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# The area of an isosceles right triangle is

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The area of an isosceles right triangle is  [#permalink]

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19 Aug 2015, 01:40
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Difficulty:

65% (hard)

Question Stats:

68% (01:58) correct 32% (03:07) wrong based on 131 sessions

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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

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Re: The area of an isosceles right triangle is  [#permalink]

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19 Aug 2015, 03:35
Gnpth wrote:
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

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.
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Re: The area of an isosceles right triangle is  [#permalink]

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13 Oct 2018, 04:17
Can someone please post the solution to this question
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The area of an isosceles right triangle is  [#permalink]

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Updated on: 18 Oct 2018, 08:40
1
GittinGud wrote:
Can someone please post the solution to this question

GittinGud

If area 18x^2 + 6x + 1/2

Area of right angle isosceles = 1/2 a^2

So a^2 = 36x^2 + 12x + 1

a^2 = (6x+1)^2
a = 6x+1

An isosceles in terms 45:45:90 is a:a:a sqrt2

So the perimeter would be 2a+a sqrt of 2

Take a as a common factor

(2 + sqrt 2) (a)

It is (2 + sqrt 2) (6x + 1)

Please let me know if this helps.

Posted from my mobile device

Originally posted by Salsanousi on 13 Oct 2018, 04:46.
Last edited by Salsanousi on 18 Oct 2018, 08:40, edited 1 time in total.
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Re: The area of an isosceles right triangle is  [#permalink]

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13 Oct 2018, 04:50
1
GittinGud wrote:
Can someone please post the solution to this question

It's a right isosceles triangle (45-45-90). Twp sides are equal.
Let perpendicular and base be "k" and hypo be k√2.

$$1/2 *k*k = 18x^2 + 6x + 1/2$$
$$k^2 = 36x^2 + 12x + 1 = (6x+1)^2$$
$$k = 6x+1 ; k√2 = (6x+1)√2$$

Perimeter = k + k + k√2 = 6x+1 + 6x+1 + 6x+1√2
= (6x+1)(2+√2)
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Re: The area of an isosceles right triangle is  [#permalink]

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12 Jan 2019, 04:51
Could someone provide more insight about how C is the correct answer (in terms of steps of the problem)? I am getting stuck in the middle. Thank you!
Gnpth wrote:
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
Re: The area of an isosceles right triangle is &nbs [#permalink] 12 Jan 2019, 04:51
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