Maths功課

Let x be the side length of the small square and y be the side length of the large square
4x+4y = 32
=> x + y = 8 ----(1)
x^2 + y^2 = 34 ----(2)

Sub Eq(1) int Eq(2):
x^2 + (8 - x)^2 = 34
2x^2 - 16x + 30 = 0
x^2 - 8x + 15 = 0
(x-3) (x-5) = 0
x = 3 or x = 5
When x = 3, from Eq(1): y= 8-3 = 5
When x = 5, from Eq(2): y=8-5 = 3
The iron wire segment length of the large square = 4x = 4(5) = 20cm
The iron wire segment length of the small square = 4y = 4(3) = 12cm

X^2 + Y^2 = 34 ---------(1)
4X + 4Y = 32 -----------(2)

From (2),
X + Y = 8
X = 8 - Y -------(3)

Substitute (3) into (1),
(8-Y)^2 +Y^2 = 34
64 -16Y + Y^2 + Y^2 = 34
32 - 8Y + Y^2 = 17
Y^2 -8Y -15 = 0
(Y-5)(Y-3) = 0

Y= 5 or 3

When Y=5,
Substitute Y into (3)
X = 8-Y = 3

When Y=3,
Substitute Y into (3)
X = 8-Y = 5


即係兩個 square 邊長分別係 3 同 5 cm

2 iron wire segments 就係 12cm 同 20cm !!

4x + 4y = 32

x^2 + y^2 = 34

2008-01-15 23:43:43 補充：
4x 4y =32x = 8 - y(8-y)^2 y^2 = 34(y-3)(y-5)=0so lengths of two iron wire segments are 3 and 5