Question

如图2-23,等腰梯形ABCD的内切圆圆心为点O,点EH、G为切点,AH、BO的长度分别为关于x的方程x²-

 mx + 9m =0的两根,又cos∠BAO =.

图2-23

(1)求证:AO⊥BO;

(2)求梯形ABCD的面积.

Answer 1

思路分析:(1)运用切线长定理;(2)依据一元二次方程“根与系数的关系”求出m的值,进一步求得AD、BC、EF的长,求出梯形面积.

(1)证明:由切线长定理知OA平分∠DAB,OB平分∠ABC,又∵AD∥BC,∴∠AOB=90°,即AO⊥BO.

(2)解:连结OH,则OH⊥AB.∵cos∠BAO =,?

∴设OA =3k,AB =5k,OB =4k,?

在Rt△AOH中, , , ,?

由韦达定理得OB +AH = m,OB·AH =9m,?

即=,4k·5k =9m.?

解之,得k₁=5,k₂ =0(舍去).?

∴BO =20,AH =9.?

又∵AD =2AH =18,BC =2BF =2BH =32,EF =2OH =24,

即梯形的高EF =24.?

∴S_梯形ABCD =×(18+32)×24=600.