To determine if a triangle can be formed with sides of lengths \( \frac{1}{2} \), \( \frac{1}{4} \), and \( \frac{7}{8} \), we can use the triangle inequality theorem. This theorem states that for any triangle with sides of lengths \( a \), \( b \), and \( c \), the following inequalities must hold:
- \( a + b > c \)
- \( a + c > b \)
- \( b + c > a \)
Let's define the sides:
- \( a = \frac{1}{2} \)
- \( b = \frac{1}{4} \)
- \( c = \frac{7}{8} \)
Now we will check each of the inequalities:
-
\( a + b > c \): \[ \frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4} \quad vs \quad \frac{7}{8} \] \( \frac{3}{4} = \frac{6}{8} < \frac{7}{8} \) (fails)
-
\( a + c > b \): \[ \frac{1}{2} + \frac{7}{8} = \frac{4}{8} + \frac{7}{8} = \frac{11}{8} \quad vs \quad \frac{1}{4} \] \( \frac{11}{8} > \frac{1}{4} \) (passes)
-
\( b + c > a \): \[ \frac{1}{4} + \frac{7}{8} = \frac{2}{8} + \frac{7}{8} = \frac{9}{8} \quad vs \quad \frac{1}{2} \] \( \frac{9}{8} > \frac{4}{8} \) (passes)
Since the first inequality fails, a triangle cannot be formed with these side lengths.
Therefore, the answer is \( 2 \) for No.