The area of an isoscales traingle is `60 cm^(2).` The length of equal sides is 13 cm find length of its base.

Answer

To find the length of the base of the isosceles triangle given that the area is \(60 \, \text{cm}^2\) and the length of the equal sides is \(13 \, \text{cm}\), we can follow these steps:

### Step-by-Step Solution:

1. **Draw the Triangle**:
Let's denote the isosceles triangle as \(ABC\) where \(AB = AC = 13 \, \text{cm}\) (the equal sides) and \(BC\) is the base we need to find.

2. **Draw the Height**:
Draw a perpendicular line from vertex \(A\) to the base \(BC\) at point \(D\). This height \(AD\) will split the base \(BC\) into two equal segments, \(BD\) and \(CD\).

3. **Define Variables**:
Let the length of \(BD = CD = a\). Therefore, the length of the base \(BC = BD + CD = 2a\).

4. **Apply Pythagorean Theorem**:
In the right triangle \(ABD\), we can apply the Pythagorean theorem:
\[ AB^2 = AD^2 + BD^2 \]
Substituting the known values:
\[ 13^2 = h^2 + a^2 \]
This simplifies to:
\[ 169 = h^2 + a^2 \quad \text{(Equation 1)} \]

5. **Use Area Formula**:
The area of triangle \(ABC\) can be expressed as:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BC \times AD \]
Substituting the known values:
\[ 60 = \frac{1}{2} \times (2a) \times h \]
This simplifies to:
\[ 60 = ah \quad \text{(Equation 2)} \]

6. **Substitute \(h\)**:
From Equation 2, we can express \(h\) in terms of \(a\):
\[ h = \frac{60}{a} \]

7. **Substitute \(h\) into Equation 1**:
Now substitute \(h\) in Equation 1:
\[ 169 = \left(\frac{60}{a}\right)^2 + a^2 \]
Simplifying this gives:
\[ 169 = \frac{3600}{a^2} + a^2 \]
Multiplying through by \(a^2\) to eliminate the fraction:
\[ 169a^2 = 3600 + a^4 \]
Rearranging gives us a quadratic equation:
\[ a^4 - 169a^2 + 3600 = 0 \]

8. **Let \(x = a^2\)**:
Let \(x = a^2\), then the equation becomes:
\[ x^2 - 169x + 3600 = 0 \]

9. **Solve the Quadratic Equation**:
We can solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = -169\), and \(c = 3600\):
\[ x = \frac{169 \pm \sqrt{(-169)^2 - 4 \cdot 1 \cdot 3600}}{2 \cdot 1} \]
\[ x = \frac{169 \pm \sqrt{28561 - 14400}}{2} \]
\[ x = \frac{169 \pm \sqrt{14161}}{2} \]
\[ x = \frac{169 \pm 119}{2} \]
This gives us two solutions:
\[ x = \frac{288}{2} = 144 \quad \text{and} \quad x = \frac{50}{2} = 25 \]

10. **Find \(a\)**:
Since \(x = a^2\):
\[ a^2 = 144 \Rightarrow a = 12 \quad \text{and} \quad a^2 = 25 \Rightarrow a = 5 \]

11. **Calculate the Base**:
The base \(BC = 2a\):
- If \(a = 12\), then \(BC = 2 \times 12 = 24 \, \text{cm}\).
- If \(a = 5\), then \(BC = 2 \times 5 = 10 \, \text{cm}\).

### Final Answer:
The lengths of the base of the isosceles triangle can be \(10 \, \text{cm}\) or \(24 \, \text{cm}\).