求解幾何(請用國中方法)?
回答 (1)
2015-12-28 10:45 pm
1)
易知∠ACD=∠BCD=∠DAB=∠DBA=45°,以D為旋轉中心,把△ACD順時針轉到AD重合BD,記C新位置是C' , 則
△CDC' 為等腰直角△,CD = DC', 故四邊形ADBC = CD×DC'/2 = CD²/2.
2a)
由1)的結論知 DP²/2 = 四邊形APBD = (AP×PB + AD×BD)/2
DP² = AP×PB + AD×BD
DP² = 2√3 ×2 + 2√2 ×2√2
DP = √(8 + 4√3) = √2 + √6
b)
△PBD - △PCA = 四邊形APBD - 四邊形ACPD = DP²/2 - AP²/2 = (4 + 2√3) - (2√3)²/2 = 2√3 - 2.
易知∠ACD=∠BCD=∠DAB=∠DBA=45°,以D為旋轉中心,把△ACD順時針轉到AD重合BD,記C新位置是C' , 則
△CDC' 為等腰直角△,CD = DC', 故四邊形ADBC = CD×DC'/2 = CD²/2.
2a)
由1)的結論知 DP²/2 = 四邊形APBD = (AP×PB + AD×BD)/2
DP² = AP×PB + AD×BD
DP² = 2√3 ×2 + 2√2 ×2√2
DP = √(8 + 4√3) = √2 + √6
b)
△PBD - △PCA = 四邊形APBD - 四邊形ACPD = DP²/2 - AP²/2 = (4 + 2√3) - (2√3)²/2 = 2√3 - 2.
收錄日期: 2021-04-24 23:38:34
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https://hk.answers.yahoo.com/question/index?qid=20151228073311AAuIIXy