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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
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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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
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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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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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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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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hiranmay
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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zhanbo
hiranmay
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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