1.已知b分之a=c分之d=e分之f=2∕3且2a-c+5e=18求2b-d+5f 2.如图AB∕AD=AC∕AE﹦3C∕DE﹦3∕2且三角形ABC为3b求三角形ABE的周长3.三角形ABC的三边a、b、c满足(a-b):(c-b):(a+c)﹦7:1:18判断ABC的形状.
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![1.已知b分之a=c分之d=e分之f=2∕3且2a-c+5e=18求2b-d+5f 2.如图AB∕AD=AC∕AE﹦3C∕DE﹦3∕2且三角形ABC为3b求三角形ABE的周长3.三角形ABC的三边a、b、c满足(a-b):(c-b):(a+c)﹦7:1:18判断ABC的形状.](/uploads/image/z/8781613-61-3.jpg?t=1.%E5%B7%B2%E7%9F%A5b%E5%88%86%E4%B9%8Ba%3Dc%E5%88%86%E4%B9%8Bd%3De%E5%88%86%E4%B9%8Bf%3D2%E2%88%953%E4%B8%942a-c%2B5e%3D18%E6%B1%822b-d%2B5f+2.%E5%A6%82%E5%9B%BEAB%E2%88%95AD%3DAC%E2%88%95AE%EF%B9%A63C%E2%88%95DE%EF%B9%A63%E2%88%952%E4%B8%94%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%BA3b%E6%B1%82%E4%B8%89%E8%A7%92%E5%BD%A2ABE%E7%9A%84%E5%91%A8%E9%95%BF3.%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E4%B8%89%E8%BE%B9a%E3%80%81b%E3%80%81c%E6%BB%A1%E8%B6%B3%28a-b%29%EF%BC%9A%28c-b%29%EF%BC%9A%28a%2Bc%29%EF%B9%A67%EF%BC%9A1%EF%BC%9A18%E5%88%A4%E6%96%ADABC%E7%9A%84%E5%BD%A2%E7%8A%B6.)
1.已知b分之a=c分之d=e分之f=2∕3且2a-c+5e=18求2b-d+5f 2.如图AB∕AD=AC∕AE﹦3C∕DE﹦3∕2且三角形ABC为3b求三角形ABE的周长3.三角形ABC的三边a、b、c满足(a-b):(c-b):(a+c)﹦7:1:18判断ABC的形状.
1.已知b分之a=c分之d=e分之f=2∕3且2a-c+5e=18求2b-d+5f 2.如图AB∕AD=AC∕AE﹦3C∕DE﹦3∕2且三角形ABC为3b求三角形ABE的周长
3.三角形ABC的三边a、b、c满足(a-b):(c-b):(a+c)﹦7:1:18判断ABC的形状.
1.已知b分之a=c分之d=e分之f=2∕3且2a-c+5e=18求2b-d+5f 2.如图AB∕AD=AC∕AE﹦3C∕DE﹦3∕2且三角形ABC为3b求三角形ABE的周长3.三角形ABC的三边a、b、c满足(a-b):(c-b):(a+c)﹦7:1:18判断ABC的形状.
1)
因为a分之b=c分之d=e分之f=2∕3
所以(2b-d+5f)/(2a-c+5e)=b/a=d/c=f/e=2/3
又2a-c+5e=18
所以(2b-d+5f)/18=2/3
解得2b-d+5f=12
2)因为AB∕AD=AC∕AE=BC∕DE﹦3∕2
所以△ABC∽△ADE,且相似比为3/2
因为△ABC周长为36
所以△ABC周长:△ADE周长=36:△ADE周长=3:2
解得△ADE周长为24
3)因为(a-b):(c-b):(a+c)﹦7:1:18
所以(a-b)/7=(c-b)/1=(a+c)/18
设(a-b)/7=(c-b)/1=(a+c)/18=t,
则a-b=7t,
c-b=t
a+c=18t
解得a=12t,b=5t,c=6t
b+c