Two sides of a triangle are 5 cm and 12 cm long. The measure of third sides is an integer in cm. if the tirangles is an obtuse tirangle, then how many such triangles are possible ?

Options

A. 9
B. 8
C. 7
D. 6

Explanation

To determine how many obtuse triangles can be formed with two sides measuring 5 cm and 12 cm, and an integer length for the third side, we can follow these steps:

### Step 1: Determine the range for the third side
For any triangle with sides \( a \), \( b \), and \( c \), the triangle inequality states:
1. \( a + b > c \)
2. \( a + c > b \)
3. \( b + c > a \)

In our case, let \( a = 5 \) cm, \( b = 12 \) cm, and \( c \) be the third side.

From the triangle inequalities:
1. \( 5 + 12 > c \) → \( c < 17 \)
2. \( 5 + c > 12 \) → \( c > 7 \)
3. \( 12 + c > 5 \) → This inequality is always satisfied since \( c \) is positive.

Thus, we have:
\[ 7 < c < 17 \]
Since \( c \) must be an integer, the possible values for \( c \) are:
\[ c = 8, 9, 10, 11, 12, 13, 14, 15, 16 \]

### Step 2: Check for obtuse triangles
An obtuse triangle has one angle greater than 90 degrees. For a triangle with sides \( a \), \( b \), and \( c \), if \( c \) is the longest side, the triangle is obtuse if:
\[ c^2 > a^2 + b^2 \]

If \( 12 \) is the longest side, we check:
\[ 12^2 > 5^2 + c^2 \]
\[ 144 > 25 + c^2 \]
\[ c^2 < 119 \]
Thus, \( c < \sqrt{119} \approx 10.91 \). Therefore, the integer values for \( c \) in this case are:
\[ c = 8, 9, 10 \]

Now, if \( c \) is the longest side, we check:
\[ c^2 > 5^2 + 12^2 \]
\[ c^2 > 25 + 144 \]
\[ c^2 > 169 \]
Thus, \( c > \sqrt{169} = 13 \). Therefore, the integer values for \( c \) in this case are:
\[ c = 14, 15, 16 \]

### Step 3: Count the possible values of \( c \)
From the first case (where \( 12 \) is the longest side), we have:
- Possible values: \( 8, 9, 10 \) (3 values)

From the second case (where \( c \) is the longest side), we have:
- Possible values: \( 14, 15, 16 \) (3 values)

### Final Count
Adding both cases together:
- Total possible obtuse triangles = \( 3 + 3 = 6 \)

Thus, the total number of obtuse triangles possible is **6**.