In a right angled triangle ABC, right angled at B, AB=3, BC=x+2 and AC=x+3. Then find the value of x:

Options

A. 4
B. 2
C. 1
D. 3

Answer

To solve the problem, we will use the Pythagorean theorem, which states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Given:
- Triangle ABC is right-angled at B.
- AB = 3 (one leg of the triangle)
- BC = x + 2 (the other leg of the triangle)
- AC = x + 3 (the hypotenuse)

According to the Pythagorean theorem:
\[ AC^2 = AB^2 + BC^2 \]

Substituting the values we have:
\[ (x + 3)^2 = 3^2 + (x + 2)^2 \]

Now, let's expand both sides of the equation.

1. Expand the left side:
\[ (x + 3)^2 = x^2 + 6x + 9 \]

2. Expand the right side:
\[ 3^2 = 9 \]
\[ (x + 2)^2 = x^2 + 4x + 4 \]
So, the right side becomes:
\[ 9 + (x^2 + 4x + 4) = x^2 + 4x + 13 \]

Now, we can set the two sides of the equation equal to each other:
\[ x^2 + 6x + 9 = x^2 + 4x + 13 \]

3. Subtract \(x^2\) from both sides:
\[ 6x + 9 = 4x + 13 \]

4. Now, subtract \(4x\) from both sides:
\[ 2x + 9 = 13 \]

5. Next, subtract 9 from both sides:
\[ 2x = 4 \]

6. Finally, divide both sides by 2:
\[ x = 2 \]

Thus, the value of \(x\) is \(2\).