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If a right angled isosceles triangle has an area of 2x^2 + 2x + 1/2.

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Bunuel
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If a right angled isosceles triangle has an area of 2x^2 + 2x + 1/2. [#permalink] New post  03 Nov 2019, 01:44
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If a right angled isosceles triangle has an area of \(2x^2 + 2x + \frac{1}{2}\). What is its perimeter?


A. \(3(x + 1)\)

B. \(\sqrt{2}(x^2 + 1)\)

C. \((\sqrt{2}x - 1)(2x + 1)\)

D. \((2 + \sqrt{2})(2x + 1)\)

E. \(2x + \sqrt{2}\)


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Re: If a right angled isosceles triangle has an area of 2x^2 + 2x + 1/2. [#permalink] New post  03 Nov 2019, 01:54
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Bunuel wrote:
If a right angled isosceles triangle has an area of \(2x^2 + 2x + \frac{1}{2}\). What is its perimeter?


A. \(3(x + 1)\)

B. \(\sqrt{2}(x^2 + 1)\)

C. \((\sqrt{2}x - 1)(2x + 1)\)

D. \((2 + \sqrt{2})(2x + 1)\)

E. \(2x + \sqrt{2}\)


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The sides of an isosceles right angled triangle will be a, a, \(\sqrt{2}\)a
Perimeter = a + a + \(\sqrt{2}\)a = (2 + \(\sqrt{2}\))a

Area = \(\frac{1}{2}*a*a\) = \(\frac{a^2}{2}\)

Given Area = \(2x^2 + 2x + \frac{1}{2}\)
--> \(\frac{a^2}{2}\) = \(2x^2 + 2x + \frac{1}{2}\)
--> \(a^2 = 4x^2 + 4x + 1\)
--> \(a^2 = (2x + 1)^2\)
--> \(a = 2x + 1\)

Required Perimeter = (2 + \(\sqrt{2}\))a = \((2 + \sqrt{2})(2x + 1)\)

IMO Option D
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Re: If a right angled isosceles triangle has an area of 2x^2 + 2x + 1/2. [#permalink] New post  03 Nov 2019, 01:52
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D let sides b y , y and sqrty
Then
(Y^2)/2= 2x^2+2X+ 1/2
Y= 2x+1
Y + y + sqrt(y) = perimeter = D

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Re: If a right angled isosceles triangle has an area of 2x^2 + 2x + 1/2. [#permalink] New post  03 Nov 2019, 02:00
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Assume equal sides of right angled isosceles triangle= y
hypotenuse= \(\sqrt{2}y\)
Perimeter= y+y+\(\sqrt{2}y\)= \((2+\sqrt{2})y\)
Area of such triangle is \(\frac{1}{2}*y^2\)......(1)


\(2x^2 + 2x + \frac{1}{2}\)
=\(\frac{1}{2}\)* \(4x^2 + 4x + 1\)

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

Comparing (1) and (2)

y=2x+1

Perimeter of triangle = \((2+\sqrt{2})(2x+1)\)



Bunuel wrote:
If a right angled isosceles triangle has an area of \(2x^2 + 2x + \frac{1}{2}\). What is its perimeter?


A. \(3(x + 1)\)

B. \(\sqrt{2}(x^2 + 1)\)

C. \((\sqrt{2}x - 1)(2x + 1)\)

D. \((2 + \sqrt{2})(2x + 1)\)

E. \(2x + \sqrt{2}\)


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Re: If a right angled isosceles triangle has an area of 2x^2 + 2x + 1/2. [#permalink] New post  03 Nov 2019, 02:50
isosceled ∆ right angled ; sides x:x:x√2
for area
1/2 * a^2 = \(2x^2 + 2x + \frac{1}{2}\).

a=2x+1
the perimeter of side = 4x+2+√2.2x+√2
IMO D \((2 + \sqrt{2})(2x + 1)\)

Bunuel wrote:
If a right angled isosceles triangle has an area of \(2x^2 + 2x + \frac{1}{2}\). What is its perimeter?


A. \(3(x + 1)\)

B. \(\sqrt{2}(x^2 + 1)\)

C. \((\sqrt{2}x - 1)(2x + 1)\)

D. \((2 + \sqrt{2})(2x + 1)\)

E. \(2x + \sqrt{2}\)


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Re: If a right angled isosceles triangle has an area of 2x^2 + 2x + 1/2. [#permalink] New post  03 Nov 2019, 08:27
If a right angled isosceles triangle has an area of \(2x^2 + 2x + \frac{1}{2}\). What is its perimeter?


A. \(3(x + 1)\)

B. \(\sqrt{2}(x^2 + 1)\)

C. \((\sqrt{2}x - 1)(2x + 1)\)

D. \((2 + \sqrt{2})(2x + 1)\) --> correct

E. \(2x + \sqrt{2}\)

Solution:
length of the each two equal side = a
So area = a*a/2 = \(2x^2 + 2x + \frac{1}{2}\)
=> a^2=4x^2+4x+1=(2x+1)^2
=> a = 2x+1
So perimeter of the right angled isosceles triangle = a+a+ a \(\sqrt{2}\) = a (2+\(\sqrt{2}\) ) = (2+\(\sqrt{2}\) )*(2x+1)
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If a right angled isosceles triangle has an area of 2x^2 + 2x + 1/2. [#permalink] New post  03 Nov 2019, 11:13
The area, A of a right angled isosceles triangle is \(\frac{1}{2}\)\(i^{2}\), where i represents the equal legs

Also, the sides of a right angled isosceles triangle are i, i, i\(\sqrt{2}\)

Therefore, \(\frac{1}{2}\)\(i^{2}\) = 2\(x^{2}\)+2x+\(\frac{1}{2}\) ==> \(i^{2}\) = 4\(x^{2}\)+4x+1 ==> i = 2x + 1 (since the sides are positive)

Therefore, perimeter = 2x + 1 + 2x + 1 + \(\sqrt{2}\)(2x+1) ==> (2x +1)(2+\(\sqrt{2}\)) ==> the answer is D
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Re: If a right angled isosceles triangle has an area of 2x^2 + 2x + 1/2. [#permalink] New post  22 Apr 2020, 07:10
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Assume X=1
Area of the triangle would be 9/2

let a be the side and h of the isoceles right triangle
a^2/2=9/2 , a=3
Perimeter will be a+a+sqrt(a) [Sides for isoceles triangle] = 6 + 3sqrt(2)
Substitute x=1 in the answer options and
IMO D would match the answer.

Simpler way and could be done in a minute.
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Re: If a right angled isosceles triangle has an area of 2x^2 + 2x + 1/2. [#permalink] New post  22 Mar 2021, 12:59
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hiranmay wrote:
If a right angled isosceles triangle has an area of \(2x^2 + 2x + \frac{1}{2}\). What is its perimeter?


A. \(3(x + 1)\)

B. \(\sqrt{2}(x^2 + 1)\)

C. \((\sqrt{2}x - 1)(2x + 1)\)

D. \((2 + \sqrt{2})(2x + 1)\) --> correct

E. \(2x + \sqrt{2}\)

Solution:
length of the each two equal side = a
So area = a*a/2 = \(2x^2 + 2x + \frac{1}{2}\)
=> a^2=4x^2+4x+1=(2x+1)^2
=> a = 2x+1
So perimeter of the right angled isosceles triangle = a+a+ a \(\sqrt{2}\) = a (2+\(\sqrt{2}\) ) = (2+\(\sqrt{2}\) )*(2x+1)


Because x can be negative such as -2,
With a^2=4x^2+4x+1=(2x+1)^2
We should get: a = |2x+1|

There is no such option in answers though.
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Re: If a right angled isosceles triangle has an area of 2x^2 + 2x + 1/2. [#permalink] New post  01 May 2021, 04:29
zhanbo wrote:
hiranmay wrote:
If a right angled isosceles triangle has an area of \(2x^2 + 2x + \frac{1}{2}\). What is its perimeter?


A. \(3(x + 1)\)

B. \(\sqrt{2}(x^2 + 1)\)

C. \((\sqrt{2}x - 1)(2x + 1)\)

D. \((2 + \sqrt{2})(2x + 1)\) --> correct

E. \(2x + \sqrt{2}\)

Solution:
length of the each two equal side = a
So area = a*a/2 = \(2x^2 + 2x + \frac{1}{2}\)
=> a^2=4x^2+4x+1=(2x+1)^2
=> a = 2x+1
So perimeter of the right angled isosceles triangle = a+a+ a \(\sqrt{2}\) = a (2+\(\sqrt{2}\) ) = (2+\(\sqrt{2}\) )*(2x+1)


Because x can be negative such as -2,
With a^2=4x^2+4x+1=(2x+1)^2
We should get: a = |2x+1|

There is no such option in answers though.


Bunuel can you please post a reply for this??
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Re: If a right angled isosceles triangle has an area of 2x^2 + 2x + 1/2. [#permalink] New post  25 Dec 2022, 18:12
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Re: If a right angled isosceles triangle has an area of 2x^2 + 2x + 1/2. [#permalink]
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