【答案】
分析:證明△BO′A≌△BOC,又∠OBO′=60°,所以△BO′A可以由△BOC繞點(diǎn)B逆時(shí)針旋轉(zhuǎn)60°得到,故結(jié)論①正確;
由△OBO′是等邊三角形,可知結(jié)論②正確;
在△AOO′中,三邊長(zhǎng)為3,4,5,這是一組勾股數(shù),故△AOO′是直角三角形;進(jìn)而求得∠AOB=150°,故結(jié)論③正確;

=S
△AOO′+S
△OBO′=6+4

,故結(jié)論④錯(cuò)誤;
如圖②,將△AOB繞點(diǎn)A逆時(shí)針旋轉(zhuǎn)60°,使得AB與AC重合,點(diǎn)O旋轉(zhuǎn)至O″點(diǎn).利用旋轉(zhuǎn)變換構(gòu)造等邊三角形與直角三角形,將S
△AOC+S
△AOB轉(zhuǎn)化為S
△COO″+S
△AOO″,計(jì)算可得結(jié)論⑤正確.
解答:
解:由題意可知,∠1+∠2=∠3+∠2=60°,∴∠1=∠3,
又∵OB=O′B,AB=BC,
∴△BO′A≌△BOC,又∵∠OBO′=60°,
∴△BO′A可以由△BOC繞點(diǎn)B逆時(shí)針旋轉(zhuǎn)60°得到,
故結(jié)論①正確;
如圖①,連接OO′,
∵OB=O′B,且∠OBO′=60°,
∴△OBO′是等邊三角形,
∴OO′=OB=4.
故結(jié)論②正確;
∵△BO′A≌△BOC,∴O′A=5.
在△AOO′中,三邊長(zhǎng)為3,4,5,這是一組勾股數(shù),
∴△AOO′是直角三角形,∠AOO′=90°,
∴∠AOB=∠AOO′+∠BOO′=90°+60°=150°,
故結(jié)論③正確;

=S
△AOO′+S
△OBO′=

×3×4+

×4
2=6+4

,

故結(jié)論④錯(cuò)誤;
如圖②所示,將△AOB繞點(diǎn)A逆時(shí)針旋轉(zhuǎn)60°,使得AB與AC重合,點(diǎn)O旋轉(zhuǎn)至O″點(diǎn).
易知△AOO″是邊長(zhǎng)為3的等邊三角形,△COO″是邊長(zhǎng)為3、4、5的直角三角形,
則S
△AOC+S
△AOB=S
四邊形AOCO″=S
△COO″+S
△AOO″=

×3×4+

×3
2=6+

,
故結(jié)論⑤正確.
綜上所述,正確的結(jié)論為:①②③⑤.
故選A.
點(diǎn)評(píng):本題考查了旋轉(zhuǎn)變換中等邊三角形,直角三角形的性質(zhì).利用勾股定理的逆定理,判定勾股數(shù)3、4、5所構(gòu)成的三角形是直角三角形,這是本題的要點(diǎn).在判定結(jié)論⑤時(shí),將△AOB向不同方向旋轉(zhuǎn),體現(xiàn)了結(jié)論①-結(jié)論④解題思路的拓展應(yīng)用.