Choice B is correct. It's given that the right triangle is isosceles. In an isosceles right triangle, the two legs have equal lengths, and the length of the hypotenuse is $StartRoot 2 EndRoot$ times the length of one of the legs. Let $l$ represent the length, in inches, of each leg of the isosceles right triangle. It follows that the length of the hypotenuse is $script l StartRoot 2 EndRoot$ inches. The perimeter of a figure is the sum of the lengths of the sides of the figure. Therefore, the perimeter of the isosceles right triangle is $l + l + l \sqrt{2}$ inches. It's given that the perimeter of the triangle is $94 plus 94 StartRoot 2 EndRoot$ inches. It follows that $l + l + l \sqrt{2} = 94 + 94 \sqrt{2}$ . Factoring the left-hand side of this equation yields $(1 + 1 + \sqrt{2}) l = 94 + 94 \sqrt{2}$ , or $(2 + \sqrt{2}) l = 94 + 94 \sqrt{2}$ . Dividing both sides of this equation by $2 plus StartRoot 2 EndRoot$ yields $l = \frac{94 + 94 \sqrt{2}}{2 + \sqrt{2}}$ . Rationalizing the denominator of the right-hand side of this equation by multiplying the right-hand side of the equation by $\frac{2 - \sqrt{2}}{2 - \sqrt{2}}$ yields $l = \frac{(94 + 94 \sqrt{2}) (2 - \sqrt{2})}{(2 + \sqrt{2}) (2 - \sqrt{2})}$ . Applying the distributive property to the numerator and to the denominator of the right-hand side of this equation yields $l = \frac{188 - 94 \sqrt{2} + 188 \sqrt{2} - 94 \sqrt{4}}{4 - 2 \sqrt{2} + 2 \sqrt{2} - \sqrt{4}}$ . This is equivalent to $l = \frac{94 \sqrt{2}}{2}$ , or $script l equals 47 StartRoot 2 EndRoot$ . Therefore, the length, in inches, of one leg of the isosceles right triangle is $47 StartRoot 2 EndRoot$ .

Choice A is incorrect and may result from conceptual or calculation errors.

Choice C is incorrect. This is the length, in inches, of the hypotenuse.

Choice D is incorrect and may result from conceptual or calculation errors.