8如图长方体ABCD—A1BlClD1中AD=3AA1=4AB=5则从A点沿表面到C1的最短距离为______
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![8如图长方体ABCD—A1BlClD1中AD=3AA1=4AB=5则从A点沿表面到C1的最短距离为______](/uploads/image/z/4105492-52-2.jpg?t=8%E5%A6%82%E5%9B%BE%26%2357349%3B%E9%95%BF%E6%96%B9%E4%BD%93ABCD%E2%80%94A1BlClD1%E4%B8%AD%26%2357349%3BAD%3D3%26%2357349%3BAA1%3D4%26%2357349%3BAB%3D5%26%2357349%3B%E5%88%99%E4%BB%8EA%E7%82%B9%E6%B2%BF%E8%A1%A8%E9%9D%A2%E5%88%B0C1%E7%9A%84%E6%9C%80%E7%9F%AD%E8%B7%9D%E7%A6%BB%E4%B8%BA______)
8如图长方体ABCD—A1BlClD1中AD=3AA1=4AB=5则从A点沿表面到C1的最短距离为______
8如图长方体ABCD—A1BlClD1中AD=3AA1=4AB=5则从A点沿表面到C1的最短距离为______
8如图长方体ABCD—A1BlClD1中AD=3AA1=4AB=5则从A点沿表面到C1的最短距离为______
A1B1=AB
AD=B1C1
根据勾股定理,得AC1²=AA1²+(A1B1+B1C1)²=4²+(5+3)²=80,
所以AC1=√80=4√5,
答:从 A 点沿表面到 Cl的最短距离为4√5
相当于把长方体展开,A C1在同一平面
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