Question

如图,平行四边形ABCD,∠ABC=3∠A,F在CB的延长线上,FE⊥DC与点E,CF=CD,EF=1,求DE的长.

Answer 1

设菱形对角线交于点O.

(1)∠DAB+∠ABC=180°;∠DAB:∠ABC=1:2.

则∠DAB=60°;又AD=AB.故⊿ABD为等边三角形.

∴DB=AB=48/4=12(cm);

AC=2AO=2*(√3/2)*12=12√3(cm).

(2)S菱形=AC*DB/2=(12√3)*12/2=72√3(cm^2).