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Patricia purchased x meters of fencing. She originally intended to use all of the fencing to enclose a square region, but later decided to use all of the fencing to enclose a rectangular region with length y meters greater than its width. In square meters, what is the positive difference between the area of the square region and the area of the rectangular region?
(1)xy=256
(2)y=4
分析A选项
分析B选项
分析C选项
分析D选项
分析E选项
条件二:
①因为长方形和矩形的周长相等
所以设长方形的宽为a,则长为a+y,周长=4a+2y=正方形的周长
②正方形的边=(4a+2y)/4=a+y/2,面积=a^2+ay+y^2/4
③长方形的面积=a^+ay
④因为题目要我们求的是difference between the area of the square region and the area of the rectangular region
所以就是(a^2+ay+y^2/4)-(a^+ay)=y^2/4,所以只需要知道y的值即可,充分。
fugace.s
太坑了 题目里设的x完全没用啊
Ryan_x97
x刚好够用,没有剩余吗
willowX
length y meters greater than its width意思是长比宽多y!设数只能设短为n,长为n+y,再使用周长为x来将n表示出来。最后正方形面积(x/4)^2,而长方形面积为(x-2y)/4乘(x-2y+4y)/4。二者做差,化简得y^2/4,即只需要知道y即可
66miao
x米围栏,正方形边长就是x/4, 面积是 (x/4)^2; 长方形长是 x/4 + y/2, 宽是 x/4 - y/2,面积是 (x/4 + y/2)*(x/4 - y/2)= (x/4)^2 -(y/2)^2; 正方形长方形面积差 = (x/4)^2 -(y/2)^2 - (x/4)^2 = (y/2)^2;所以只要知道y就能解。 B
睿旋必须得行
看成圆形了 我傻了
221555tvgl回复睿旋必须得行
加一
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2021-11-28 21:27:44
水寒528
S1-S2=y^2/4所以B正确
173993qp
两个图形的周长相等,设长方形的宽为a,则长为a+y,周长=4a+2y=正方形的周长。正方形的边=a+y/2,面积=a^2+ay+y^2/4,而长方形的面积=a^+ay,差y^2/4,所以只需要知道y的值即可
凡人L
有第二个条件最后得出的表达式为:仅含有x的表达式;而这里的x作为题干中所给的已知量,因此条件二sufficient
416348tqhf回复 凡人L
如果X是题目已给的已知量,条件一也是sufficient,y=256/x, 乱答一通
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2021-01-17 12:19:11
wordy要你吗回复凡人L
不是的bro
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2022-09-22 18:08:59
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数据充分DS
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