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Sagot :
Alright, let's go through the process of proving that a triangle with side lengths [tex]\( a = x^2 - 1 \)[/tex], [tex]\( b = 2x \)[/tex], and [tex]\( c = x^2 + 1 \)[/tex] is a right triangle when [tex]\( x > 1 \)[/tex]. We'll use the Pythagorean theorem to do this.
### Step-by-Step Solution:
1. State the Pythagorean theorem:
The Pythagorean theorem states that in a right triangle with legs [tex]\( a \)[/tex] and [tex]\( b \)[/tex], and hypotenuse [tex]\( c \)[/tex], the following equation holds:
[tex]\[ a^2 + b^2 = c^2 \][/tex]
2. Insert given expressions for [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
The side lengths given are [tex]\( a = x^2 - 1 \)[/tex], [tex]\( b = 2x \)[/tex], and [tex]\( c = x^2 + 1 \)[/tex]. Let's compute [tex]\( a^2 \)[/tex], [tex]\( b^2 \)[/tex], and [tex]\( c^2 \)[/tex].
3. Calculate [tex]\( a^2 \)[/tex]:
[tex]\[ a = x^2 - 1 \][/tex]
[tex]\[ a^2 = (x^2 - 1)^2 \][/tex]
Expanding the square:
[tex]\[ a^2 = (x^2 - 1)(x^2 - 1) = x^4 - 2x^2 + 1 \][/tex]
4. Calculate [tex]\( b^2 \)[/tex]:
[tex]\[ b = 2x \][/tex]
[tex]\[ b^2 = (2x)^2 = 4x^2 \][/tex]
5. Calculate [tex]\( c^2 \)[/tex]:
[tex]\[ c = x^2 + 1 \][/tex]
[tex]\[ c^2 = (x^2 + 1)^2 \][/tex]
Expanding the square:
[tex]\[ c^2 = (x^2 + 1)(x^2 + 1) = x^4 + 2x^2 + 1 \][/tex]
6. Compare [tex]\( a^2 + b^2 \)[/tex] with [tex]\( c^2 \)[/tex]:
Now, let's add [tex]\( a^2 \)[/tex] and [tex]\( b^2 \)[/tex]:
[tex]\[ a^2 + b^2 = (x^4 - 2x^2 + 1) + 4x^2 \][/tex]
Simplify the expression:
[tex]\[ a^2 + b^2 = x^4 - 2x^2 + 1 + 4x^2 = x^4 + 2x^2 + 1 \][/tex]
We have computed [tex]\( c^2 \)[/tex] as:
[tex]\[ c^2 = x^4 + 2x^2 + 1 \][/tex]
Notice that:
[tex]\[ a^2 + b^2 = c^2 \][/tex]
7. Conclusion:
Since [tex]\( a^2 + b^2 = c^2 \)[/tex] holds true, we have shown that the triangle with sides [tex]\( a = x^2 - 1 \)[/tex], [tex]\( b = 2x \)[/tex], and [tex]\( c = x^2 + 1 \)[/tex] satisfies the Pythagorean theorem.
Therefore, it is proven that for [tex]\( x > 1 \)[/tex], a triangle with side lengths [tex]\( a = x^2 - 1 \)[/tex], [tex]\( b = 2x \)[/tex], and [tex]\( c = x^2 + 1 \)[/tex] is indeed a right triangle.
### Step-by-Step Solution:
1. State the Pythagorean theorem:
The Pythagorean theorem states that in a right triangle with legs [tex]\( a \)[/tex] and [tex]\( b \)[/tex], and hypotenuse [tex]\( c \)[/tex], the following equation holds:
[tex]\[ a^2 + b^2 = c^2 \][/tex]
2. Insert given expressions for [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
The side lengths given are [tex]\( a = x^2 - 1 \)[/tex], [tex]\( b = 2x \)[/tex], and [tex]\( c = x^2 + 1 \)[/tex]. Let's compute [tex]\( a^2 \)[/tex], [tex]\( b^2 \)[/tex], and [tex]\( c^2 \)[/tex].
3. Calculate [tex]\( a^2 \)[/tex]:
[tex]\[ a = x^2 - 1 \][/tex]
[tex]\[ a^2 = (x^2 - 1)^2 \][/tex]
Expanding the square:
[tex]\[ a^2 = (x^2 - 1)(x^2 - 1) = x^4 - 2x^2 + 1 \][/tex]
4. Calculate [tex]\( b^2 \)[/tex]:
[tex]\[ b = 2x \][/tex]
[tex]\[ b^2 = (2x)^2 = 4x^2 \][/tex]
5. Calculate [tex]\( c^2 \)[/tex]:
[tex]\[ c = x^2 + 1 \][/tex]
[tex]\[ c^2 = (x^2 + 1)^2 \][/tex]
Expanding the square:
[tex]\[ c^2 = (x^2 + 1)(x^2 + 1) = x^4 + 2x^2 + 1 \][/tex]
6. Compare [tex]\( a^2 + b^2 \)[/tex] with [tex]\( c^2 \)[/tex]:
Now, let's add [tex]\( a^2 \)[/tex] and [tex]\( b^2 \)[/tex]:
[tex]\[ a^2 + b^2 = (x^4 - 2x^2 + 1) + 4x^2 \][/tex]
Simplify the expression:
[tex]\[ a^2 + b^2 = x^4 - 2x^2 + 1 + 4x^2 = x^4 + 2x^2 + 1 \][/tex]
We have computed [tex]\( c^2 \)[/tex] as:
[tex]\[ c^2 = x^4 + 2x^2 + 1 \][/tex]
Notice that:
[tex]\[ a^2 + b^2 = c^2 \][/tex]
7. Conclusion:
Since [tex]\( a^2 + b^2 = c^2 \)[/tex] holds true, we have shown that the triangle with sides [tex]\( a = x^2 - 1 \)[/tex], [tex]\( b = 2x \)[/tex], and [tex]\( c = x^2 + 1 \)[/tex] satisfies the Pythagorean theorem.
Therefore, it is proven that for [tex]\( x > 1 \)[/tex], a triangle with side lengths [tex]\( a = x^2 - 1 \)[/tex], [tex]\( b = 2x \)[/tex], and [tex]\( c = x^2 + 1 \)[/tex] is indeed a right triangle.
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