Adv Alegbra HW Solutions 143

Adv Alegbra HW Solutions 143 - angles 1 O SE and 1 OC E are...

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143 5.5 There is a circular castle whose diameter is unknown; it is pro- vided with four gates, and two lengths out of the north gate there is a large tree, which is visible from a point six lengths east of the south gate. What is the length of the diameter? S E N C O T 2 r r r a 6 Figure 5.7 The Castle Problem (i) Prove that the radius r of the castle is a root of the cubic X 3 + X 2 36. Solution. We compute the area of 1 TSE in two ways. On the one hand, area (1 TSE ) = 1 2 ( 2 + 2 r ) 6 = 6 + 6 r . On the other hand, this area is the sum of the areas of the three smaller triangles: 1 OSE ; 1 OCE ; 1 OCT . Now the right tri-
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Unformatted text preview: angles 1 O SE and 1 OC E are congruent (they have same hy-potenuse and a leg of equal length, namely, r ), and so they have the same area: 1 2 6 r = 3 r ; also, area (1 O SE ) = 1 2 ra , where a = | CT | . We conclude that 6 + 6 r = 3 r + 3 r + 1 2 ra ; that is, 12 = ra . The Pythagorean theorem (applied to 1 OCT ) gives r 2 + a 2 = ( r + 2 ) 2 = r 2 + 4 r + 4 , and so a 2 = 4 r + 4. Since 12 = ra , we have 144 r 2 = 4 r + 4 , and this simpli f es to r 3 + r 2 36 = 0....
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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