Math question

2006-10-28 2:18 pm
The perimeter of a rectangular lawn is 40m. If the length is increased by 4m and the breadth by 2m, the area will be (5/3) of its fomer value. Find the original dimensions of the lawn.

回答 (3)

2006-10-28 2:49 pm
✔ 最佳答案
希望唔好係你自己既功課,如果係,咁我答你即係害o左你@@

let x be the original length and y be the original breadth,

original area=xy, original perimeter = 2x+2y = 40...........Eq(1a)
x+y = 20..........Eq(1b)

now, length=x+4, breadth = y+2,
area = (x+4)(y+2) = xy+2x+4y+8 = (5/3)xy
:
:
(2/3)xy-2x-4y-8 = 0.............Eq(2)
(1a)+(2),
u will get (2/3)xy-2y-48 = 0.............Eq(3)

Put (1b) into (3), (2/3)(20-y)y-2y-48 = 0
:
:
y^2-17y+72 = 0
(y-8)(y-9) = 0
y= 8 or 9
For y=8, x= 20 -8 = 12,i.e. length=12,breadth=8
For y=9,x = 20-9 = 11, i.e. length = 11,breadth = 9.......................End

應該o岩,不過錯o左我唔負責^^

2006-10-31 21:15:34 補充:
嗯嗯,不過呢個問題都幾好,我差d都答錯o左@@
參考: myself
2006-10-30 3:43 pm
the length of the original lawn is x
the breadth of the original lawn is y
then the length of the new lawn is x+4
the breadth of the original lawn is y+2
According the question
2(x+y)=40....(1)
(x+4)(y+2)=(5/3)xy....(2)
from (1)
x+y=20
y=20-x
from (2)
(x+4)(y+2)=(5/3)xy
xy+2x+4y+8=(5/3)xy
2x+4y+8=(2/3)xy
x+2y+4=(1/3)xy
x+2(20-x)+4=(1/3)x(20-x)
3(44-x)=20x-x^2
132-3x=20x-x^2
x^2-23x+132=0
(x-11)(x-12)=0
x=11 or x=12
y=9 or y=8
the original dimensions of the lawn are
length 11m breadth 9m
length 12m breadth 8m
參考: myself
2006-10-28 2:31 pm
let
the length of the original lawn is x
the breadth of the original lawn is y
then
the length of the new lawn is x+4
the breadth of the original lawn is y+2
According the question
2(x+y)=40....(1)
(x+4)(y+2)=(5/3)xy....(2)
from (1)
x+y=20
y=20-x
from (2)
(x+4)(y+2)=(5/3)xy
xy+2x+4y+8=(5/3)xy
2x+4y+8=(2/3)xy
x+2y+4=(1/3)xy
x+2(20-x)+4=(1/3)x(20-x)
3(44-x)=20x-x^2
132-3x=20x-x^2
x^2-23x+132=0
(x-11)(x-12)=0
x=11 or x=12
y=9 or y=8
the original dimensions of the lawn are
length 11m breadth 9m
length 12m breadth 8m


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