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Sagot :
To prove that in a [tex]\(45^{\circ}-45^{\circ}-90^{\circ}\)[/tex] triangle, the hypotenuse is [tex]\(\sqrt{2}\)[/tex] times the length of each leg, follow these steps:
1. Start with the Pythagorean theorem for the right triangle:
In any right triangle, the side lengths satisfy the Pythagorean theorem: [tex]\(a^2 + b^2 = c^2\)[/tex].
2. Apply the Pythagorean theorem to our specific isosceles right triangle:
Since it is a [tex]\(45^{\circ}-45^{\circ}-90^{\circ}\)[/tex] triangle, the legs are equal in length. Let's denote each leg as [tex]\(a\)[/tex]. Thus, the equation becomes [tex]\(a^2 + a^2 = c^2\)[/tex].
3. Combine like terms:
This simplifies to:
[tex]\[ 2a^2 = c^2 \][/tex]
4. Isolate [tex]\(a^2\)[/tex] by dividing both sides by 2:
[tex]\[ a^2 = \frac{c^2}{2} \][/tex]
5. Determine the principal square root of both sides:
Take the square root of both sides of the equation to solve for [tex]\(a\)[/tex]:
[tex]\[ a = \sqrt{\frac{c^2}{2}} \][/tex]
6. Simplify the square root expression:
[tex]\[ a = \frac{c}{\sqrt{2}} \][/tex]
7. Solve for [tex]\(c\)[/tex] in terms of [tex]\(a\)[/tex]:
Multiply both sides by [tex]\(\sqrt{2}\)[/tex]:
[tex]\[ c = a \sqrt{2} \][/tex]
Therefore, the hypotenuse [tex]\(c\)[/tex] is [tex]\(\sqrt{2}\)[/tex] times the length of each leg [tex]\(a\)[/tex] in a [tex]\(45^{\circ}-45^{\circ}-90^{\circ}\)[/tex] triangle. This completes the proof.
1. Start with the Pythagorean theorem for the right triangle:
In any right triangle, the side lengths satisfy the Pythagorean theorem: [tex]\(a^2 + b^2 = c^2\)[/tex].
2. Apply the Pythagorean theorem to our specific isosceles right triangle:
Since it is a [tex]\(45^{\circ}-45^{\circ}-90^{\circ}\)[/tex] triangle, the legs are equal in length. Let's denote each leg as [tex]\(a\)[/tex]. Thus, the equation becomes [tex]\(a^2 + a^2 = c^2\)[/tex].
3. Combine like terms:
This simplifies to:
[tex]\[ 2a^2 = c^2 \][/tex]
4. Isolate [tex]\(a^2\)[/tex] by dividing both sides by 2:
[tex]\[ a^2 = \frac{c^2}{2} \][/tex]
5. Determine the principal square root of both sides:
Take the square root of both sides of the equation to solve for [tex]\(a\)[/tex]:
[tex]\[ a = \sqrt{\frac{c^2}{2}} \][/tex]
6. Simplify the square root expression:
[tex]\[ a = \frac{c}{\sqrt{2}} \][/tex]
7. Solve for [tex]\(c\)[/tex] in terms of [tex]\(a\)[/tex]:
Multiply both sides by [tex]\(\sqrt{2}\)[/tex]:
[tex]\[ c = a \sqrt{2} \][/tex]
Therefore, the hypotenuse [tex]\(c\)[/tex] is [tex]\(\sqrt{2}\)[/tex] times the length of each leg [tex]\(a\)[/tex] in a [tex]\(45^{\circ}-45^{\circ}-90^{\circ}\)[/tex] triangle. This completes the proof.
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