Triangles must satisfy the triangle inequality theorem, which states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. Given the side lengths of triangle ABC as \( AB = 2x + 4 \), \( BC = 6x \), and \( AC = 18 \), we need to set up the inequalities based on the triangle inequality theorem:
1. \( AB + BC > AC \)
2. \( AB + AC > BC \)
3. \( BC + AC > AB \)
Let's solve each of these in turn.
### 1. \( AB + BC > AC \)
\( (2x + 4) + 6x > 18 \)
Simplify:
\( 8x + 4 > 18 \)
Subtract 4 from both sides:
\( 8x > 14 \)
Divide by 8:
\( x > \frac{7}{4} \)
### 2. \( AB + AC > BC \)
\( (2x + 4) + 18 > 6x \)
Simplify:
\( 2x + 22 > 6x \)
Subtract 2x from both sides:
\( 22 > 4x \)
Divide by 4:
\( \frac{22}{4} > x \)
Simplify:
\( \frac{11}{2} > x \)
or
\( x < \frac{11}{2} \)
### 3. \( BC + AC > AB \)
\( 6x + 18 > 2x + 4 \)
Simplify:
\( 6x + 18 > 2x + 4 \)
Subtract 2x from both sides:
\( 4x + 18 > 4 \)
Subtract 18 from both sides:
\( 4x > -14 \)
Divide by 4:
\( x > -\frac{7}{4} \)
However, the condition \( x > -\frac{7}{4} \) is redundant because \( x > \frac{7}{4} \) is a stricter requirement.
Combining \( x > \frac{7}{4} \) and \( x < \frac{11}{2} \), we get:
\[ \frac{7}{4} < x < \frac{11}{2} \]
So, the correct answer is:
A. \( \frac{7}{4} < x < \frac{11}{2} \)
Select the correct answer.
Which inequality represents the values of x that ensure triangle ABC exists?
Line AB = 2x + 4
Line BC = 6x
Line AC = 18
A. 7/4 < x < 11/2
B. -7/4 < x< 11/2
C. 1 < x < 5
D. 2 < x < 6
1 answer