an isosceles triangle has perimeter of 15m. Find all integral possibilities for the lengths of the side in meters. Hint: the sum of the lengths of any two sides of a triangle must exceed the third side.
let each of the equal sides be x
let the third side by y
2x + y = 15
y = 15-2x , clearly 15-2x >0 ----> x < 7.5 **
also x+x > y ----> y < 2x
x+y > x ----> y > 0 ----------> 0 < y < 2x ***
form a table of values satisfying both ** and ***
x y
7 1 yes
6 3 yes
5 5 yes
4 7 yes
3 9 NO
state your conclusion
Well, it's no triangle's business to be a party pooper and break the rules, especially when it comes to side lengths. So, let's get cracking!
Since an isosceles triangle has two sides of equal length, let's call the length of those sides "x" and the length of the remaining side "y". According to the hint, we know that x + x > y, and we also know their perimeter is 15m, so x + x + y = 15m.
Combining these two pieces of information, we can rewrite the equation as 2x + y = 15m. Now, let's list out the possible integral values for x and y:
For x = 1: 2(1) + y = 15
2 + y = 15
y = 13
For x = 2: 2(2) + y = 15
4 + y = 15
y = 11
For x = 3: 2(3) + y = 15
6 + y = 15
y = 9
For x = 4: 2(4) + y = 15
8 + y = 15
y = 7
For x = 5: 2(5) + y = 15
10 + y = 15
y = 5
For x = 6: 2(6) + y = 15
12 + y = 15
y = 3
For x = 7: 2(7) + y = 15
14 + y = 15
y = 1
So, the integral possibilities for the lengths of the sides of the isosceles triangle are:
(1, 13), (2, 11), (3, 9), (4, 7), (5, 5), (6, 3), and (7, 1).
Remember, no clowning around with the triangle inequality rule!
Let's assume the lengths of the two equal sides of the isosceles triangle are x, and the length of the remaining side is y.
According to the given hint, the sum of any two sides of a triangle must be greater than the third side. So, we have the following inequalities:
x + x > y (1)
x + y > x (2)
x + y > x (3)
Simplifying these inequalities, we get:
2x > y (4)
x + y > x (5)
x + x > y (6)
Now, let's consider the perimeter of the triangle, which is given as 15m. According to the formula for the perimeter of a triangle, it is equal to the sum of all three sides:
x + x + y = 15 (7)
Simplifying equation (7), we get:
2x + y = 15 (8)
Now, we can combine equations (4) and (8) to find the possible values of x and y:
2x > y (4)
2x + y = 15 (8)
To find the integral possibilities for x and y, we can start by assuming x = 1 and solve for y:
2(1) > y (4)
2 + y = 15 (8)
This gives us y = 13 as a possible value for y when x = 1. However, y should be greater than 2x, so y > 2(1) = 2. Therefore, y = 13 is valid.
Using the same approach, we can continue to find other integral possibilities for x and y:
x = 1, y = 13
x = 2, y = 11
x = 3, y = 9
x = 4, y = 7
x = 5, y = 5
Therefore, the integral possibilities for the lengths of the sides of the isosceles triangle with a perimeter of 15m are:
(x, y) = (1, 13), (2, 11), (3, 9), (4, 7), (5, 5).
To find the integral possibilities for the lengths of the sides of an isosceles triangle with a perimeter of 15m, we can use the given hint:
"The sum of the lengths of any two sides of a triangle must exceed the third side."
In an isosceles triangle, two sides have the same length, while the third side is different. Let's assume the lengths of the two equal sides are both 'x', and the length of the third side is 'y'.
According to the hint, we have three inequalities to consider:
1) x + x > y (sum of the two equal sides is greater than the third side)
2) x + y > x (sum of one equal side and the third side is greater than the other equal side)
3) x + y > x (sum of one equal side and the third side is greater than the other equal side)
Simplifying these inequalities, we get:
1) 2x > y
2) y > 0
3) y > 0
Now, let's solve these inequalities step by step to find the integral possibilities for 'x' and 'y'.
From inequality 1, we have 2x > y. Rearranging, we get y < 2x.
From inequality 2, we have y > 0, which means y must be a positive value.
From inequality 3, we have y > 0, which again confirms that y must be positive.
Considering these conditions, let's find the integral possibilities for 'y' based on different values of 'x'.
1) When x = 1, the possible values for y are y = 1, 2, 3, 4, ...
2) When x = 2, the possible values for y are y = 1, 2, 3, 4, ...
3) When x = 3, the possible values for y are y = 1, 2, 3, 4, ...
Following this pattern, we can see that for any positive integer value of 'x', the possible values for 'y' are also positive integers starting from 1.
Hence, the integral possibilities for the lengths of the sides of the isosceles triangle with a perimeter of 15m are:
{(1, 1, 13), (2, 2, 11), (3, 3, 9), (4, 4, 7), (5, 5, 5)}
These are the five possible combinations of side lengths for the isosceles triangle.