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如圖10.平行四邊形ABCD中.AB=5.BC=10.BC邊上的高AM=4.E為 BC邊上的一個動點(不與B.C重合).過E作直線AB的垂線.垂足為F. FE與DC的延長線相交于點G.連結(jié)DE.DF.. (1) 求證:ΔBEF ∽ΔCEG. (2) 當(dāng)點E在線段BC上運(yùn)動時.△BEF和△CEG的周長之間有什么關(guān)系?并說明你的理由. (3)設(shè)BE=x.△DEF的面積為 y.請你求出y和x之間的函數(shù)關(guān)系式.并求出當(dāng)x為何值時,y有最大值.最大值是多少? (1) 因為四邊形ABCD是平行四邊形. 所以 1分 所以 所以 ···························································································· 3分 (2)的周長之和為定值.····························································· 4分 理由一: 過點C作FG的平行線交直線AB于H . 因為GF⊥AB.所以四邊形FHCG為矩形.所以 FH=CG.FG=CH 因此.的周長之和等于BC+CH+BH 由 BC=10.AB=5.AM=4.可得CH=8.BH=6. 所以BC+CH+BH=24 ···························································································· 6分 理由二: 由AB=5.AM=4.可知 在Rt△BEF與Rt△GCE中.有: . 所以.△BEF的周長是. △ECG的周長是 又BE+CE=10.因此的周長之和是24.·········································· 6分 (3)設(shè)BE=x.則 所以 ···································· 8分 配方得:. 所以.當(dāng)時.y有最大值.············································································· 9分 最大值為.··········································································································· 10分 查看更多

 

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如圖,在等腰梯形ABCD中,AD∥BC.O是CD邊的中點,以O(shè)為圓心,OC長為半徑作圓,交BC邊于點E.過E作EH⊥AB,垂足為H.已知⊙O與AB邊相切,切點為F.
(1)求證:OE∥AB;
(2)求證:EH=
12
AB;
(3)若AD與⊙O也相切,如圖二,已知BE(BC)=5,BH=3,求⊙O的半徑.
(江蘇蘇州10年中考27題改編)
精英家教網(wǎng)

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如圖,在等腰梯形ABCD中,AD∥BC.O是CD邊的中點,以O(shè)為圓心,OC長為半徑作圓,交BC邊于點E.過E作EH⊥AB,垂足為H.已知⊙O與AB邊相切,切點為F.
(1)求證:OE∥AB;
(2)求證:EH=AB;
(3)若AD與⊙O也相切,如圖二,已知BE(BC)=5,BH=3,求⊙O的半徑.
(江蘇蘇州10年中考27題改編)

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已知如27題圖,點B、E分別是在AC、DF上的點,且BD、CE均與AF相交,若∠1=∠2,∠C=∠D,試問∠A與∠F相等嗎?仿照27題說明理由.

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如圖,在等腰梯形ABCD中,AD∥BC.O是CD邊的中點,以O(shè)為圓心,OC長為半徑作圓,交BC邊于點E.過E作EH⊥AB,垂足為H.已知⊙O與AB邊相切,切點為F.
(1)求證:OE∥AB;
(2)求證:EH=AB;
(3)若AD與⊙O也相切,如圖二,已知BE(BC)=5,BH=3,求⊙O的半徑.
(江蘇蘇州10年中考27題改編)

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如圖,在等腰梯形ABCD中,AD∥BC.O是CD邊的中點,以O(shè)為圓心,OC長為半徑作圓,交BC邊于點E.過E作EH⊥AB,垂足為H.已知⊙O與AB邊相切,切點為F.
(1)求證:OE∥AB;
(2)求證:EH=AB;
(3)若AD與⊙O也相切,如圖二,已知BE(BC)=5,BH=3,求⊙O的半徑.
(江蘇蘇州10年中考27題改編)

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