Question

以▱ABCD的四條邊為邊，在其形外分別作正方形，如圖，連接EF、GH、IJ、KL．若▱ABCD的面積為5，則圖中陰影部分四個三角形的面積和為（　　）                                                              

                                                                         

A．5                            B．10                           C．15                          D．20

Answer 1

B【考點】平行四邊形的性質．                                                                

【分析】過D作DN⊥AB于N，過E作EM⊥FA交FA延長線于M，連接AC，BD，求出∠EAM=∠BAD，根據銳角三角形函數定義求出EM=DN，求出△AEF和△ABD面積相等，同理求出S_△BHG=S_△ABC，S_△CIJ=S_△CBD，S_△DLK=S_△DAC，代入S=S_△AEF+S_△BGH+S_△CIJ+S_△DLK得出S=2S_{平行四邊形ABCD}，代入求出即可．                

【解答】解：過D作DN⊥AB于N，過E作EM⊥FA交FA延長線于M，連接AC，BD，                  

∵四邊形ABGF和四邊形ADLE是正方形，                                             

∴AE=AD，AF=AB，∠FAB=∠EAD=90°，                                               

∴∠EAF+∠BAD=360°﹣90°﹣90°=180°，                                                

∵∠EAF+∠EAM=180°，                                                                         

∴∠EAM=∠DAN，                                                                           

∴sin∠EAM=，sin∠DAN=，                                                          

∵AE=AD，                                                                                        

∴EM=DN，                                                                                       

∵S_△AEF=AF×EM，S_△ADB=AB×DN，                                                  

∴S_△AEF=S_△ABD，                                                                              

同理S_△BHG=S_△ABC，S_△CIJ=S_△CBD，S_△DLK=S_△DAC，                                     

∴陰影部分的面積S=S_△AEF+S_△BGH+S_△CIJ+S_△DLK=2S_{平行四邊形ABCD}=2×5=10．                

故選B．                                                                                            

                                                                         

【點評】本題考查了平行四邊形的性質，銳角三角函數的定義，三角形的面積等知識點的應用，主要考查學生運用定理進行推理和計算的能力，題目比較好，但有一定的難度．