The lengths of the sides of a right-angled triangle are [tex]\((2x+1) \, \text{cm}\)[/tex], [tex]\(2x \, \text{cm}\)[/tex], and [tex]\((x-1) \, \text{cm}\)[/tex]. Find the value of [tex]\(x\)[/tex].



Answer :

To find the value of [tex]\( x \)[/tex] given the sides of a right-angled triangle, we will use the Pythagorean theorem. The Pythagorean theorem states that for a right-angled triangle with sides [tex]\( a \)[/tex], [tex]\( b \)[/tex] (the legs), and [tex]\( c \)[/tex] (the hypotenuse), the relationship must be:

[tex]\[ a^2 + b^2 = c^2 \][/tex]

Given:
- Side 1 ([tex]\( a \)[/tex]) = [tex]\( 2x + 1 \)[/tex]
- Side 2 ( [tex]\( b \)[/tex]) = [tex]\( 2x \)[/tex]
- Hypotenuse ( [tex]\( c \)[/tex]) = [tex]\( x - 1 \)[/tex]

Since the hypotenuse is the longest side in a right-angled triangle, we can equate:

[tex]\[ (2x + 1)^2 + (2x)^2 = (x - 1)^2 \][/tex]

Let's expand and simplify the equation step-by-step:

1. Expand each term:

[tex]\[ (2x + 1)^2 = (2x + 1)(2x + 1) = 4x^2 + 4x + 1 \][/tex]

[tex]\[ (2x)^2 = 4x^2 \][/tex]

[tex]\[ (x - 1)^2 = (x - 1)(x - 1) = x^2 - 2x + 1 \][/tex]

So our equation becomes:

[tex]\[ 4x^2 + 4x + 1 + 4x^2 = x^2 - 2x + 1 \][/tex]

2. Combine like terms:

[tex]\[ 8x^2 + 4x + 1 = x^2 - 2x + 1 \][/tex]

3. Move all terms to one side to set the equation to zero:

[tex]\[ 8x^2 + 4x + 1 - x^2 + 2x - 1 = 0 \][/tex]

4. Combine like terms:

[tex]\[ 7x^2 + 6x = 0 \][/tex]

5. Factor out the common factor [tex]\( x \)[/tex]:

[tex]\[ x(7x + 6) = 0 \][/tex]

This equation has two possible solutions:

[tex]\[ x = 0 \quad \text{or} \quad 7x + 6 = 0 \][/tex]

6. Solve for [tex]\( x \)[/tex] from [tex]\( 7x + 6 = 0 \)[/tex]:

[tex]\[ 7x + 6 = 0 \][/tex]

[tex]\[ 7x = -6 \][/tex]

[tex]\[ x = -\frac{6}{7} \][/tex]

So, the possible values of [tex]\( x \)[/tex] are 0 or [tex]\( -\frac{6}{7} \)[/tex].

Since the lengths of the sides of a triangle must be positive, [tex]\( x = 0 \)[/tex] or [tex]\( x = -\frac{6}{7} \)[/tex] is not valid because they would lead to negative or zero lengths.

Therefore, there is actually no valid solution for [tex]\( x \)[/tex] that will result in positive side lengths for the given triangle with these expressions.

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