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# (a 0(b 1(c 2(d 3(e 4(f 5 solution key 2.8 solution

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Unformatted text preview: (a) 0 (b) 1 (c) 2 (d) 3 (e) 4 (f) 5 Solution Key: 2.8 Solution: 3.8 Question 9: Label the consequtive edges of an Euler circuit in the following graph: Solution Key: 2.9 Solution: 3.9 8 Question 10: True or False. Give a reason or a counter-example. (1) There exists a planar graph with 6 vertices, 10 edges, and 6 faces. (2) A planar graph consisting of 4 connected pieces must have Euler characteristic 5. (3) If a connected planar graph has 6 edges and splits the plane into 7 regions, then it must have 1 vertex. (a) (1) (2) (3) T T T (b) (1) (2) (3) T F T (c) (1) (2) (3) F T F (d) (1) (2) (3) F F F (e) (1) (2) (3) T F F (f) (1) (2) (3) T T F Solution Key: 2.10 Solution: 3.10 9 2 Solution key (1) (d) (2) (a) (3) (e) (4) (a) (5) (e) (6) (f) (7) (c) (8) (c) (9) see solution (10) (a) 10 3 Solutions Solution of problem 1.1: The altitude splits teh hypothenuse into two segments. If we denote the length of the shorter segment by x , then the length of the other segment will be 25- x . x 12 25- x a b If we denote the corresponding sides of the triangle by a and b respec- tively we can apply the Pythagorean theorem to two small triangles and to the big triangle: x 2 + 12 2 = a 2 (25- x ) 2 + 12 2 = b 2 a 2 + b 2 = 25 2 Substituting the first two identities into the third one we get x 2 + (25- x ) 2 + 288 = 625 or x 2 + 625- 50 x + x 2 + 288 = 625 . After cancelling 625 from both sides and dividing both sides by 2 we get the equation x 2- 25 x + 144 = 0 . By the quadratic formula we have x = 25 ± √ 25 2- 4 · 144 2 = 25 ± √ 49 2 = 25 ± 7 2 . 11 Thus x must be equal to either 9 or 16 , and since x was the shorter of the two segments we must have x = 9. Substituting this in the two identities involving a and b above we get a 2 = x 2 + 12 2 = 81 + 144 = 225 , b 2 = (25- x ) 2 + 12 2 = 256 + 144 = 400 . Therefore we have a = 15, b = 20, and a + b = 35. The correct answer is (d) . square Solution of problem 1.2: Denote the longer side of R 1 by x . Since R 1 is a golden rectangle we have that x/ 4 must be equal to the golden ratio ϕ , i.e. x = 4 ϕ. In particular R 1 has sides 4 and 4 ϕ , and R 2 has sides 4 ϕ and 4 + 4 ϕ . We can now compute the ratio of areas: area( R 2 ) area( R 1 ) = 4 ϕ · (4 + 4 ϕ ) 4 · 4 ϕ = 16 ϕ (1 + ϕ ) 16 ϕ = 1 + ϕ. The correct answer is (a) . square Solution of problem 1.3: Slicing off a vertex of cube creates a new trian- gular face. Also, when we slice off a vertex all the original edges remain edges and we introduce three new edges - the sides of teh newly cre- ated triangular face. Therefore, when we slice off the four top edges we introduce 4 · 3 = 12 new edges. Adding these to the original 12 edges the cube has we get a total of 24 edges for the new solid. The correctthe cube has we get a total of 24 edges for the new solid....
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(a 0(b 1(c 2(d 3(e 4(f 5 Solution Key 2.8 Solution 3.8...

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