Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Join our Q&A platform and get accurate answers to all your questions from professionals across multiple disciplines. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
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.
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.
