A triangle has sides x,x+7 , x+ 8. The value of x must fall between what boundaries to make this an obtuse triangle?
To determine the boundaries for the value of x that would make the given triangle an obtuse triangle, we need to use the concept of the triangle inequality theorem and the Pythagorean theorem.
According to the triangle inequality theorem, in any triangle, the sum of the lengths of any two sides must be greater than the length of the third side. In this case, we have three sides: x, x + 7, and x + 8.
So, using the triangle inequality theorem, we can write:
x + (x + 7) > (x + 8)
2x + 7 > x + 8
2x - x > 8 - 7
x > 1
This tells us that x must be greater than 1 for the triangle to exist.
Now, to determine if the triangle is obtuse, we need to use the Pythagorean theorem. In an obtuse triangle, the square of the length of the longest side must be greater than the sum of the squares of the other two sides.
Let's assume x + 8 is the longest side. According to the Pythagorean theorem, we have:
(x + 8)^2 > x^2 + (x + 7)^2
Expanding both sides and simplifying, we get:
x^2 + 16x + 64 > x^2 + x^2 + 14x + 49
Simplifying further:
16x + 64 > 2x^2 + 14x + 49
0 > 2x^2 - 2x - 15
Now, we can solve this quadratic inequality to find the range of x values that satisfy it.
First, let's set the quadratic inequality to zero:
2x^2 - 2x - 15 = 0
Next, factorizing the quadratic equation, we have:
(2x - 5)(x + 3) > 0
Solving for x, we find two possible values:
x > 5/2 or x < -3
Therefore, the value of x must be greater than 1 and either greater than 5/2 or less than -3 for the triangle to be both obtuse and exist.
In summary, the boundaries for the value of x that make the triangle an obtuse triangle are x > 1 and either x > 5/2 or x < -3.
any side of a triangle is less than the sum of the other two sides.
x < x+7 + x+8
-15 < x
no problem there
x+7 < x + x+8
-1 < x
no problem there
x+8 < x + x+7
1 < x
so, as long as x is at least 1, the three values could form a triangle.