xy = 2004
x =21y
21y * y = 2004
21y^2 =2004
y^2 = 2004/21
y=sqrt 2004/21
plug in to get x
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.
x =21y
21y * y = 2004
21y^2 =2004
y^2 = 2004/21
y=sqrt 2004/21
plug in to get x
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.