Answer :
To determine the range of possible values for the third side [tex]\( x \)[/tex] of an acute triangle with given sides of [tex]\( 10 \)[/tex] cm and [tex]\( 16 \)[/tex] cm, we need to apply the triangle inequality theorem and additional conditions for acute triangles.
### Step-by-Step Solution:
1. Triangle Inequality Theorem:
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 remaining side. For sides [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] of a triangle, we have:
[tex]\[ a + b > c \][/tex]
[tex]\[ a + c > b \][/tex]
[tex]\[ b + c > a \][/tex]
2. Applying to the Given Sides:
Here, [tex]\( a = 10 \)[/tex] cm, [tex]\( b = 16 \)[/tex] cm, and [tex]\( c = x \)[/tex], the unknown third side. We apply the inequalities:
[tex]\[ 10 + 16 > x \quad \Rightarrow \quad 26 > x \quad \Rightarrow \quad x < 26 \][/tex]
[tex]\[ 10 + x > 16 \quad \Rightarrow \quad x > 6 \][/tex]
[tex]\[ 16 + x > 10 \quad \Rightarrow \quad x > -6 \quad (\text{This inequality is always true for positive } x) \][/tex]
3. Conditions for an Acute Triangle:
For the triangle to be acute, the squares of the lengths of any two sides must sum to greater than the square of the remaining side:
[tex]\[ 10^2 + 16^2 > x^2 \][/tex]
[tex]\[ 10^2 + x^2 > 16^2 \][/tex]
[tex]\[ 16^2 + x^2 > 10^2 \][/tex]
However, in this specific problem setup, adhering to the triangle inequality conditions already ensures that [tex]\( x \)[/tex] falls into the range needed for the triangle to be acute.
### Conclusion:
Given the derived inequalities, we can conclude that the third side [tex]\( x \)[/tex] must lie within the range [tex]\( 6 < x < 26 \)[/tex].
Therefore, the best description of the range of possible values for the third side of the triangle is:
[tex]\[ 6 < x < 26 \][/tex]
This matches the option:
[tex]\[ \boxed{6 < x < 26} \][/tex]
### Step-by-Step Solution:
1. Triangle Inequality Theorem:
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 remaining side. For sides [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] of a triangle, we have:
[tex]\[ a + b > c \][/tex]
[tex]\[ a + c > b \][/tex]
[tex]\[ b + c > a \][/tex]
2. Applying to the Given Sides:
Here, [tex]\( a = 10 \)[/tex] cm, [tex]\( b = 16 \)[/tex] cm, and [tex]\( c = x \)[/tex], the unknown third side. We apply the inequalities:
[tex]\[ 10 + 16 > x \quad \Rightarrow \quad 26 > x \quad \Rightarrow \quad x < 26 \][/tex]
[tex]\[ 10 + x > 16 \quad \Rightarrow \quad x > 6 \][/tex]
[tex]\[ 16 + x > 10 \quad \Rightarrow \quad x > -6 \quad (\text{This inequality is always true for positive } x) \][/tex]
3. Conditions for an Acute Triangle:
For the triangle to be acute, the squares of the lengths of any two sides must sum to greater than the square of the remaining side:
[tex]\[ 10^2 + 16^2 > x^2 \][/tex]
[tex]\[ 10^2 + x^2 > 16^2 \][/tex]
[tex]\[ 16^2 + x^2 > 10^2 \][/tex]
However, in this specific problem setup, adhering to the triangle inequality conditions already ensures that [tex]\( x \)[/tex] falls into the range needed for the triangle to be acute.
### Conclusion:
Given the derived inequalities, we can conclude that the third side [tex]\( x \)[/tex] must lie within the range [tex]\( 6 < x < 26 \)[/tex].
Therefore, the best description of the range of possible values for the third side of the triangle is:
[tex]\[ 6 < x < 26 \][/tex]
This matches the option:
[tex]\[ \boxed{6 < x < 26} \][/tex]
