题目内容
如图,△ABC的面积为S,在BC上有点A′,且BA′:A′C=m(m>0);在CA的延长线有点B′,且CB′:AB′=n(n>1);在AB的延长线有点C′,且AC′:BC′=k(k>1).则S△A′B′C′=______.
连接BB′,C′C,则S△A′B′C′=S△A′B′B+S△A′BC′+S△BB′C′,
∵BA′:A′C=m,CB′:AB′=n,AC′:BC′=k,
∴B′A:AC=1:(n-1),BA′:A′C=m:1,C′B:BA=1:(k-1),
∴
S△C′BA′ |
S△C′BC |
m |
m+1 |
∴S△C′BA′=
m |
m+1 |
同理S△C′BC=
1 |
K |
∴S△C′BA′=
m |
m+1 |
1 |
k |
同理:S△B′C′B=
1 |
k-1 |
1 |
k-1 |
1 |
n-1 |
S△B′BA′=
m |
m+1 |
m |
m+1 |
n |
n-1 |
∴①+②+③得:S△A′B′C′=S△C′BA′+S△B′C′B+S△B′BA′=
mnk+1 |
(m+1)(n-1)(k-1) |
故答案为:
mnk+1 |
(m+1)(n-1)(k-1) |
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