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数学
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如图,⊙O的直径AB与弦EF相交于点P,交角为45°,若PE
2
+PF
2
=8,则AB等于______.
人气:478 ℃ 时间:2019-12-15 02:56:04
解答
作OG⊥EF于G,连接OE,
根据垂径定理,可设EG=FG=x,则PE=x+PG,PF=x-PG,
又∵PE
2
+PF
2
=8,
∴(x+PG)
2
+(x-PG)
2
=8,
整理得2x
2
+2PG
2
=8,x
2
+PG
2
=4,
∵交角为45°,
∴OG=PG,
∴OE
2
=OG
2
+EG
2
=4,
即圆的半径是2,
∴直径是4.
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