1.解:过点D作BC的垂线,交BC于E,所以角ADE=90°DE=AB=5√3/2.
cos角CDE=(5√3/2)/5=√3/2
角CDE=30°,所以角D=角CDE+90°=30+90=120°2.解:角A=90°,即梯形ABCD是直角梯形,
过点D作BC的垂线,交BC于E,则CE=√2,BE=AD,BC=CE+BE=1+√2
S=(AD+BC)*AB/2=(2+√2)*1/2=1+√2/2(cm^2)3.解:设梯形的高为x,则有x*(tan 35°+tan20)=5,解得x≈13.736
CD=x/cos 20°≈13.83
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