the side of a triangle are ` 1/2 ( a+b),1/2(a-b) and sqrt(ab)` state the nature of triangle.

Answer

To determine the nature of the triangle with sides given as \( \frac{1}{2}(a+b) \), \( \frac{1}{2}(a-b) \), and \( \sqrt{ab} \), we will follow these steps:

### Step 1: Identify the sides
Let:
- Side 1 = \( \frac{1}{2}(a+b) \)
- Side 2 = \( \frac{1}{2}(a-b) \)
- Side 3 = \( \sqrt{ab} \)

### Step 2: Determine the longest side
To determine the nature of the triangle, we first need to identify which of the three sides is the longest.

We compare:
- \( \frac{1}{2}(a+b) \) and \( \frac{1}{2}(a-b) \): Since \( b \) is subtracted in the second term, \( \frac{1}{2}(a+b) \) is greater than \( \frac{1}{2}(a-b) \) when \( b > 0 \).
- Now we compare \( \frac{1}{2}(a+b) \) with \( \sqrt{ab} \).

To check if \( \frac{1}{2}(a+b) \) is greater than \( \sqrt{ab} \):
\[ \left(\frac{1}{2}(a+b)\right)^2 = \frac{1}{4}(a^2 + 2ab + b^2) \]
\[ \sqrt{ab}^2 = ab \]
We need to check if:
\[ \frac{1}{4}(a^2 + 2ab + b^2) > ab \]
This simplifies to:
\[ a^2 + 2ab + b^2 > 4ab \]
\[ a^2 - 2ab + b^2 > 0 \]
\[ (a-b)^2 > 0 \]
This is true for \( a \neq b \).

### Step 3: Use the Pythagorean theorem
Since \( \frac{1}{2}(a+b) \) is the longest side, we will check if it satisfies the Pythagorean theorem:
\[ \left(\frac{1}{2}(a+b)\right)^2 = \left(\frac{1}{2}(a-b)\right)^2 + (\sqrt{ab})^2 \]
Substituting the values:
\[ \frac{1}{4}(a^2 + 2ab + b^2) = \frac{1}{4}(a^2 - 2ab + b^2) + ab \]
This simplifies to:
\[ \frac{1}{4}(a^2 + 2ab + b^2) = \frac{1}{4}(a^2 - 2ab + b^2) + ab \]
Multiplying through by 4 to eliminate the fraction:
\[ a^2 + 2ab + b^2 = a^2 - 2ab + b^2 + 4ab \]
Simplifying gives:
\[ a^2 + 2ab + b^2 = a^2 + 2ab + b^2 \]
This is true, indicating that the triangle satisfies the Pythagorean theorem.

### Conclusion
Since the longest side squared equals the sum of the squares of the other two sides, the triangle is a **right triangle**.

---