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An isosceles right triangle has a perimeter of $94+94\\sqrt{2}$ inches. What is the length, in inches, of one leg of this triangle?

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An isosceles right triangle has a perimeter of $94+94\\sqrt{2}$ inches. What is the length, in inches, of one leg of this triangle?

","comment":{"@type":"Comment","text":"

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 $\\sqrt{2}$ 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 $l\\sqrt{2}$ 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 $94+94\\sqrt{2}$ inches. It follows that $l+l+l\\sqrt{2}=94+94\\sqrt{2}$. Factoring the left-hand side of this equation yields $\\left(1+1+\\sqrt{2}\\right)l=94+94\\sqrt{2}$, or $\\left(2+\\sqrt{2}\\right)l=94+94\\sqrt{2}$. Dividing both sides of this equation by $2+\\sqrt{2}$ 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 $l=\\frac{\\left(94+94\\sqrt{2}\\right)\\left(2-\\sqrt{2}\\right)}{\\left(2+\\sqrt{ 2}\\right)\\left(2-\\sqrt{2}\\right)}$. 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 $l=47\\sqrt{2}$. Therefore, the length, in inches, of one leg of the isosceles right triangle is $47\\sqrt{2}$.

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.

"},"encodingFormat":"text/markdown","text":"

An isosceles right triangle has a perimeter of $94+94\\sqrt{2}$ inches. What is the length, in inches, of one leg of this triangle?

","suggestedAnswer":[{"@type":"Answer","position":0,"encodingFormat":"text/html","text":"

$47$

"},{"@type":"Answer","position":2,"encodingFormat":"text/html","text":"

$94$

"},{"@type":"Answer","position":3,"encodingFormat":"text/html","text":"

$94\\sqrt{2}$

"}],"acceptedAnswer":{"@type":"Answer","position":1,"encodingFormat":"text/html","text":"

$47\\sqrt{2}$

","comment":{"@type":"Comment","text":"

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"}}},"url":"https://classnotes.xyz/mcqs/sat-math-geometry-and-trigonometry-multiple-choice-question-28"}

An isosceles right triangle has a perimeter of $94+94\sqrt{2}$ inches. What is the length, in inches, of one leg of this triangle?