Question

如圖，在直角坐標(biāo)系中，直線AB經(jīng)點P（3，4），與坐標(biāo)軸正半軸相交于A，B兩點，當(dāng)△AOB的面積最小時，△AOB的內(nèi)切圓的半徑是（　�。�                                                                         

                                                                          

A．2                            B．3.5                          C．              D．4

　                                                                                                      

Answer 1

A【考點】三角形的內(nèi)切圓與內(nèi)心；坐標(biāo)與圖形性質(zhì)．                                   

【專題】壓軸題；探究型．                                                                     

【分析】設(shè)直線AB的解析式是y=kx+b，把P（3，4）代入求出直線AB的解析式是y=kx+4﹣3k，求出OA=4﹣3k，OB=，求出△AOB的面積是OBOA=12﹣=12﹣（9k+），根據(jù)﹣9k﹣≥2=24和當(dāng)且僅當(dāng)﹣9k=﹣時，取等號求出k=﹣，求出OA=4﹣3k=8，OB==6，設(shè)三角形AOB的內(nèi)切圓的半徑是R，由三角形面積公式得：×6×8=×6R+×8R+×10R，求出即可．                                               

【解答】解：設(shè)直線AB的解析式是y=kx+b，                                         

把P（3，4）代入得：4=3k+b，                                                              

b=4﹣3k，                                                                                          

即直線AB的解析式是y=kx+4﹣3k，                                                       

當(dāng)x=0時，y=4﹣3k，                                                                        

當(dāng)y=0時，x=，                                                                            

即A（0，4﹣3k），B（，0），                                                    

△AOB的面積是OBOA=（4﹣3k）=12﹣=12﹣（9k+），                   

∵要使△AOB的面積最小，                                                                     

∴必須最大，                                                                       

∵k＜0，                                                                                            

∴﹣k＞0，                                                                                        

∵﹣9k﹣≥2=2×12=24，                                                     

當(dāng)且僅當(dāng)﹣9k=﹣時，取等號，解得：k=±，                                            

∵k＜0，                                                                                            

∴k=﹣，                                                                                         

即OA=4﹣3k=8，OB==6，                                                            

根據(jù)勾股定理得：AB=10，                                                                      

設(shè)三角形AOB的內(nèi)切圓的半徑是R，                                                      

由三角形面積公式得：×6×8=×6R+×8R+×10R，                                    

R=2，                                                                                                

故選A．                                                                                            

【點評】本題考查了勾股定理，取最大值，三角形的面積，三角形的內(nèi)切圓等知識點的應(yīng)用，關(guān)鍵是求OA和OB的值，本題比較好，但是有一定的難度．