Fig. 7-36, (a) find the length of hypotenuse c if a (c) find b if a 4 23 and c 8.
12 and b
9; (b) find a if b
6 and c
Fig. 7-36
Solutions
(a) c2 (b) a2 (c) b2 a2 c2 c2 b2 b2 a2 122 82 82 92 62 225 and c 28 and a 64 15. 2 27. 48 16 and b 4.
(4 23)2
7.34 Ratios in a right triangle In a right triangle, the hypotenuse has length 20 and the ratio of the two arms is 3:4. Find each arm.
Solution
Let the lengths of the two arms be denoted by 3x and 4x. Then 202 (3x)2 (4x)2. Multiplying out, we get 400 9x2 16x2 or 400 25x2, and x 4; hence, the arms have lengths 12 and 16.
7.35 Applying the pythagorean theorem to an isosceles triangle Find the length of the altitude to the base of an isosceles triangle if the base is 8 and the equal sides are 12.
Solution
The altitude h of an isosceles triangle bisects the base (Fig. 7-37). Then h2 h 8 22. a2 (1b)2 2 122 42 128 and
7.36 Applying the pythagorean theorem to a rhombus In a rhombus, find (a) the length of a side s if the diagonals are 30 and 40; (b) the length of a diagonal d if a side is 26 and the other diagonal is 20.
Solution
The diagonals of a rhombus are perpendicular bisectors of each other; hence, 1 1 s2 (2d)2 (2d )2 in Fig. 7-38. (a) If d (b) If s 30 and d 26 and d 40, then s2 20, then 262 152 (2d)2
625 or s
25. (2d)2. Thus, 2d
Fig. 7-37
102 or 576
24 or d
Fig. 7-38
CHAPTER 7 Similarity
7.37 Applying the pythagorean theorem to a trapezoid Find x in each part of Fig. 7-39 if ABCD is a trapezoid.
Fig. 7-39
Solutions
The dashed perpendiculars in the diagrams are additional segments needed only for the solutions. Note how rectangles are formed by these added segments. (a) EF (b) b
BC 25
12 and AE 7
5. Then x2 b
144 or x
12. 242 102 or x 26.
576 or b
24; also, BE
24 and CE
17 7
10. Then x2
7.38 Applying the pythagorean theorem to a circle (a) Find the distance d from the center of a circle of radius 17 to a chord whose length is 30 [Fig. 7-40(a)]. (b) Find the length of a common external tangent to two externally tangent circles with radii 4 and 9 [Fig. 7-40(b)].
Fig. 7-40
Solutions
(a) BC
1 2(30)
15. Then d2
64 and d 9 4
8. 5. Then in right nOSQ, (OS)2 132 52 144 so
(b) OS > PR, RS 4, OQ 13, and SQ OS 12; hence PR 12.
7.10 Special Right Triangles
7.10A The 30 -60 -90 Triangle
A 30 -60 -90 triangle is one-half an equilateral triangle. Thus, in right nABC (Fig. 7-41), a that c 2; then a 1, and the Pythagorean Theorem gives b2 c2 a2 22 12 3 or b 23
c. Consider
CHAPTER 7 Similarity
The ratio of the sides is then a:b:c
1: 23:2
Fig. 7-41
Principles of the 30 -60 -90 Triangle
PRINCIPLE 1:
The length of the leg opposite the 30 angle equals one-half the length of the hypotenuse.
1 2c.
In Fig. 7-41, a
PRINCIPLE 2:
The length of the leg opposite the 60 angle equals one-half the length of the hypotenuse times the square root of 3.
1 2c 23.
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