An isosceles triangle has a perimeter of 15 m. Find all the integral possibilities for the lengths of the sides in meters. Hint the sum of the lengths of any two sides of a triangle must exceed the third side.
If the 3rd side has length 1, then the sides are
1,7,7
similarly, if it grows in steps of 2, the sides are
3,6,6
5,5,5,
7,4,4
But then two sides would not exceed the length of the third side so I am thinking
1, 1, 13
2, 2, 11
3, 3, 9
Let's start solving the problem step-by-step.
Step 1: Recall the properties of an isosceles triangle.
In an isosceles triangle, two sides are equal in length. Let's assume the two equal sides have length x, and the remaining side has length y.
Step 2: Use the perimeter information to form an equation.
The perimeter of a triangle is the sum of the lengths of all its sides. In this case, we have:
x + x + y = 15
Step 3: Simplify the equation.
Adding the lengths of the two equal sides, we have:
2x + y = 15
Step 4: Apply the triangle inequality.
According to the triangle inequality theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. In our case, we can apply this condition:
x + x > y
2x > y
Step 5: Find the possible values for x and y.
To find the integral possibilities for x and y, we need to find integer solutions that satisfy both equations from Step 3 and Step 4. We can analyze the values of x and y separately.
For x:
The value of x can be any positive integer greater than zero since the length of a side cannot be zero or negative.
For y:
Considering the values of y that satisfy the equation 2x + y = 15 while also satisfying 2x > y, we find that y can take any value greater than 2x and less than 15.
Step 6: Calculate the integral possibilities for x and y.
For x = 1:
Since 2x = 2, y can take values between 2 and 15. So, the possible values for y are 3, 4, 5, ..., 14.
For x = 2:
Since 2x = 4, y can take values between 4 and 15. So, the possible values for y are 5, 6, 7, ..., 14.
For x = 3:
Since 2x = 6, y can take values between 6 and 15. So, the possible values for y are 7, 8, 9, ..., 14.
Continuing this pattern, we can determine all the integral possibilities for the lengths of the sides in meters for the isosceles triangle with a perimeter of 15 m.
Thus, the possible integral lengths of the sides are:
(1, 3), (1, 4), (1, 5), ..., (1, 14)
(2, 5), (2, 6), (2, 7), ..., (2, 14)
(3, 7), (3, 8), (3, 9), ..., (3, 14)
...
(14, 14)
(Note: This list may not contain all possible combinations, but it covers the general pattern of integral possibilities based on the given constraints.)
To find the possible lengths of the sides of an isosceles triangle with a perimeter of 15 m, we first need to understand the properties of an isosceles triangle.
An isosceles triangle has two sides of equal length. Let's assume the two equal sides have length 'x' and the remaining side has length 'y'. Since the perimeter is the sum of all three sides, we can write the equation:
x + x + y = 15
Simplifying this equation, we have:
2x + y = 15
Next, we need to consider the hint provided: the sum of the lengths of any two sides of a triangle must exceed the third side. In this case, the two equal sides (x and x) must be greater than the remaining side (y). Mathematically, this can be represented as:
2x > y
Now we have two equations:
2x + y = 15 (from the perimeter)
2x > y (from the triangle inequality)
Let's begin by finding the possible values for x by substituting the second equation into the first equation:
2x + 2x > 15
4x > 15
x > 15/4
x > 3.75
Since x has to be an integer, the smallest possible value for x is 4.
Substituting this value of x into the first equation:
2(4) + y = 15
8 + y = 15
y = 15 - 8
y = 7
So, one possible set of side lengths for an isosceles triangle with a perimeter of 15 m is {4, 4, 7}.
To find additional possible sets of side lengths, we can now increment the value of x by 1 and repeat the same steps.
For x = 5:
2(5) + y = 15
10 + y = 15
y = 15 - 10
y = 5
So another possible set of side lengths is {5, 5, 5}.
Continuing this process, we can increment x and find more integral possibilities for the lengths of the sides of the triangle.
Therefore, the integral possibilities for the lengths of the sides in meters are {4, 4, 7} and {5, 5, 5}.