If the measures of the sides of triangle are `(x^2 - 1), (x^2 + 1)` and 2x cm, then the triangle would be

Options

A. equilateral
B. acute - angled
C. isosceles
D. right-angled

Answer

To determine the type of triangle formed by the sides \( (x^2 - 1) \), \( (x^2 + 1) \), and \( 2x \) cm, we will follow these steps:

### Step 1: Identify the sides of the triangle
The sides of the triangle are given as:
- Side 1: \( a = x^2 - 1 \)
- Side 2: \( b = x^2 + 1 \)
- Side 3: \( c = 2x \)

### Step 2: Determine the longest side
To apply the Pythagorean theorem, we need to identify the longest side. We compare the three sides:
- \( x^2 + 1 \) is greater than \( x^2 - 1 \) since \( 1 > -1 \).
- To compare \( x^2 + 1 \) and \( 2x \), we need to analyze the expression \( x^2 + 1 \) vs. \( 2x \).

We can rearrange the inequality:
\[ x^2 + 1 \geq 2x \implies x^2 - 2x + 1 \geq 0 \implies (x - 1)^2 \geq 0 \]
This inequality holds for all \( x \), and it is equal to zero when \( x = 1 \). Therefore, \( x^2 + 1 \) is the longest side for \( x \geq 1 \).

### Step 3: Apply the Pythagorean theorem
Now we will check if the triangle is a right triangle by applying the Pythagorean theorem:
\[ \text{Hypotenuse}^2 = \text{Perpendicular}^2 + \text{Base}^2 \]
Let’s assume \( c = x^2 + 1 \) (the longest side), and check if:
\[ (x^2 + 1)^2 = (x^2 - 1)^2 + (2x)^2 \]

Calculating each side:
1. Left side (hypotenuse):
\[ (x^2 + 1)^2 = x^4 + 2x^2 + 1 \]

2. Right side (perpendicular and base):
\[ (x^2 - 1)^2 = x^4 - 2x^2 + 1 \]
\[ (2x)^2 = 4x^2 \]
Combining these:
\[ (x^2 - 1)^2 + (2x)^2 = (x^4 - 2x^2 + 1) + 4x^2 = x^4 + 2x^2 + 1 \]

### Step 4: Compare both sides
Now we compare both sides:
\[ x^4 + 2x^2 + 1 = x^4 + 2x^2 + 1 \]
Since both sides are equal, the triangle satisfies the Pythagorean theorem.

### Conclusion
Thus, the triangle with sides \( (x^2 - 1) \), \( (x^2 + 1) \), and \( 2x \) is a right triangle.