- 试题详情及答案解析
- 如图,BD是∠ABC的平分线,DE⊥AB于点E,DF⊥AC于点F,S△ABC=36 cm2,AB=18 cm,BC=12 cm,则DE=________cm.

- 答案:2.4
- 试题分析:首先过点D作DF⊥BC于点F,由BD是∠ABC的平分线,DE⊥AB,根据角平分线的性质,可得DE=DF,然后由S△ABC=S△ABD+S△BCD=
AB•DE+
BC•DF,求得答案.
试题解析:过点D作DF⊥BC于点F,

∵BD是∠ABC的平分线,DE⊥AB,
∴DE=DF,
∵AB=18cm,BC=12cm,
∴S△ABC=S△ABD+S△BCD=
AB•DE+
BC•DF=
DE•(AB+BC)=36cm2,
∴DE=2.4(cm).
考点:角平分线的性质.