Question

在邊長(zhǎng)為2cm的正方形ABCD中，點(diǎn)Q為BC邊的中點(diǎn)，點(diǎn)P為對(duì)角線AC上一動(dòng)點(diǎn)，連結(jié)PB.PQ，則△PBQ周長(zhǎng)的最小值為_(kāi)__cm(結(jié)果不取近似值)．

Answer 1

+1

由題,連接PD,由正方形的對(duì)稱性知PD=PB,所以△PBQ周長(zhǎng)=BQ+PB+PQ=PD+PQ+BQ,當(dāng)點(diǎn)PDQ共線時(shí), △PBQ周長(zhǎng)最短,連接DQ與AC相交于點(diǎn)P,因?yàn)锽C="2cm," 點(diǎn)Q為BC邊的中點(diǎn),所以CQ=1,在Rt△DCQ中,CD=2,CQ=1,由勾股定理知DQ=cm,所以△PBQ周長(zhǎng)的最小值為(+1)cm.

試題分析：求兩條線段和的最小值,一般是利用對(duì)稱性將兩條線段化成一條折線段,當(dāng)折線段變成直線段時(shí),此時(shí)兩條線段的和最短,由題,連接PD,由正方形的對(duì)稱性知PD=PB,所以△PBQ周長(zhǎng)=BQ+PB+PQ=PD+PQ+
BQ,當(dāng)點(diǎn)PDQ共線時(shí), △PBQ周長(zhǎng)最短,連接DQ與AC相交于點(diǎn)P,因?yàn)锽C="2cm," 點(diǎn)Q為BC邊的中點(diǎn),所以CQ=1,在Rt△DCQ中,CD=2,CQ=1,由勾股定理知DQ=cm,所以△PBQ周長(zhǎng)的最小值為(+1)cm.