题目内容
如图,在△ABC中,EF∥BC,EF=
BC=
cm,△AEF的周长为10
cm.
(1)求梯形BCFE的周长;
(2)S△AEF:S梯形BCFE等于多少?
| ||
3 |
2 |
2 |
(1)求梯形BCFE的周长;
(2)S△AEF:S梯形BCFE等于多少?
考点:相似三角形的判定与性质
专题:
分析:(1)首先得出△AEF∽△ABC,即可得出C△AEF:C△ABC=EF:BC,进而求出C△ABC=30cm,再利用C梯形BCFE=C△ABC-C△AEF+2EF得出答案;
(2)由(1)知,△AEF∽△ABC,利用EF:BC=
:3,得出S△AEF:S△ABC=2:9,再利用S梯形BCFE=S△ABC-S△AEF,则S△AEF:S梯形BCFE=S△AEF:(S△ABC-S△AEF)即可得出答案.
(2)由(1)知,△AEF∽△ABC,利用EF:BC=
2 |
解答:解:(1)∵EF∥BC,
∴△AEF∽△ABC,
∴C△AEF:C△ABC=EF:BC=
,
又∵△AEF的周长为10
cm,
∴C△ABC=30cm,
∴C梯形BCFE=C△ABC-C△AEF+2EF=30-10
+2
=(30-8
)cm;
(2)由(1)知,△AEF∽△ABC,且EF:BC=
:3,
∴S△AEF:S△ABC=2:9,
又S梯形BCFE=S△ABC-S△AEF,
∴S△AEF:S梯形BCFE=S△AEF:(S△ABC-S△AEF)=2:(9-2)=2:7.
∴△AEF∽△ABC,
∴C△AEF:C△ABC=EF:BC=
| ||
3 |
又∵△AEF的周长为10
2 |
∴C△ABC=30cm,
∴C梯形BCFE=C△ABC-C△AEF+2EF=30-10
2 |
2 |
2 |
(2)由(1)知,△AEF∽△ABC,且EF:BC=
2 |
∴S△AEF:S△ABC=2:9,
又S梯形BCFE=S△ABC-S△AEF,
∴S△AEF:S梯形BCFE=S△AEF:(S△ABC-S△AEF)=2:(9-2)=2:7.
点评:此题主要考查了相似三角形的判定与性质以及相似三角形中相似比与周长比和面积比的关系,正确分割图形得出是解题关键.
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