高中數(shù)學(xué)第一章相似三角形定理與圓冪定理1.1.2相似三角形的性質(zhì)課件新人教B版選修4_1

1、-1-1.1.2相似三角形的性質(zhì)ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航1.掌握相似三角形的性質(zhì).2.能利用相似三角形的性質(zhì)解決有關(guān)問(wèn)題.ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航相似三角形的性質(zhì)(1)相似三角形對(duì)應(yīng)角相等,對(duì)應(yīng)邊成比例;(2)相似三角形對(duì)應(yīng)高的比、對(duì)應(yīng)中線的比和對(duì)應(yīng)角平分線的比都等于相似比;(3)相似三角形周長(zhǎng)的比等于

2、相似比;(4)相似三角形面積的比等于相似比的平方;(5)相似三角形外接(內(nèi)切)圓的直徑比、周長(zhǎng)比等于相似比,外接(內(nèi)切)圓的面積比等于相似比的平方.ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航【做一做1】 ABCABC,AB=4,AB=3,那么BC和BC上對(duì)應(yīng)中線的比等于()答案:A ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航答案:B ZH

3、ISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航答案:C ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航相似三角形性質(zhì)和全等三角形性質(zhì)的比較剖析如下表所示. ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航ZHISHI SHULI

4、知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航題型一題型二題型三【例1】 ABCABC,ABC的周長(zhǎng)為60 cm,ABC的周長(zhǎng)為72 cm,AB=15 cm,BC=24 cm.求:(1)BC,AB;(2)AC,AC.分析先由相似三角形周長(zhǎng)的比得到相似比,再利用相似比求解.ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航題型一題型二題型三ZHISHI SHULI知識(shí)梳理ZH

5、ONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航題型一題型二題型三分析由于四邊形BCDE是不規(guī)那么四邊形,直接求其面積有困難,因此轉(zhuǎn)化為求ABC與ADE面積的差.ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航題型一題型二題型三ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHA

6、NG目標(biāo)導(dǎo)航題型一題型二題型三【例3】 某施工隊(duì)在道路拓寬施工時(shí)遇到這樣一個(gè)問(wèn)題,馬路旁邊原有一塊三角形綠化地ABC,由于馬路拓寬,綠化地被削去了一個(gè)角,變成了一個(gè)面積為84 m2,周長(zhǎng)為48 m的梯形BCED,原綠化地一邊AB的長(zhǎng)由原來(lái)的30 m縮短成18 m.那么被削去的局部面積有多大?分析利用相似三角形的性質(zhì),列方程進(jìn)展求解即可.ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航題型一題型二題型三反思反思此類問(wèn)題是利用數(shù)學(xué)模型解決實(shí)際問(wèn)題,關(guān)鍵在于認(rèn)真分析題意轉(zhuǎn)化成數(shù)

7、學(xué)問(wèn)題,利用相似三角形求解. ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航12345解析:ABC和ABC對(duì)應(yīng)角平分線的比等于它們內(nèi)切圓直徑的比,應(yīng)選D.答案:DZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航12345ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI

8、典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航123453.一條河的兩岸是平行的一條河的兩岸是平行的,在河的這一岸每隔在河的這一岸每隔5 m有一棵樹(shù)有一棵樹(shù),在河的在河的對(duì)岸每隔對(duì)岸每隔50 m有一根電線桿有一根電線桿,在這岸離岸邊在這岸離岸邊25 m處看對(duì)岸處看對(duì)岸,看到對(duì)看到對(duì)岸相鄰的兩根電線桿恰好被這岸的兩棵樹(shù)遮住岸相鄰的兩根電線桿恰好被這岸的兩棵樹(shù)遮住,并且在這兩棵樹(shù)之并且在這兩棵樹(shù)之間還有兩棵樹(shù)間還有兩棵樹(shù),那么河的寬度為那么河的寬度為m. 解析解析:如圖如圖,A,B是相鄰兩電線桿的底部是相鄰兩電線桿的底部,F,G中間還有兩棵樹(shù)中間還有兩棵樹(shù),那么那么AB=50 m,FG=35=15(m

9、),EC=25 m,CDAB,ABFG,ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航123454.如圖如圖,在在 ABCD中中,AE EB=1 2,AEF的面積為的面積為6,那么那么ADF的面積為的面積為. 答案答案:18ZHISHI SHULI知識(shí)梳理ZHONGNAN JVJIAO重難聚焦SUITANGYANLIAN隨堂演練DIANLI TOUXI典例透析MUBIAODAOHANG目標(biāo)導(dǎo)航123455.如圖如圖,D是是ABC中中AB邊上一點(diǎn)邊上一點(diǎn),DEBC,且交且交AC于點(diǎn)于點(diǎn)E,EFAB,且交且交BC于點(diǎn)于點(diǎn)F,且且SADE=1,SEFC=4,那么四邊形那么四邊形BFED的面積等于多少的面