If a right angled isosceles triangle has an area of 2x^2 + 2x + 1/2. : Problem Solving (PS)

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

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

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03 Nov 2019, 02:44

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77% (02:47) correct

23% (02:52) wrong

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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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03 Nov 2019, 02:52

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]

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03 Nov 2019, 02:54

Bunuel wrote:

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 = \((2 + \sqrt{2})(2x + 1)\)

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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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03 Nov 2019, 03:00

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 = \((2+\sqrt{2})(2x+1)\)

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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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03 Nov 2019, 03:50

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

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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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03 Nov 2019, 09:27

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]

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03 Nov 2019, 12:13

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

If a right angled isosceles triangle has an area of 2x^2 + 2x + 1/2. [#permalink] 03 Nov 2019, 12:13

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

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