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Sagot :
To find the length of one leg of a [tex]\(45^\circ-45^\circ-90^\circ\)[/tex] triangle given the hypotenuse, you can make use of the properties of this special type of right triangle.
In a [tex]\(45^\circ-45^\circ-90^\circ\)[/tex] triangle, both legs are of equal length, and the hypotenuse is [tex]\(\sqrt{2}\)[/tex] times the length of one leg. Let's denote the length of one leg by [tex]\(x\)[/tex].
The relationship between the hypotenuse and the leg in a [tex]\(45^\circ-45^\circ-90^\circ\)[/tex] triangle is given by:
[tex]\[ \text{Hypotenuse} = x \sqrt{2} \][/tex]
Given that the hypotenuse is 18 cm, we can set up the equation:
[tex]\[ 18 = x \sqrt{2} \][/tex]
To solve for [tex]\(x\)[/tex], we divide both sides of the equation by [tex]\(\sqrt{2}\)[/tex]:
[tex]\[ x = \frac{18}{\sqrt{2}} \][/tex]
To rationalize the denominator (if needed) and simplify, multiply the numerator and the denominator by [tex]\(\sqrt{2}\)[/tex]:
[tex]\[ x = \frac{18 \sqrt{2}}{2} \][/tex]
[tex]\[ x = 9 \sqrt{2} \][/tex]
Thus, the length of one leg of the triangle is:
[tex]\[ 9 \sqrt{2} \, \text{cm} \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{9 \sqrt{2} \text{ cm}} \][/tex]
In a [tex]\(45^\circ-45^\circ-90^\circ\)[/tex] triangle, both legs are of equal length, and the hypotenuse is [tex]\(\sqrt{2}\)[/tex] times the length of one leg. Let's denote the length of one leg by [tex]\(x\)[/tex].
The relationship between the hypotenuse and the leg in a [tex]\(45^\circ-45^\circ-90^\circ\)[/tex] triangle is given by:
[tex]\[ \text{Hypotenuse} = x \sqrt{2} \][/tex]
Given that the hypotenuse is 18 cm, we can set up the equation:
[tex]\[ 18 = x \sqrt{2} \][/tex]
To solve for [tex]\(x\)[/tex], we divide both sides of the equation by [tex]\(\sqrt{2}\)[/tex]:
[tex]\[ x = \frac{18}{\sqrt{2}} \][/tex]
To rationalize the denominator (if needed) and simplify, multiply the numerator and the denominator by [tex]\(\sqrt{2}\)[/tex]:
[tex]\[ x = \frac{18 \sqrt{2}}{2} \][/tex]
[tex]\[ x = 9 \sqrt{2} \][/tex]
Thus, the length of one leg of the triangle is:
[tex]\[ 9 \sqrt{2} \, \text{cm} \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{9 \sqrt{2} \text{ cm}} \][/tex]
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