Pre-Calc Exam Notes 62

Pre-Calc Exam Notes 62 - b a c A B C O D r E F Figure 2.5.6...

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62 Chapter 2 General Triangles §2.5 Theorem 2.5 can be used to derive another formula for the area of a triangle: Theorem 2.8. For a triangle ABC , let K be its area and let R be the radius of its circum- scribed circle. Then K = abc 4 R (and hence R = abc 4 K ) . (2.36) To prove this, note that by Theorem 2.5 we have 2 R = a sin A = b sin B = c sin C sin A = a 2 R , sin B = b 2 R , sin C = c 2 R . Substitute those expressions into formula (2.26) from Section 2.4 for the area K : K = a 2 sin B sin C 2 sin A = a 2 · b 2 R · c 2 R 2 · a 2 R = abc 4 R QED Combining Theorem 2.8 with Heron’s formula for the area of a triangle, we get: Corollary 2.9. For a triangle ABC , let s = 1 2 ( a + b + c ). Then the radius R of its circum- scribed circle is R = abc 4 r s ( s a )( s b )( s c ) . (2.37) In addition to a circumscribed circle, every triangle has an inscribed circle , i.e. a circle to which the sides of the triangle are tangent, as in Figure 2.5.6.
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Unformatted text preview: b a c A B C O D r E F Figure 2.5.6 Inscribed circle for ABC Let r be the radius of the inscribed circle, and let D , E , and F be the points on AB , BC , and AC , respectively, at which the circle is tangent. Then OD AB , OE BC , and OF AC . Thus, OAD and OAF are equivalent triangles, since they are right triangles with the same hypotenuse OA and with corresponding legs OD and OF of the same length r . Hence, OAD = OAF , which means that OA bisects the angle A . Similarly, OB bisects B and OC bisects C . We have thus shown: For any triangle, the center of its inscribed circle is the intersection of the bisectors of the angles....
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This note was uploaded on 01/21/2012 for the course MAC 1130 taught by Professor Dr.cheun during the Fall '11 term at FSU.

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