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Bridge arch box mathematics
Solution: suppose that the center of AB is point O, connecting OA, OM, OD,

∵AB=24m,CD=8m,AB⊥CD,

∴ad=db= 12ab= 12×24= 12m.

Let OA=r, then OD=r-CD,

At Rt△AOD,

OA2=OD2+AD2, that is, r2=(r-8)2+ 122, and the solution is r= 13m.

∴OD= 13-8=5m,

In Rt△MOE, suppose ME=5m,

Then OM2=OE2+ME2, that is, 132=OE2+52, and the solution is OE= 12m.

∴DE=OE-OD= 12-5=7m>6m.

The freighter can cross the bridge smoothly.