In `DeltaABC, AB = BC = k, AC = sqrt(2) k`, then `DeltaABC` is a:

Options

A. Isosceles triangle
B. Right-angled triangle
C. Equilateral triangle
D. Right isosceles triangle

Answer

To determine the type of triangle \( \Delta ABC \) given that \( AB = BC = k \) and \( AC = \sqrt{2} k \), we can follow these steps:

### Step 1: Identify the sides of the triangle
We have:
- \( AB = k \)
- \( BC = k \)
- \( AC = \sqrt{2} k \)

### Step 2: Check if the triangle is isosceles
Since \( AB \) and \( BC \) are equal, \( \Delta ABC \) is an isosceles triangle.

### Step 3: Check if the triangle is a right triangle using the Pythagorean theorem
To determine if \( \Delta ABC \) is a right triangle, we can apply the Pythagorean theorem, which states:
\[ c^2 = a^2 + b^2 \]
where \( c \) is the length of the hypotenuse (the longest side), and \( a \) and \( b \) are the lengths of the other two sides.

In our case:
- Let \( AC \) be the hypotenuse, so \( c = AC = \sqrt{2} k \)
- Let \( AB = k \) and \( BC = k \), so \( a = k \) and \( b = k \)

### Step 4: Calculate the squares of the sides
Now we calculate:
\[ c^2 = (\sqrt{2} k)^2 = 2k^2 \]
\[ a^2 + b^2 = k^2 + k^2 = 2k^2 \]

### Step 5: Compare the results
Since \( c^2 = a^2 + b^2 \):
\[ 2k^2 = 2k^2 \]
This confirms that the triangle satisfies the Pythagorean theorem.

### Conclusion
Since \( \Delta ABC \) is isosceles (two sides are equal) and also satisfies the Pythagorean theorem (indicating it has a right angle), we conclude that \( \Delta ABC \) is a right isosceles triangle.

### Final Answer
The triangle \( \Delta ABC \) is a right isosceles triangle.

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