if triangle MNO is an acute triangle and MN=x-1, NO=x+1, and MO=8, then what are the possible solutions of x?
well, MN+NO>MO
x-1+x>8
2x>9
so any x greater than 4.5 is a possible solution. Check my thinking.
x>4
x-1+x+1>8, 2x>8, x>4
Someone check my thinking
To find the possible solutions for x, we can make use of the properties of an acute triangle.
In an acute triangle, the sum of the squares of the two shorter sides (legs) should be greater than the square of the longest side (hypotenuse).
Let's apply this property to the given triangle MNO:
In triangle MNO, we have:
MN = x - 1
NO = x + 1
MO = 8
According to the property mentioned above, we can write the following inequality:
(MN)^2 + (NO)^2 > (MO)^2
Substituting the given values, we have:
(x - 1)^2 + (x + 1)^2 > 8^2
Expanding and simplifying the inequality:
x^2 - 2x + 1 + x^2 + 2x + 1 > 64
2x^2 + 2 > 64
2x^2 > 62
Dividing both sides of the inequality by 2:
x^2 > 31
To find the possible values of x, we take the square root of both sides:
x > √31 or x < -√31
Therefore, the possible solutions for x are x > √31 or x < -√31.