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Answer :
To solve the problem of finding the length of one leg of a [tex]\(45^\circ-45^\circ-90^\circ\)[/tex] triangle where the hypotenuse is given as [tex]\(2f\)[/tex] inches, you can use the properties of this special type of triangle.
In a [tex]\(45^\circ-45^\circ-90^\circ\)[/tex] triangle, the legs are of equal length and the hypotenuse is [tex]\(\sqrt{2}\)[/tex] times the length of one leg. Therefore, if the hypotenuse is [tex]\(2f\)[/tex], we can express it as:
[tex]\[ \text{Hypotenuse} = \text{Leg} \times \sqrt{2} \][/tex]
Given:
[tex]\[ \text{Hypotenuse} = 2f \][/tex]
We can set up the equation:
[tex]\[ 2f = \text{Leg} \times \sqrt{2} \][/tex]
To solve for the length of one leg, divide both sides by [tex]\(\sqrt{2}\)[/tex]:
[tex]\[ \text{Leg} = \frac{2f}{\sqrt{2}} \][/tex]
Simplify this by multiplying the numerator and the denominator by [tex]\(\sqrt{2}\)[/tex] to rationalize the denominator:
[tex]\[ \text{Leg} = \frac{2f \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} \][/tex]
[tex]\[ \text{Leg} = \frac{2f \times \sqrt{2}}{2} \][/tex]
[tex]\[ \text{Leg} = f \times \sqrt{2} \][/tex]
However, since initially, the hypotenuse was interpreted as 2 (and "f" was treated as a unit factor), let’s consider:
[tex]\[ \text{Leg} = \frac{2}{\sqrt{2}} \][/tex]
This simplifies to approximately:
[tex]\[ \text{Leg} \approx 1.414 \][/tex]
This shows that if the hypotenuse is numerically treated as 2, the length of one leg would roughly be 1.414 inches.
In a [tex]\(45^\circ-45^\circ-90^\circ\)[/tex] triangle, the legs are of equal length and the hypotenuse is [tex]\(\sqrt{2}\)[/tex] times the length of one leg. Therefore, if the hypotenuse is [tex]\(2f\)[/tex], we can express it as:
[tex]\[ \text{Hypotenuse} = \text{Leg} \times \sqrt{2} \][/tex]
Given:
[tex]\[ \text{Hypotenuse} = 2f \][/tex]
We can set up the equation:
[tex]\[ 2f = \text{Leg} \times \sqrt{2} \][/tex]
To solve for the length of one leg, divide both sides by [tex]\(\sqrt{2}\)[/tex]:
[tex]\[ \text{Leg} = \frac{2f}{\sqrt{2}} \][/tex]
Simplify this by multiplying the numerator and the denominator by [tex]\(\sqrt{2}\)[/tex] to rationalize the denominator:
[tex]\[ \text{Leg} = \frac{2f \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} \][/tex]
[tex]\[ \text{Leg} = \frac{2f \times \sqrt{2}}{2} \][/tex]
[tex]\[ \text{Leg} = f \times \sqrt{2} \][/tex]
However, since initially, the hypotenuse was interpreted as 2 (and "f" was treated as a unit factor), let’s consider:
[tex]\[ \text{Leg} = \frac{2}{\sqrt{2}} \][/tex]
This simplifies to approximately:
[tex]\[ \text{Leg} \approx 1.414 \][/tex]
This shows that if the hypotenuse is numerically treated as 2, the length of one leg would roughly be 1.414 inches.
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