P为矩形ABCD内任意一点,若PA=3,PB=4,PC=5,则PD长是多少?
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![P为矩形ABCD内任意一点,若PA=3,PB=4,PC=5,则PD长是多少?](/uploads/image/z/7225953-33-3.jpg?t=P%E4%B8%BA%E7%9F%A9%E5%BD%A2ABCD%E5%86%85%E4%BB%BB%E6%84%8F%E4%B8%80%E7%82%B9%2C%E8%8B%A5PA%3D3%2CPB%3D4%2CPC%3D5%2C%E5%88%99PD%E9%95%BF%E6%98%AF%E5%A4%9A%E5%B0%91%3F)
P为矩形ABCD内任意一点,若PA=3,PB=4,PC=5,则PD长是多少?
P为矩形ABCD内任意一点,若PA=3,PB=4,PC=5,则PD长是多少?
P为矩形ABCD内任意一点,若PA=3,PB=4,PC=5,则PD长是多少?
过P做两边的垂线,交AB、BC、CD、DA于EFGH
ABCD是矩形,所以PE=BF,PF=BE,PG=CF,DF=AE
AP^2=AE^2+BF^2.①
BP^2=BE^2+BF^2.②
CP^2=BE^2+CF^2.③
DP^2=AE^2+CF^2.④
①-②+③
AP^2-BP^2+CP^2=AE^2+BF^2-(BE^2+BF^2)+BE^2+CF^2
=AE^2+CF^2=DP^2
所以DP^2=AP^2-BP^2+CP^2=9-16+25=18
DP=3√2
设AB//DC,AD//BC,A与C,B与D为对角,过P点作直线EG⊥AB、DC,交AB于E,DC于G;作HF⊥AD、BC,交AD于H,BC于F,则根据勾股定理,得
PA^2=AE^2+BF^2......(1)
PB^2=BE^2+BF^2......(2)
PC^2=BE^2+CF^2......(3)
(1)-(2)+(3),得
AE^2+CF^2=P...
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设AB//DC,AD//BC,A与C,B与D为对角,过P点作直线EG⊥AB、DC,交AB于E,DC于G;作HF⊥AD、BC,交AD于H,BC于F,则根据勾股定理,得
PA^2=AE^2+BF^2......(1)
PB^2=BE^2+BF^2......(2)
PC^2=BE^2+CF^2......(3)
(1)-(2)+(3),得
AE^2+CF^2=PA^2-PB^2+PC^2=3^2-4^2+5^2=18
PD^2=AE^2+CF^2=18
PD=3√2
答:PD=3√2
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