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11. 如图,一架云梯$AB$斜靠在墙上,底端$B到墙角C的距离是7m$,若云梯上端下滑$4m$,则底端$B沿水平方向向外滑动8m$,试求云梯$AB$的长度。
解:设AC = xm,则DC = (x - 4)m,在Rt△ABC中,AB² = AC² + BC² = x² + 7²,在Rt△DEC中,DE² = DC² + CE² = (x - 4)² + (7 + 8)²,
故x² + 7² = (x - 4)² + (7 + 8)²,解得x = 24,所以AB = $\sqrt{24² + 7²}$ =
解:设AC = xm,则DC = (x - 4)m,在Rt△ABC中,AB² = AC² + BC² = x² + 7²,在Rt△DEC中,DE² = DC² + CE² = (x - 4)² + (7 + 8)²,
故x² + 7² = (x - 4)² + (7 + 8)²,解得x = 24,所以AB = $\sqrt{24² + 7²}$ =
25
(m),即云梯AB的长度为25m.答案
解:设AC = xm,则DC = (x - 4)m,在Rt△ABC中,AB² = AC² + BC² = x² + 7²,在Rt△DEC中,DE² = DC² + CE² = (x - 4)² + (7 + 8)²,
故x² + 7² = (x - 4)² + (7 + 8)²,解得x = 24,所以AB = $\sqrt{24² + 7²}$ = 25(m),即云梯AB的长度为25m.
故x² + 7² = (x - 4)² + (7 + 8)²,解得x = 24,所以AB = $\sqrt{24² + 7²}$ = 25(m),即云梯AB的长度为25m.
