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19 Aug 2015, 02:40
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The area of an isosceles right triangle is \(18x^2 + 6x + \frac{1}{2}\). Find the perimeter of the triangle.
(A) \(36x^2 + 12x + 2\)
(B) \(9x^2 + 3x + \frac{1}{4}\)
(C) \((2 + 2^\frac{1}{2}) (6x+1)\)
(D) \((1 + 2^\frac{1}{2}) (6x+1)\)
(E) \(18x+ 3\)
Source : Aristotle Prep
(A) \(36x^2 + 12x + 2\)
(B) \(9x^2 + 3x + \frac{1}{4}\)
(C) \((2 + 2^\frac{1}{2}) (6x+1)\)
(D) \((1 + 2^\frac{1}{2}) (6x+1)\)
(E) \(18x+ 3\)
Source : Aristotle Prep
ENGRTOMBA2018
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Concentration: Finance, Strategy
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19 Aug 2015, 04:35
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Gnpth
The area of an isosceles right triangle is \(18x^2 + 6x + \frac{1}{2}\). Find the perimeter of the triangle.
(A) \(36x^2 + 12x + 2\)
(B) \(9x^2 + 3x + \frac{1}{4}\)
(C) \((2 + 2^\frac{1}{2}) (6x+1)\)
(D) \((1 + 2^\frac{1}{2}) (6x+1)\)
(E) \(18x+ 3\)
Source : Aristotle Prep
(A) \(36x^2 + 12x + 2\)
(B) \(9x^2 + 3x + \frac{1}{4}\)
(C) \((2 + 2^\frac{1}{2}) (6x+1)\)
(D) \((1 + 2^\frac{1}{2}) (6x+1)\)
(E) \(18x+ 3\)
Source : Aristotle Prep
Let a be the perpendicular and base for the isosceles right triangle and b be the hypotenuse.
Thus area of the triangle = 0.5*a^2 = 18x^2 + 6x + 1/2 ---> a = (6x+1) ---> 2a = 2(6x+1)
b = a\(\sqrt{2}\) ---> b =\(\sqrt{2}\) (6x+1)
Thus the perimeter = 2a+b = (6x+1)(2+\(\sqrt{2}\)). C is the correct answer.
GittinGud
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Schools: Ross '22 (A) Johnson '22 (A$) Anderson '22 (A$) Marshall '22 (A) Kenan-Flagler '22 (A$) Mendoza '22 (A$)
GMAT 1: 750 Q50 V41
Schools: Ross '22 (A) Johnson '22 (A$) Anderson '22 (A$) Marshall '22 (A) Kenan-Flagler '22 (A$) Mendoza '22 (A$)
GMAT 1: 750 Q50 V41
Posts: 94
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29
13 Oct 2018, 05:17
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Can someone please post the solution to this question
