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Answer :
Let each leg of the isosceles right triangle be [tex]$L$[/tex]. Then, the hypotenuse is given by [tex]$L\sqrt{2}$[/tex] (by the Pythagorean theorem). The perimeter [tex]$P$[/tex] of the triangle is the sum of the two legs and the hypotenuse. Therefore, we have
[tex]$$
P = L + L + L\sqrt{2} = 2L + L\sqrt{2} = L(2+\sqrt{2}).
$$[/tex]
We are given that the perimeter is
[tex]$$
94 + 94\sqrt{2}.
$$[/tex]
Setting the two expressions equal, we write
[tex]$$
L(2+\sqrt{2}) = 94 + 94\sqrt{2}.
$$[/tex]
Notice that the given perimeter can be factored as
[tex]$$
94 + 94\sqrt{2} = 94(1+\sqrt{2}).
$$[/tex]
Thus, the equation becomes
[tex]$$
L(2+\sqrt{2}) = 94(1+\sqrt{2}).
$$[/tex]
To solve for [tex]$L$[/tex], divide both sides by [tex]$(2+\sqrt{2})$[/tex]:
[tex]$$
L = \frac{94(1+\sqrt{2})}{2+\sqrt{2}}.
$$[/tex]
To simplify this expression, multiply both the numerator and the denominator by the conjugate of the denominator, which is [tex]$(2-\sqrt{2})$[/tex]:
[tex]$$
L = \frac{94(1+\sqrt{2})(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})}.
$$[/tex]
Calculate the denominator using the difference of squares:
[tex]$$
(2+\sqrt{2})(2-\sqrt{2}) = 2^2 - (\sqrt{2})^2 = 4 - 2 = 2.
$$[/tex]
Now, simplify the numerator:
[tex]$$
(1+\sqrt{2})(2-\sqrt{2}) = 1\cdot2 + 1\cdot(-\sqrt{2}) + \sqrt{2}\cdot2 - \sqrt{2}\cdot\sqrt{2}.
$$[/tex]
Simplify each term:
[tex]$$
= 2 - \sqrt{2} + 2\sqrt{2} - 2 = (\sqrt{2}).
$$[/tex]
The numerator now becomes:
[tex]$$
94 \cdot \sqrt{2}.
$$[/tex]
Thus, we have
[tex]$$
L = \frac{94\sqrt{2}}{2} = 47\sqrt{2}.
$$[/tex]
Therefore, the length of one leg of the triangle is
[tex]$$
\boxed{47\sqrt{2}} \text{ inches}.
$$[/tex]
[tex]$$
P = L + L + L\sqrt{2} = 2L + L\sqrt{2} = L(2+\sqrt{2}).
$$[/tex]
We are given that the perimeter is
[tex]$$
94 + 94\sqrt{2}.
$$[/tex]
Setting the two expressions equal, we write
[tex]$$
L(2+\sqrt{2}) = 94 + 94\sqrt{2}.
$$[/tex]
Notice that the given perimeter can be factored as
[tex]$$
94 + 94\sqrt{2} = 94(1+\sqrt{2}).
$$[/tex]
Thus, the equation becomes
[tex]$$
L(2+\sqrt{2}) = 94(1+\sqrt{2}).
$$[/tex]
To solve for [tex]$L$[/tex], divide both sides by [tex]$(2+\sqrt{2})$[/tex]:
[tex]$$
L = \frac{94(1+\sqrt{2})}{2+\sqrt{2}}.
$$[/tex]
To simplify this expression, multiply both the numerator and the denominator by the conjugate of the denominator, which is [tex]$(2-\sqrt{2})$[/tex]:
[tex]$$
L = \frac{94(1+\sqrt{2})(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})}.
$$[/tex]
Calculate the denominator using the difference of squares:
[tex]$$
(2+\sqrt{2})(2-\sqrt{2}) = 2^2 - (\sqrt{2})^2 = 4 - 2 = 2.
$$[/tex]
Now, simplify the numerator:
[tex]$$
(1+\sqrt{2})(2-\sqrt{2}) = 1\cdot2 + 1\cdot(-\sqrt{2}) + \sqrt{2}\cdot2 - \sqrt{2}\cdot\sqrt{2}.
$$[/tex]
Simplify each term:
[tex]$$
= 2 - \sqrt{2} + 2\sqrt{2} - 2 = (\sqrt{2}).
$$[/tex]
The numerator now becomes:
[tex]$$
94 \cdot \sqrt{2}.
$$[/tex]
Thus, we have
[tex]$$
L = \frac{94\sqrt{2}}{2} = 47\sqrt{2}.
$$[/tex]
Therefore, the length of one leg of the triangle is
[tex]$$
\boxed{47\sqrt{2}} \text{ inches}.
$$[/tex]
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