Quadratic Equations
2009-12-15 5:23 am
The height and the perimeter of a right-angled triangle are x cm and 40 cm respectively. If the length of its base is 1 cm less than twice its height, find the value of x.
回答 (2)
2009-12-15 5:29 am
✔ 最佳答案
length of base=2x-1x+2x-1+√[x^2+(2x-1)^2]=40
x^2+(2x-1)^2=(41-3x)^2
5x^2-4x+1=1681-246x+9x^2
4x^2-242x+1680=0
2x^2-121x+840=0
X=52.5 (rejected) or x=8
So the value of x is 8
2009-12-15 5:30 am
The height of the triangle is x cm
So, the length of the base is (2x - 1) cm
And the length of the slant = sqrt[x^2 + (2x - 1)^2] = sqrt(5x^2 - 4x + 1) cm
Perimeter of the triangle = 40 cm
So, x + (2x - 1) + sqrt(5x^2 - 4x + 1) = 40
sqrt(5x^2 - 4x + 1) = 41 - 3x
5x^2 - 4x + 1 = (41 - 3x)^2
2009-12-14 21:30:34 補充：
5x^2 - 4x + 1 = 1681 - 246x + 9x^2
4x^2 - 242x + 1680 = 0
2x^2 - 121x + 840 = 0
(2x - 105)(x - 8) = 0
x = 8 or 52.5 (rejected, since it is even longer than the perimeter)
So, x = 8
So, the length of the base is (2x - 1) cm
And the length of the slant = sqrt[x^2 + (2x - 1)^2] = sqrt(5x^2 - 4x + 1) cm
Perimeter of the triangle = 40 cm
So, x + (2x - 1) + sqrt(5x^2 - 4x + 1) = 40
sqrt(5x^2 - 4x + 1) = 41 - 3x
5x^2 - 4x + 1 = (41 - 3x)^2
2009-12-14 21:30:34 補充：
5x^2 - 4x + 1 = 1681 - 246x + 9x^2
4x^2 - 242x + 1680 = 0
2x^2 - 121x + 840 = 0
(2x - 105)(x - 8) = 0
x = 8 or 52.5 (rejected, since it is even longer than the perimeter)
So, x = 8
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