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發問:

1.)A cake is divided into four pieces in the ratio 2 : 3 : 4 : 6 by weight. If the cake weights 900 g, find the weight of the heaviest piece. 2.) The three sides of △ABC are in the ratio AB : BC : CA = 5 : 7 : 4. If AB = 20 cm, find the permeter of △ABC.

最佳解答:

(1) Because the heaviest one is 6/(2+3+4+6) of the whole. Therefore the weight of the heaviest one is 900x6/(2+3+4+6) =900x6/15 =360(kg) (2) AB:BC:CA=5:7:4 where AB=20cm Therefore 5y=20 y=4 Therefore BC=4x7cm=28cm CA=4x4cm=16cm Therefore the perimeter is (28+16+20)cm=64cm

其他解答:

1.)the weight of the heaviest piece =900×6 / (2+3+4+6) =900×6 / 15 =900×2 / 5 =360g 2.)Let y be the permeter of △ABC y×5 / (5+7+4) = 20 y×5 / 16 = 20 y×5 = 20×16 y×5 = 320 y = 320 / 5 y = 64 Therefore the permeter(周界) of △ABC is 64cm.|||||1.) 900g x 6 / 2+3+4+6 = 900g x 6/15 = 360g 2.) Let X be the permeter △ABC X x 5/5+7+4 = 20cm X = 20cm x 16/5 X = 64cm Therefore, the permeter of △ABC is 64cm|||||900*[6/(2+3+4+6)]=900*[6/15]=360g 20/5=4(1 unit) there are total 5+7+4 units=15units, therefore permeter is 15*4=60 cm|||||1.)120:180:240:360|||||1. 900 x 6 / (2+3+4+6) = 900 x 6 / 15 = 360 g