THE Perimeter of a triangle is 2004.
One Side of a triangle = 21 Times the other.
the shortest side is integral length.
SOLVE For lengths of the sides of the traingle in every possible case.
To solve this problem, we'll use the information given and set up equations to find the lengths of the sides of the triangle.
Let's assume that the lengths of the sides of the triangle are a, b, and c, with a being the shortest side.
We know that the perimeter of the triangle is 2004, so we can write the equation:
a + b + c = 2004 --------- Equation 1
We are also given that one side of the triangle is 21 times the other. This can be written as:
b = 21a --------- Equation 2
Now, let's substitute Equation 2 into Equation 1:
a + (21a) + c = 2004
Simplifying this equation:
22a + c = 2004 --------- Equation 3
We are looking for integral lengths, which means all sides should have whole number lengths.
Now, we need to find the possible values of a, b, and c that satisfy the conditions.
One approach to solve this is to use trial and error. We can start by trying different values for a and calculating the lengths of b and c.
Let's begin with the smallest possible value for a, which is 1. Substituting this into Equation 2, we get:
b = 21(1) = 21
Substituting the values of a and b into Equation 1, we get:
1 + 21 + c = 2004
22 + c = 2004
c = 1982
So, when a = 1, b = 21, and c = 1982, we have a possible solution.
Now, let's try the next possible value for a, which is 2. Substituting this into Equation 2, we get:
b = 21(2) = 42
Substituting the values of a and b into Equation 1, we get:
2 + 42 + c = 2004
44 + c = 2004
c = 1960
So, when a = 2, b = 42, and c = 1960, we have another possible solution.
We can continue this process by checking other values of a until we find all the possible cases that satisfy the given conditions.
In summary, the lengths of the sides of the triangle in every possible case, where a is the shortest side, are as follows:
1) a = 1, b = 21, c = 1982
2) a = 2, b = 42, c = 1960
Note: It is also possible to solve this problem algebraically by manipulating the equations further, but trial and error is a more straightforward approach in this case.
xy = 2004
x =21y
21y * y = 2004
21y^2 =2004
y^2 = 2004/21
y=sqrt 2004/21
plug in to get x