a triangle have lengths that are consecutive whole numbers and its perimeter is greater than 2008 cm.If the least possible perimeter of the triangle is x cm, find the value of x.
let the sides be x-1, x, and x+1
x-1+x+x+1 > 2008
3x > 2008
x > 669.333
but x must be a whole number
if x = 669, the 3 sides would be 668 669 and 670 for a perimeter of 2007 , no good
if x = 670, the 3 sides would be 669 670 and 671 for a perimeter of 2010 , ok
so x = 670
To find the least possible perimeter of the triangle, we need to consider the condition that the lengths of the triangle sides are consecutive whole numbers.
Let's assume the three consecutive whole numbers as n, n+1, and n+2.
The perimeter of the triangle is the sum of the lengths of its sides. So, the perimeter of the triangle would be:
Perimeter = n + (n+1) + (n+2)
We need to find the value of n for which the perimeter is greater than 2008 cm.
Substituting the perimeter equation into the inequality:
n + (n+1) + (n+2) > 2008
Simplifying the equation, we get:
3n + 3 > 2008
Subtracting 3 from both sides:
3n > 2005
Dividing both sides by 3, we get:
n > 2005 / 3
Since n is a whole number, we need to round up the result to the nearest whole number:
n > 668.33
So, the least possible value for n is 669.
Now, we can find the perimeter by substituting this value back into the perimeter equation:
Perimeter = 669 + (669+1) + (669+2)
Perimeter = 669 + 670 + 671
Perimeter = 2010 cm
Therefore, the value of x, which is the least possible perimeter of the triangle, is 2010 cm.