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Intern  Joined: 21 Apr 2014
Posts: 38
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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So we know that the proportions of a isosceles right triangle are x:x:x\sqrt{2}, since the perimeter is 16+16*sqrt(2), then
2x+x*sqrt(2)=16+16\sqrt{2}. we can solve for x and use that information to figure out what the hypotenuse is

First, let's isolate x, so we get x=16+16*sqrt(2)/(2+sqrt(2))
we can eliminate the root from the denominator by multiplying both the numerator and the denominator by 2-sqrt(2), and we end up with:
16*sqrt(2)/2=x or x=8*sqrt(2)

This is one of the options, because the GMAT is trying to trick you into picking it. However, we are looking for the hypotenuse and this is one of the sides, so we have to multiply it by sqrt(2) and we get 16 (B)
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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Flexxice wrote:
The perimeter of a certain isosceles triangle is 16 + 16 * square root of 2. What is the length of the hypotenuse of the triangle?

A) 8

B) 16

C) 4 * square root of 2

D) 8 * square root of 2

E) 16 * square root of 2

Firstly you have missed out onright angle in "certain isosceles triangle"..
secondly..
16 +16$$\sqrt{2}$$ should tell us that the two equal sides are either 8 or 8$$\sqrt{2}$$..
If it is 8, perimeter = 8+8+8$$\sqrt{2}$$, which is not the case here..
if it is 8$$\sqrt{2}$$, P = 8$$\sqrt{2}$$+8$$\sqrt{2}$$+8$$\sqrt{2}*\sqrt{2}$$ = 16 + 16$$\sqrt{2}$$.. which is the P
so side is 8$$\sqrt{2}$$ and HYP = 16
B
OA is being edited accordingly
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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coffeeloverfreak wrote:
The perimeter of a certain isosceles right triangle is $$16 + 16 \sqrt{2}$$. What is the length of the hypotenuse of the triangle?

$$(A) \hspace{5} 8$$
$$(B) \hspace{5} 16$$
$$(C) \hspace{5} 4\sqrt{2}$$
$$(D) \hspace{5} 8\sqrt{2}$$
$$(E) \hspace{5} 16\sqrt{2}$$

Hence the answer is B) 16
Attachments triangle1.PNG [ 14.17 KiB | Viewed 3042 times ]

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Joined: 20 Sep 2016
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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I am little confuse here. given that this is a Isosceles right angle triangle which can be either 30-60-90 triangle or 45-45-90.
I don't understand,why it should be intuitive to consider 45-45-90 triangle ?

would appreciate your response. thanks !
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Posts: 60460
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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jyotipes21@gmail.com wrote:
I am little confuse here. given that this is a Isosceles right angle triangle which can be either 30-60-90 triangle or 45-45-90.
I don't understand,why it should be intuitive to consider 45-45-90 triangle ?

would appreciate your response. thanks !

An isosceles triangle is a triangle with at least two sides of the same length. An isosceles triangle also has at least two angles of the same measure; namely, the angles opposite to the two sides of the same length. So, 30-60-90 is NOT an isosceles triangle.

You should go through the basics once more.

Basics to Brush-Up Fundamentals

For more check Ultimate GMAT Quantitative Megathread
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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coffeeloverfreak wrote:
The perimeter of a certain isosceles right triangle is $$16 + 16 \sqrt{2}$$. What is the length of the hypotenuse of the triangle?

$$(A) \hspace{5} 8$$

$$(B) \hspace{5} 16$$

$$(C) \hspace{5} 4\sqrt{2}$$

$$(D) \hspace{5} 8\sqrt{2}$$

$$(E) \hspace{5} 16\sqrt{2}$$

Main Idea: Equate the expression for the sum of the sides to the value given

Details: Let the equal sides be x. So hypotenuse is x*sqrt(2)

Perimeter = 2x+x*sqrt(2)= 16 + 16 *sqrt(2)

Simplifying we have x= 16(1+ sqrt(2)) /(2+sqrt(2))

So x*sqrt(2) = 16.

Hence B.
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The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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coffeeloverfreak wrote:
The perimeter of a certain isosceles right triangle is $$16 + 16 \sqrt{2}$$. What is the length of the hypotenuse of the triangle?

(A) $$8$$

(B) $$16$$

(C) $$4\sqrt{2}$$

(D) $$8\sqrt{2}$$

(E) $$16\sqrt{2}$$

If the legs are x and x, then Hypotenuse = $$\sqrt{2}x$$
$$x+x+\sqrt{2}x=16+16 \sqrt{2}$$
$$x(2+\sqrt{2})=16(1+\sqrt{2})$$
$$x(\sqrt{2}*\sqrt{2} +\sqrt{2})=16(1+\sqrt{2})$$
$$x\sqrt{2}(\sqrt{2} +1)=16(1+\sqrt{2})$$
=>$$\sqrt{2}*x =16$$
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The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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coffeeloverfreak wrote:
The perimeter of a certain isosceles right triangle is $$16 + 16 \sqrt{2}$$. What is the length of the hypotenuse of the triangle?

(A) $$8$$

(B) $$16$$

(C) $$4\sqrt{2}$$

(D) $$8\sqrt{2}$$

(E) $$16\sqrt{2}$$

x+x+$$x\sqrt{2}$$=16+$$16\sqrt{2}$$
2x+$$x\sqrt{2}$$= 16+$$16\sqrt{2}$$
x(2+$$\sqrt{2}$$)=8(2+$$2\sqrt{2}$$)
x=8 and x=$$8\sqrt{2}$$
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Posts: 60460
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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rishabhmishra1993 wrote:
coffeeloverfreak wrote:
The perimeter of a certain isosceles right triangle is $$16 + 16 \sqrt{2}$$. What is the length of the hypotenuse of the triangle?

(A) $$8$$

(B) $$16$$

(C) $$4\sqrt{2}$$

(D) $$8\sqrt{2}$$

(E) $$16\sqrt{2}$$

x+x+$$x\sqrt{2}$$=16+$$16\sqrt{2}$$
2x+$$x\sqrt{2}$$= 16+$$16\sqrt{2}$$
x(2+$$\sqrt{2}$$)=8(2+$$2\sqrt{2}$$)
x=8 and x=$$8\sqrt{2}$$

$$x\sqrt{2}+x+x=16+16\sqrt{2}$$;

$$x\sqrt{2}+2x=16+16\sqrt{2}$$;

Factor out $$x\sqrt{2}$$ from the left hand side: $$x\sqrt{2}(1+\sqrt{2})=16+16\sqrt{2}$$;

Factor out 16 from the right hand side: $$x\sqrt{2}(1+\sqrt{2})=16(1+\sqrt{2})$$;

Reduce by $$1+\sqrt{2}$$: $$x\sqrt{2}=16$$;

Hope it's clear.
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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Bunuel wrote:
rishabhmishra1993 wrote:
coffeeloverfreak wrote:
The perimeter of a certain isosceles right triangle is $$16 + 16 \sqrt{2}$$. What is the length of the hypotenuse of the triangle?

(A) $$8$$

(B) $$16$$

(C) $$4\sqrt{2}$$

(D) $$8\sqrt{2}$$

(E) $$16\sqrt{2}$$

x+x+$$x\sqrt{2}$$=16+$$16\sqrt{2}$$
2x+$$x\sqrt{2}$$= 16+$$16\sqrt{2}$$
x(2+$$\sqrt{2}$$)=8(2+$$2\sqrt{2}$$)
x=8 and x=$$8\sqrt{2}$$

$$x\sqrt{2}+x+x=16+16\sqrt{2}$$;

$$x\sqrt{2}+2x=16+16\sqrt{2}$$;

Factor out $$x\sqrt{2}$$ from the left hand side: $$x\sqrt{2}(1+\sqrt{2})=16+16\sqrt{2}$$;

Factor out 16 from the right hand side: $$x\sqrt{2}(1+\sqrt{2})=16(1+\sqrt{2})$$;

Reduce by $$1+\sqrt{2}$$: $$x\sqrt{2}=16$$;

Hope it's clear.

but what's wrong with what i did in this question?
Math Expert V
Joined: 02 Sep 2009
Posts: 60460
Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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rishabhmishra1993 wrote:
Bunuel wrote:
rishabhmishra1993 wrote:
The perimeter of a certain isosceles right triangle is $$16 + 16 \sqrt{2}$$. What is the length of the hypotenuse of the triangle?

(A) $$8$$

(B) $$16$$

(C) $$4\sqrt{2}$$

(D) $$8\sqrt{2}$$

(E) $$16\sqrt{2}$$

x+x+$$x\sqrt{2}$$=16+$$16\sqrt{2}$$
2x+$$x\sqrt{2}$$= 16+$$16\sqrt{2}$$
x(2+$$\sqrt{2}$$)=8(2+$$2\sqrt{2}$$)
x=8 and x=$$8\sqrt{2}$$

$$x\sqrt{2}+x+x=16+16\sqrt{2}$$;

$$x\sqrt{2}+2x=16+16\sqrt{2}$$;

Factor out $$x\sqrt{2}$$ from the left hand side: $$x\sqrt{2}(1+\sqrt{2})=16+16\sqrt{2}$$;

Factor out 16 from the right hand side: $$x\sqrt{2}(1+\sqrt{2})=16(1+\sqrt{2})$$;

Reduce by $$1+\sqrt{2}$$: $$x\sqrt{2}=16$$;

Hope it's clear.

but what's wrong with what i did in this question?

First of all, we have a linear (first degrees) equation and it cannot give two solutions. Next, $$x=8$$ does not satisfy $$x(2+\sqrt{2})=8(2+2\sqrt{2})$$.
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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coffeeloverfreak wrote:
The perimeter of a certain isosceles right triangle is $$16 + 16 \sqrt{2}$$. What is the length of the hypotenuse of the triangle?

(A) $$8$$

(B) $$16$$

(C) $$4\sqrt{2}$$

(D) $$8\sqrt{2}$$

(E) $$16\sqrt{2}$$

Iso right triangle ratio is x : x : x√2
We are given P = 16 + 16√2 = 2x + x√2
In a triangle 2 sides must be more than the 3rd side, so 2x= 16√2 and x√2 = 16
You can solve the first one... x=8√2 * √2 for hyp length = 16, ... second one is the answer directly.
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Re: The perimeter of a certain isosceles right triangle is 16 +  [#permalink]

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