凸五边形ABCDE中,∠A=角∠B=120度,AB=BC=AE=2,CD=DE=4,求它的面积?看了题目的解析,它的解析是这样的,连接CE,延长CB、EA并于点P,∴∠EAB=∠ABC=120度,∴∠1=∠2=60度,∴△ABP为等边三角形,∴PB=PA=AB=2,∠
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![凸五边形ABCDE中,∠A=角∠B=120度,AB=BC=AE=2,CD=DE=4,求它的面积?看了题目的解析,它的解析是这样的,连接CE,延长CB、EA并于点P,∴∠EAB=∠ABC=120度,∴∠1=∠2=60度,∴△ABP为等边三角形,∴PB=PA=AB=2,∠](/uploads/image/z/5031003-3-3.jpg?t=%E5%87%B8%E4%BA%94%E8%BE%B9%E5%BD%A2ABCDE%E4%B8%AD%2C%E2%88%A0A%3D%E8%A7%92%E2%88%A0B%EF%BC%9D120%E5%BA%A6%2CAB%3DBC%3DAE%3D2%2CCD%3DDE%3D4%2C%E6%B1%82%E5%AE%83%E7%9A%84%E9%9D%A2%E7%A7%AF%3F%E7%9C%8B%E4%BA%86%E9%A2%98%E7%9B%AE%E7%9A%84%E8%A7%A3%E6%9E%90%2C%E5%AE%83%E7%9A%84%E8%A7%A3%E6%9E%90%E6%98%AF%E8%BF%99%E6%A0%B7%E7%9A%84%2C%E8%BF%9E%E6%8E%A5CE%2C%E5%BB%B6%E9%95%BFCB%E3%80%81EA%E5%B9%B6%E4%BA%8E%E7%82%B9P%2C%E2%88%B4%E2%88%A0EAB%3D%E2%88%A0ABC%3D120%E5%BA%A6%2C%E2%88%B4%E2%88%A01%EF%BC%9D%E2%88%A02%EF%BC%9D60%E5%BA%A6%2C%E2%88%B4%E2%96%B3ABP%E4%B8%BA%E7%AD%89%E8%BE%B9%E4%B8%89%E8%A7%92%E5%BD%A2%2C%E2%88%B4PB%3DPA%3DAB%3D2%2C%E2%88%A0)
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