2次根式的解答/0.25〔a*2b*2-〔(a*2+b*2+c*2)0.5〕*2〕如何推导出/p(p-a)(p-b)(p-c) {p=〔a+b+c〕0.5}kkkkkkkkkkkkkk
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![2次根式的解答/0.25〔a*2b*2-〔(a*2+b*2+c*2)0.5〕*2〕如何推导出/p(p-a)(p-b)(p-c) {p=〔a+b+c〕0.5}kkkkkkkkkkkkkk](/uploads/image/z/15066077-5-7.jpg?t=2%E6%AC%A1%E6%A0%B9%E5%BC%8F%E7%9A%84%E8%A7%A3%E7%AD%94%EF%BC%8F0.25%E3%80%94a%EF%BC%8A2b%EF%BC%8A2-%E3%80%94%EF%BC%88a%EF%BC%8A2%2Bb%EF%BC%8A2%2Bc%EF%BC%8A2%EF%BC%890.5%E3%80%95%EF%BC%8A2%E3%80%95%E5%A6%82%E4%BD%95%E6%8E%A8%E5%AF%BC%E5%87%BA%EF%BC%8Fp%EF%BC%88p-a%EF%BC%89%EF%BC%88p-b%EF%BC%89%EF%BC%88p-c%EF%BC%89+%EF%BD%9Bp%EF%BC%9D%E3%80%94a%2Bb%2Bc%E3%80%950.5%EF%BD%9Dkkkkkkkkkkkkkk)
2次根式的解答/0.25〔a*2b*2-〔(a*2+b*2+c*2)0.5〕*2〕如何推导出/p(p-a)(p-b)(p-c) {p=〔a+b+c〕0.5}kkkkkkkkkkkkkk
2次根式的解答
/0.25〔a*2b*2-〔(a*2+b*2+c*2)0.5〕*2〕如何推导出/p(p-a)(p-b)(p-c) {p=〔a+b+c〕0.5}
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2次根式的解答/0.25〔a*2b*2-〔(a*2+b*2+c*2)0.5〕*2〕如何推导出/p(p-a)(p-b)(p-c) {p=〔a+b+c〕0.5}kkkkkkkkkkkkkk
设三角形的三边a、b、c的对角分别为A、B、C,则余弦定理为
cosC = (a^2+b^2-c^2)/2ab
S=1/2*ab*sinC
=1/2*ab*√(1-cos^2 C)
=1/2*ab*√[1-(a^2+b^2-c^2)^2/4a^2*b^2]
=1/4*√[4a^2*b^2-(a^2+b^2-c^2)^2]
=1/4*√[(2ab+a^2+b^2-c^2)(2ab-a^2-b^2+c^2)]
=1/4*√[(a+b)^2-c^2][c^2-(a-b)^2]
=1/4*√[(a+b+c)(a+b-c)(a-b+c)(-a+b+c)]
设p=(a+b+c)/2
则p=(a+b+c)/2,p-a=(-a+b+c)/2,p-b=(a-b+c)/2,p-c=(a+b-c)/2,
上式=√[(a+b+c)(a+b-c)(a-b+c)(-a+b+c)/16]
=√[p(p-a)(p-b)(p-c)]
所以,三角形ABC面积S=√[p(p-a)(p-b)(p-c)]