標題:
中二升中三數學題~
發問:
一輛私家車在正午十二時離開學校。這輛私家車以56km/h的速度向正東方行駛。與此同時，一輛貸車離開同一間學校。這貸車以42km/h速度向正北方行駛。問在下午三時這兩輛車相距多遠?
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最佳解答:
(42km X 3)2次方 + (56km x3)2次方 = 開方(兩輛車相距) 兩輛車相距 = 開方(15876+28224) = 210km
其他解答:
這輛私家車下午三時與學校距離 56x3=168km 貸車下午三時與學校距離 42x3=126km 設兩車相距為y y^2=168^2+126^2 y^2=28224+15876 y^2=44100 y=√44100 y=210km|||||let k be the distance between two car k^2 = (56x3)^2 + (42x3)^2 k^2 = 28224 + 15876 k = 210km|||||Let A be the location of the school Let AB be the distance of the car at 3 pm Let AC be the distance of the lorry at 3 pm Let BC be the distance between the car and the lorry at 3 pm AB = 56 * 3 = 168 km AC = 42 * 3 = 126 km Since ABC is a right-angled triange with 90 degree at angle A, the distance of the car and the lorry at 3 pm will be BC = sqrt ( AB**2 + AC**2) = sqrt (168**2 + 126**2) = 210 km
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