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四邊形題目...
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在圖中,ABCD是一個正方形,而AFED是一個菱形.CAF是一條直線,CE與DF相交於P(a)求∠PED及∠PDE(b)證明PF=PChttp://i143.photobucket.com/albums/r141/tokomon123/20080807122.jpg在圖中,正方形ABCD的對角線AC與BD相交於O,E,F分別是AB和BC上的點使OE垂直OF(a)證明∠BOE=∠COF(b)證明△BOE=△COF(c)圖中是兩個邊長相等的正方形,其中PQRS的頂點P正是GHMN兩條對角線的交點.利用(b)部的結果找出:如果正方形PQRS繞P點旋轉,兩個... 顯示更多 在圖中,ABCD是一個正方形,而AFED是一個菱形.CAF是一條直線,CE與DF相交於P (a)求∠PED及∠PDE (b)證明PF=PC http://i143.photobucket.com/albums/r141/tokomon123/20080807122.jpg 在圖中,正方形ABCD的對角線AC與BD相交於O,E,F分別是AB和BC上的點使 OE垂直OF (a)證明∠BOE=∠COF (b)證明△BOE=△COF (c)圖中是兩個邊長相等的正方形,其中PQRS的頂點P正是GHMN兩條 對角線的交點.利用(b)部的結果找出:如果正方形PQRS繞P點旋轉,兩個 正方形重疊部分的面積總是等於一個正方形面積的幾分之幾? http://i143.photobucket.com/albums/r141/tokomon123/20080807123.jpg 更新: http://i143.photobucket.com/albums/r141/tokomon123/20080807124.jpg
在圖中,ABCD是一個正方形,而AFED是一個菱形.CAF是一條直線,CE與DF相交於P (a)求∠PED及∠PDE ∠AFD=∠ADF (isso △) ∠CAD+∠CAB =90 2∠CAD =90 (∠CAD=∠CAB) ∠CAD =45 ∠AFD+∠ADF =∠CAD (exterior ∠) 2∠ADF =∠CAD ∠ADF =22.5 ∠PDE =∠AFD (alt ∠ AF//DE) ∠PDE =22.5 FE =AD AD =DC FE =ED ∴ ED =DC ∠PED =∠PCD (isso △) ∠PED =∠PCF (alt ∠ CF//DE) ∠PCD +∠PCF=45 2∠PED=45 ∠PED=22.5 (b)證明PF=PC ∠PDE =∠PED △PDE is isso △CFP~△DPE (AAA) △CFP is isso PF=PC http://i143.photobucket.com/albums/r141/tokomon123/20080807122.jpg 在圖中,正方形ABCD的對角線AC與BD相交於O,E,F分別是AB和BC上的點使 OE垂直OF (a)證明∠BOE=∠COF ∠COB=90 ∠BOE =∠FOE -∠FOB =90 -∠FOB ∠COF =∠COB -∠FOB =90 -∠FOB =∠BOE (b)證明△BOE≡△COF ∠OCF =∠OBE =45 ∠BOE=∠COF from (a) OC=OB (diagonals of square) ∴△BOE≡△COF (AAS) (c)圖中是兩個邊長相等的正方形,其中PQRS的頂點P正是GHMN兩條 對角線的交點.利用(b)部的結果找出:如果正方形PQRS繞P點旋轉,兩個 正方形重疊部分的面積總是等於一個正方形面積的幾分之幾? ket A be the intersection of HM and PS, B be the intersection of GH and PQ △MPA≡△HPB (from b) the intersecting area =area of △APH +area of △HPB suppose at first PQRS is at vertex M area of △APH=area of △MPH area of △HPB=area of △MPA=0 total area =area of △APH +area of △HPB =area of △MPH as PQRS rotates about P from vertex M let z be the changing area of △MPA area of △APH=area of △MPH -area of △MPA =area of △MPH -z total area =area of △APH +area of △HPB =(area of △MPH -z) +z (∵area of △HPB =area of △MPA) =area of △MPH so as PQRS rotates about P, the intersecting area of the 2 squares is constant and the area is equal to △MPH =1/4 X area of square
其他解答:
四邊形題目...
發問:
在圖中,ABCD是一個正方形,而AFED是一個菱形.CAF是一條直線,CE與DF相交於P(a)求∠PED及∠PDE(b)證明PF=PChttp://i143.photobucket.com/albums/r141/tokomon123/20080807122.jpg在圖中,正方形ABCD的對角線AC與BD相交於O,E,F分別是AB和BC上的點使OE垂直OF(a)證明∠BOE=∠COF(b)證明△BOE=△COF(c)圖中是兩個邊長相等的正方形,其中PQRS的頂點P正是GHMN兩條對角線的交點.利用(b)部的結果找出:如果正方形PQRS繞P點旋轉,兩個... 顯示更多 在圖中,ABCD是一個正方形,而AFED是一個菱形.CAF是一條直線,CE與DF相交於P (a)求∠PED及∠PDE (b)證明PF=PC http://i143.photobucket.com/albums/r141/tokomon123/20080807122.jpg 在圖中,正方形ABCD的對角線AC與BD相交於O,E,F分別是AB和BC上的點使 OE垂直OF (a)證明∠BOE=∠COF (b)證明△BOE=△COF (c)圖中是兩個邊長相等的正方形,其中PQRS的頂點P正是GHMN兩條 對角線的交點.利用(b)部的結果找出:如果正方形PQRS繞P點旋轉,兩個 正方形重疊部分的面積總是等於一個正方形面積的幾分之幾? http://i143.photobucket.com/albums/r141/tokomon123/20080807123.jpg 更新: http://i143.photobucket.com/albums/r141/tokomon123/20080807124.jpg
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最佳解答:在圖中,ABCD是一個正方形,而AFED是一個菱形.CAF是一條直線,CE與DF相交於P (a)求∠PED及∠PDE ∠AFD=∠ADF (isso △) ∠CAD+∠CAB =90 2∠CAD =90 (∠CAD=∠CAB) ∠CAD =45 ∠AFD+∠ADF =∠CAD (exterior ∠) 2∠ADF =∠CAD ∠ADF =22.5 ∠PDE =∠AFD (alt ∠ AF//DE) ∠PDE =22.5 FE =AD AD =DC FE =ED ∴ ED =DC ∠PED =∠PCD (isso △) ∠PED =∠PCF (alt ∠ CF//DE) ∠PCD +∠PCF=45 2∠PED=45 ∠PED=22.5 (b)證明PF=PC ∠PDE =∠PED △PDE is isso △CFP~△DPE (AAA) △CFP is isso PF=PC http://i143.photobucket.com/albums/r141/tokomon123/20080807122.jpg 在圖中,正方形ABCD的對角線AC與BD相交於O,E,F分別是AB和BC上的點使 OE垂直OF (a)證明∠BOE=∠COF ∠COB=90 ∠BOE =∠FOE -∠FOB =90 -∠FOB ∠COF =∠COB -∠FOB =90 -∠FOB =∠BOE (b)證明△BOE≡△COF ∠OCF =∠OBE =45 ∠BOE=∠COF from (a) OC=OB (diagonals of square) ∴△BOE≡△COF (AAS) (c)圖中是兩個邊長相等的正方形,其中PQRS的頂點P正是GHMN兩條 對角線的交點.利用(b)部的結果找出:如果正方形PQRS繞P點旋轉,兩個 正方形重疊部分的面積總是等於一個正方形面積的幾分之幾? ket A be the intersection of HM and PS, B be the intersection of GH and PQ △MPA≡△HPB (from b) the intersecting area =area of △APH +area of △HPB suppose at first PQRS is at vertex M area of △APH=area of △MPH area of △HPB=area of △MPA=0 total area =area of △APH +area of △HPB =area of △MPH as PQRS rotates about P from vertex M let z be the changing area of △MPA area of △APH=area of △MPH -area of △MPA =area of △MPH -z total area =area of △APH +area of △HPB =(area of △MPH -z) +z (∵area of △HPB =area of △MPA) =area of △MPH so as PQRS rotates about P, the intersecting area of the 2 squares is constant and the area is equal to △MPH =1/4 X area of square
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