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○ 相似図形では3組の辺の長さの比は等しい.
![]() だから,BE:CE の比が分かっていれば AB:DC の比が分かる. ![]() |
【例題1】
次図1の平行四辺形 ABCD において AP:PQ:QC=2:4:3 のとき, (1) AF:FD を求めなさい. ![]() == 図1 == ![]() 図1-1 (解答) 右図1-1のように△APF と△CPB は相似だから,AF:CB=AP:CP=2:7 四角形ABCD は平行四辺形だから AD=BC AF:AD=2:7 ゆえに AF:FD=2:(7-2)=2:5…(答)
(2) CE:ED を求めなさい.
![]() 図1-2 (解答) 右図1-2のように△AQB と△CQE は相似だから,AB:CE=AQ:CQ=6:3=2:1 四角形ABCD は平行四辺形だから AB=CD CD:CE=2:1 ゆえに CE:ED=(2-1):1=1:1…(答) ⇒ このように相似図形を見つけると辺の比が求められる. |
【問題1】
![]() == 図2 == ![]() 四角形ABCD は平行四辺形だからAB=DC したがって,BE:AB=4:11 ゆえに,AE:EB=(11-4):4=7:4 ![]() 四角形ABCD は平行四辺形だからBC=AD したがって,AD:FD=2:1 ゆえに,AF:FD=(2-1):1=1:1 |
【問題2】
![]() == 図3 == ![]() 四角形ABCD は平行四辺形だからAB=DC したがって,AE:AB=3:7 ゆえに,AE:EB=3:(7-3)=3:4 ![]() 四角形ABCD は平行四辺形だからCD=AB したがって,CF:CD=1:4 ゆえに,CF:FD=1:(4-1)=1:3 |
【問題3】
![]() == 図4 == ![]() 四角形ABCD は平行四辺形だからAD=BC したがって,BE:BC=5:7 ゆえに,BE:EC=5:(7-5)=5:2 ![]() 四角形ABCD は平行四辺形だからBC=AD したがって,AD:FD=3:1 ゆえに,AF:FD=(3-1):1=2:1 |
【問題4】
![]() == 図5 == ![]() 四角形ABCD は平行四辺形だからAD=BC したがって,BE:BC=2:9 ゆえに,BE:EC=2:(9-2)=2:7 ![]() 四角形ABCD は平行四辺形だからAB=CD したがって,FD:CD=4:7 ゆえに,CF:FD=(7-4):4=3:4 |
【問題5】
![]() == 図6 == ![]() 四角形ABCD は平行四辺形だからAD=BC したがって,AF:AD=1:3 ゆえに,AF:FD=1:(3-1)=1:2 ![]() 四角形ABCD は平行四辺形だからAB=CD したがって,CE:CD=4:8=1:2 ゆえに,CE:ED=(2-1):1=1:1 |
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