In an isosceles triangle, two sides are equal in length. Let’s denote the equal sides as \( a \) and the base as \( b \).
Given:
- The perimeter of the triangle \( P = 22 \) cm,
- One side length \( b = 6 \) cm (the base).
Since the triangle is isosceles, we have:
\[ P = 2a + b \]
Substituting the known values into the perimeter equation, we get:
\[ 22 = 2a + 6 \]
Now, solving for \( a \):
\[ 22 - 6 = 2a \] \[ 16 = 2a \] \[ a = \frac{16}{2} = 8 , \text{cm} \]
So, one possible configuration of the triangle is:
- Two equal sides \( a = 8 \) cm,
- Base \( b = 6 \) cm.
Now it's necessary to ensure that these side lengths satisfy the triangle inequality theorem, which states that the sum of the lengths of any two sides must be greater than the length of the third side.
In our case, we need to check the following:
-
\( a + a > b \) \[ 8 + 8 > 6 , (\text{True: } 16 > 6) \]
-
\( a + b > a \) \[ 8 + 6 > 8 , (\text{True: } 14 > 8) \]
-
\( a + b > a \) \[ 6 + 8 > 8 , (\text{True: } 14 > 8) \]
Since all conditions of the triangle inequality theorem are satisfied, the side lengths are valid.
Conclusion:
The only possible lengths of the other two sides (the equal sides) of the isosceles triangle are:
- Each side is \( 8 \) cm long.