1. Let the area, perimeter and length of one side of a rectangle be S,P and x respectively.One part of S varies jointly as P and x, and the other part varies directly as the square of x. When x=4, P=10,s=4. Whenx=5 and P=12, S=5.
Suppose the length of one side of a rectangle ABCD is 4 and its perimeter is 20. The area of another rectangle EFGH which is similar to rectangle ABCD is 42. Find the perimeter of rectangle EFGH. [Ans. 10√7]
thx~
Maths problem[急]
2009-08-04 3:36 am
回答 (2)
2009-08-04 5:01 am
✔ 最佳答案
Length of the other side of ABCD = 20/2 - 4 = 10 - 4 = 6.Let width of EFGH = x , since it is similar to ABCD, its length = (6/4)x
= 3x/2
so area of EFGH = (x)(3x/2) = 3x^2/2 which is equal to 42
therefore 3x^2/2 = 42
x^2 = 28
x = sqrt 28 = 2 sqrt 7
so length = 3x/2 = 3 sqrt 7
so perimeter of EFGH = 2( 2 sqrt 7 + 3 sqrt 7) = 10 sqrt 7.
2009-08-04 5:12 am
1)
Let S = k1 * Px + k2 * x^2
When x=4, P=10,s=4:
4 = k1 * (10)(4) + k2 * 4^2
4 = 40k1 + 16k2...(1) ;
Whenx=5 and P=12, S=5:
5 = k1 * (12)(5) + k2 * 5^2
5 = 60k1 + 25k2...(2)
(1)*3 - (2)*2 :
12 - 10 = 16*3k2 - 25*2k2
2 = - 2k2
k2 = - 1 , sub to (1) or (2),
k1 = 0.5
So S = 0.5Px - x^2
The area S of ABCD = 0.5(20)(4) - 4^2 = 24
(the perimeter of rectangle EFGH : the perimeter of rectangle ABCD)^2
= 42 : 24 = 7 : 4
(the perimeter of rectangle EFGH : 20) = √7 / 2
the perimeter of rectangle EFGH = (√7 / 2) * 20 = 10√7
Let S = k1 * Px + k2 * x^2
When x=4, P=10,s=4:
4 = k1 * (10)(4) + k2 * 4^2
4 = 40k1 + 16k2...(1) ;
Whenx=5 and P=12, S=5:
5 = k1 * (12)(5) + k2 * 5^2
5 = 60k1 + 25k2...(2)
(1)*3 - (2)*2 :
12 - 10 = 16*3k2 - 25*2k2
2 = - 2k2
k2 = - 1 , sub to (1) or (2),
k1 = 0.5
So S = 0.5Px - x^2
The area S of ABCD = 0.5(20)(4) - 4^2 = 24
(the perimeter of rectangle EFGH : the perimeter of rectangle ABCD)^2
= 42 : 24 = 7 : 4
(the perimeter of rectangle EFGH : 20) = √7 / 2
the perimeter of rectangle EFGH = (√7 / 2) * 20 = 10√7
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