Question

如图,在矩形ABCD中,BD=10,AD>AB,设∠ABD=α,∠ADB=β.已知sinα、sinβ为方程25x^2-3

Answer 1

方程25x^2-35x+12=0,即(5X-3)(5X-4)=0,两个实数根：X1=3/5,X2=4/5,

∵AD>AB,∴sinα=4/5、sinβ=3/5,

∵sinα=AD/BD,BD=10,∴AD=8,同理AB=6,

BE=8-X,SΔABE=1/2AB*BE=3(8-X),

FC=4-X,SΔCEF=1/2CE*CF=1/2X(4-X),

DF=6-(4-X)=X+2,∴SΔADF=1/2AD*DF=4(X+2),

∴Y=S矩形-SΔABE-SΔCEF-SΔADF

=48-3(8-X)-1/2X(4-X)-4(X+2)

=1/2X^2-3X+16

⑵Y=1/2(X^2-6X+9)-9/2+16

=1/2(X-3)^2+23/2,

∴当X=3时,Y最大=23/2,

这时CE=3,CF=1.