An isosceles triangle has sides of length x cm,x cm and 9 cm. It’s perimeter is less than 24 and x is a whole number. Find the lowest value of x and find the highest value of x.
Let's consider the two cases separately:
Case 1: The isosceles triangle has two sides of length x cm each.
In this case, the third side must be 9 cm. The perimeter of the triangle is given by: x + x + 9 = 2x + 9. So we need to find the largest whole number value of x such that 2x + 9 < 24. Solving this inequality:
2x + 9 < 24
2x < 24 - 9
2x < 15
x < 7.5
Since x must be a whole number, the largest value it can have is 7.
Case 2: The isosceles triangle has one side of length 9 cm.
Let's call this side "a". So the other two sides must also be of length 9 cm each. In this case, the perimeter of the triangle is given by: 9 + 9 + 9 = 27. But we need the perimeter to be less than 24, so this case is not possible.
Therefore, the lowest value of x is 7 and the highest value of x is not possible (since it would make the perimeter greater than 24).
To find the lowest and highest values of x, we need to consider the maximum and minimum perimeter of the triangle.
Given that the triangle is isosceles with sides of length x cm, x cm, and 9 cm, we can write the perimeter equation as:
perimeter = x + x + 9
Since the perimeter is less than 24, we can write this inequality:
2x + 9 < 24
To find the lowest value of x, we need to find the maximum possible value of x that satisfies this inequality. We can solve it step-by-step as follows:
2x + 9 < 24
2x < 24 - 9
2x < 15
x < 15/2
x < 7.5
Since x must be a whole number, the lowest value of x is 7.
Now let's find the highest value of x. We can rewrite the inequality as:
2x < 15
x < 15/2
x < 7.5
Since x must be a whole number, the highest possible value for x is 7.
Therefore, the lowest value of x is 7, and the highest value of x is also 7.
To find the lowest value of x, we need to determine the maximum possible perimeter of the triangle. Since it's an isosceles triangle, two sides are equal and the third side is different. Let's assume the third side length is 'y' cm.
Since the perimeter of a triangle is the sum of all its sides, we have:
Perimeter = x cm + x cm + y cm
Given that the perimeter is less than 24 cm:
2x + y < 24
Since x and y are both whole numbers, we need to find the largest possible values that satisfy this inequality.
To find the highest value of x, we need to determine the minimum possible perimeter of the triangle. Again, assuming the third side length is 'y' cm, we have:
2x + y > 0
The highest value of x should be the largest whole number that satisfies this inequality.
Let's calculate x for both cases:
Case 1: x is the smallest possible whole number
If x = 1, then the perimeter is:
2(1) + y = 2 + y < 24
y < 22
So, the lowest value of x = 1.
Case 2: x is the largest possible whole number
If x = 9, then the perimeter is:
2(9) + y = 18 + y > 0
y > -18
Since y must also be a whole number, the lowest possible value for y is 1 (since negative numbers are not whole numbers). Therefore, the highest value for y is 1.
So, the highest value of x = 9.
In conclusion:
Lowest value of x = 1.
Highest value of x = 9.