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

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Math Expert
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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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45% (medium)

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77% (02:47) correct 23% (02:52) wrong based on 96 sessions

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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]

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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
3
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. $$2x + \sqrt{2}$$

Are You Up For the Challenge: 700 Level Questions

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]

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

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. $$2x + \sqrt{2}$$

Are You Up For the Challenge: 700 Level Questions
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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 $$(2 + \sqrt{2})(2x + 1)$$

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. $$2x + \sqrt{2}$$

Are You Up For the Challenge: 700 Level Questions
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Posts: 474
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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