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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, 02:40
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65% (hard)

Question Stats:

64% (02:42) correct 36% (02:54) wrong based on 121 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, 04:35
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, 05:17
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, 09:40
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, 05:46.
Last edited by Salsanousi on 18 Oct 2018, 09:40, 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, 05:50
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, 05:51
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
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Joined: 22 Oct 2017
Posts: 19
Re: The area of an isosceles right triangle is  [#permalink]

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09 Feb 2019, 03:02
Salsanousi wrote:
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

Posted from my mobile device

How do you solve quickly that second grade equation?? kudos for help please
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Re: The area of an isosceles right triangle is  [#permalink]

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09 Feb 2019, 05:57
1
KHow wrote:
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

The area of the right isosceles triangle = (1/2) * base * height

Let base = height = k

Then area = $$\frac{1}{2} * k^2$$

Area is given as $$18x ^2 + 6x + \frac{1}{2}$$

= $$18x^2 + 3x +3x +\frac{1}{2}$$

= 6x(3x + $$\frac{1}{2}$$) + 1 (3x + $$\frac{1}{2}$$)

= (6x +1)(3x + $$\frac{1}{2}$$)

= $$\frac{1}{2} * ( 6x+1)^2$$

So, k= base = height = 6x+1

Since it is a right angled isosceles triangle

Hypotenuse = (6x+1) * $$\sqrt{2}$$

Perimeter = (6x+1) 2 + (6x+1) $$\sqrt{2}$$

Choice C
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Re: The area of an isosceles right triangle is   [#permalink] 09 Feb 2019, 05:57
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# The area of an isosceles right triangle is

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