如何在初中数学教学中培养学生合情推理与演绎推理能力
2014-12-04
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S = 1/2 * absinC
= 1/2 *ab√[1-(COSC)2]
1-(COSC)2 = 1 - [(A2 + B2-C2)/(2AB)] 2
= [(2AB)2-(A2 + B2-C2)^ 2] /(2AB)2
=(2AB + A2 + B2-C2)(2AB-A2-B2 + C2)/( 2AB)2
= [(A + B)2-C2] {C2-(AB)2] /(2AB)2
=(A + B + C)(A + BC)(C + AB )(CA + B)/〔4(AB)2]
= 4 *(A + B + C)/ 2 *(A + BC)/ 2 *(C + AB)/ 2 *(B + CA )/ 2] /(AB)2
P =(A + B + C)/ 2 --->(A + BC)/ 2 =(A + B + C)/ 2-C =件; (C + AB)/ 2)= PB; (B + CA)/ 2 = PA
S = 1/2 * AB *√[4P(PA)(PB)(PC)/(AB)2]
=√[P(PA)(PB )(PC)]
= 1/2 *ab√[1-(COSC)2]
1-(COSC)2 = 1 - [(A2 + B2-C2)/(2AB)] 2
= [(2AB)2-(A2 + B2-C2)^ 2] /(2AB)2
=(2AB + A2 + B2-C2)(2AB-A2-B2 + C2)/( 2AB)2
= [(A + B)2-C2] {C2-(AB)2] /(2AB)2
=(A + B + C)(A + BC)(C + AB )(CA + B)/〔4(AB)2]
= 4 *(A + B + C)/ 2 *(A + BC)/ 2 *(C + AB)/ 2 *(B + CA )/ 2] /(AB)2
P =(A + B + C)/ 2 --->(A + BC)/ 2 =(A + B + C)/ 2-C =件; (C + AB)/ 2)= PB; (B + CA)/ 2 = PA
S = 1/2 * AB *√[4P(PA)(PB)(PC)/(AB)2]
=√[P(PA)(PB )(PC)]
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