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maths maths
發問:
http://i270.photobucket.com/albums/jj88/ucc041087/22.jpg plz help me to ask above Qs,thz!!!!!! 更新: plz help me to ans above Qs,thz!!!!!!
a)Since OD = OC, i.e.,OD = DC / 2 OD = 10 / 2 OD = 5cm And ABCD is a square,i.e.,QO = BC QO = 10cm ∴OP = 10cm (radius) b)cos∠POD = OD / OP cos∠POD = 5 / 10 cos∠POD = 1 / 2 ∴∠POD = 60 c)the area of the shaded region = the area of sector OPR﹣the area of △OPR = [π(10)2 X (60 / 360)]﹣[(1/2)(10)(10)(sin60)] = [100πX (1/6)]﹣50sin60 = (52.36﹣43.3)cm2 = 9.06cm2
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maths maths
發問:
http://i270.photobucket.com/albums/jj88/ucc041087/22.jpg plz help me to ask above Qs,thz!!!!!! 更新: plz help me to ans above Qs,thz!!!!!!
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最佳解答:a)Since OD = OC, i.e.,OD = DC / 2 OD = 10 / 2 OD = 5cm And ABCD is a square,i.e.,QO = BC QO = 10cm ∴OP = 10cm (radius) b)cos∠POD = OD / OP cos∠POD = 5 / 10 cos∠POD = 1 / 2 ∴∠POD = 60 c)the area of the shaded region = the area of sector OPR﹣the area of △OPR = [π(10)2 X (60 / 360)]﹣[(1/2)(10)(10)(sin60)] = [100πX (1/6)]﹣50sin60 = (52.36﹣43.3)cm2 = 9.06cm2
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