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Sagot :
To show 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 for any [tex]\(x > 1\)[/tex], we can use the Pythagorean theorem. The theorem states that in a right triangle, the square of the length of the hypotenuse ([tex]\(c\)[/tex]) is equal to the sum of the squares of the lengths of the other two sides ([tex]\(a\)[/tex] and [tex]\(b\)[/tex]). That is, [tex]\(a^2 + b^2 = c^2\)[/tex].
Given:
[tex]\[ a = x^2 - 1 \][/tex]
[tex]\[ b = 2x \][/tex]
[tex]\[ c = x^2 + 1 \][/tex]
Let's compute [tex]\(a^2\)[/tex], [tex]\(b^2\)[/tex], and [tex]\(c^2\)[/tex]:
1. Compute [tex]\(a^2\)[/tex]:
[tex]\[ a^2 = (x^2 - 1)^2 = x^4 - 2x^2 + 1 \][/tex]
2. Compute [tex]\(b^2\)[/tex]:
[tex]\[ b^2 = (2x)^2 = 4x^2 \][/tex]
3. Compute [tex]\(c^2\)[/tex]:
[tex]\[ c^2 = (x^2 + 1)^2 = x^4 + 2x^2 + 1 \][/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]\[ a^2 + b^2 = x^4 - 2x^2 + 1 + 4x^2 \][/tex]
[tex]\[ a^2 + b^2 = x^4 + 2x^2 + 1 \][/tex]
We have found:
[tex]\[ a^2 + b^2 = x^4 + 2x^2 + 1 \][/tex]
Compare this with [tex]\(c^2\)[/tex]:
[tex]\[ c^2 = x^4 + 2x^2 + 1 \][/tex]
Since [tex]\(a^2 + b^2 = c^2\)[/tex], it confirms that the 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 for any [tex]\(x > 1\)[/tex].
Hence, by checking both sides of the equation [tex]\(a^2 + b^2 = c^2\)[/tex] and finding them equal, we have proven that the given lengths follow the rule describing right triangles according to the Pythagorean theorem.
Given:
[tex]\[ a = x^2 - 1 \][/tex]
[tex]\[ b = 2x \][/tex]
[tex]\[ c = x^2 + 1 \][/tex]
Let's compute [tex]\(a^2\)[/tex], [tex]\(b^2\)[/tex], and [tex]\(c^2\)[/tex]:
1. Compute [tex]\(a^2\)[/tex]:
[tex]\[ a^2 = (x^2 - 1)^2 = x^4 - 2x^2 + 1 \][/tex]
2. Compute [tex]\(b^2\)[/tex]:
[tex]\[ b^2 = (2x)^2 = 4x^2 \][/tex]
3. Compute [tex]\(c^2\)[/tex]:
[tex]\[ c^2 = (x^2 + 1)^2 = x^4 + 2x^2 + 1 \][/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]\[ a^2 + b^2 = x^4 - 2x^2 + 1 + 4x^2 \][/tex]
[tex]\[ a^2 + b^2 = x^4 + 2x^2 + 1 \][/tex]
We have found:
[tex]\[ a^2 + b^2 = x^4 + 2x^2 + 1 \][/tex]
Compare this with [tex]\(c^2\)[/tex]:
[tex]\[ c^2 = x^4 + 2x^2 + 1 \][/tex]
Since [tex]\(a^2 + b^2 = c^2\)[/tex], it confirms that the 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 for any [tex]\(x > 1\)[/tex].
Hence, by checking both sides of the equation [tex]\(a^2 + b^2 = c^2\)[/tex] and finding them equal, we have proven that the given lengths follow the rule describing right triangles according to the Pythagorean theorem.
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