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# The perimeter of a certain isosceles right triangle is 16 +

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Intern
Joined: 24 Feb 2011
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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03 Aug 2011, 14:02
No one before him mentioned that "This problem can be solved using 45:45:90 isosceles triangle rule. As per properties of 45:45:90 triangle, the sides are in the ratio of 1:1: √2"
Here i was, wondering why everyone takes a sqr of x in their answers. Thank you!
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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21 Aug 2011, 18:57
Please explain these 3 steps - I'm not getting the Algebra. I'm sorry to be a pest:

x(2+sqrt2) = 16(1+sqrt2) ( I understand how you got here....but these next step loses me)

x(sqrt2*sqrt2 + sqrt2) = 16(1+sqrt2) ??????

x(sqrt2)(sqrt2 +1) = 16(1 + sqrt2) Cancel out the (1+sqrt2)

x(sqrt2) = 16 = Hypotenuse

Thanks!!!!
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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10 Oct 2011, 04:43
1. a+2b=16(1+root2)
2. b=a/root2

solving for a in 1: a=16

PS: Sorry for the earlier mistake.

Last edited by BDSunDevil on 10 Oct 2011, 05:10, edited 2 times in total.
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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28 Sep 2013, 12:12
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?

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

I think the the best approach here is Ballparking the size of the Hypotenuse and POE:

We know that the perimeter is 16+16*root(2)=16+ (about 27 since (16^2=256))= a little more than 42. 42/3=14, so each side is about 13 and the Hypotenuse is 13*1.7=about 15 - closest answer choice is B.
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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17 Dec 2014, 04:50
Let the hypotenuse be a.
Then the sides are a/sqrt(2). Sum of the 2 sides = 2*a/sqrt(2) = a*sqrt (2)

So the equation becomes a+a*sqrt(2) = 16+16*sqrt(2)==>t a = 16.
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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07 Mar 2015, 19:36
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 + [#permalink]

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14 Jan 2016, 02:47
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?

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

isosceles right triangle = x, x, x \sqrt{2}
So we can have here only two cases, in which 2 sides are equal:
1. Sides = 8, 8 and $$Hypotenuse=16 \sqrt{2}$$ --> $$Hypotenuse^2=8^2+8^2 is not equal to 16 \sqrt{2}$$, NO
2. $$Sides = 8 \sqrt{2}, 8 \sqrt{2}$$, $$Hypotenuse=16$$--> $$Hypotenuse^2=(8 \sqrt{2})^2+(8 \sqrt{2})^2=16$$ YES
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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19 Apr 2016, 18:14
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Expert's post
Quote:
The Perimeter of a certain isosceles right triangle is 16 + 16√2. What is the length of the hypotenuse of the triangle?

A) 8
B) 16
C) 4√2
D) 8√2
E) 16√2

An IMPORTANT point to remember is that, in any isosceles right triangle, the sides have length x, x, and x√2 for some positive value of x.

Note: x√2 is the length of the hypotenuse, so our goal is to find the value of x√2

From here, we can see that the perimeter will be x + x + x√2

In the question, the perimeter is 16 + 16√2, so we can create the following equation:
x + x + x√2 = 16 + 16√2,
Simplify: 2x + x√2 = 16 + 16√2
IMPORTANT: Factor x√2 from the left side to get : x√2(√2 + 1) = 16 + 16√2
Now factor 16 from the right side to get: x√2(√2 + 1) = 16(1 + √2)
Divide both sides by (1 + √2) to get: x√2 = 16

Cheers,
Brent
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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19 Apr 2016, 21:00
1
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Expert's post
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 + [#permalink]

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06 Aug 2016, 04:32
Very efficient solution is the following:
P=2a+b=16+16*sqrt(2), then it means that either 2a=16 and b=16*sqrt(2), or 2a=16*sqrt(2) and b=16. Since in triangle b<2a, then b=16 and a=8*sqrt(2). Then plug into a^2+a^2=b^2 and check if its works.
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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18 Aug 2016, 22:57
1
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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
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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30 Dec 2016, 04:00
1
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Let the hypotenuse be x, then the length of the leg is x/root2.
x+2x/root2=16+16*root2
x+root*x=16+16*root2
So, x=16
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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05 Nov 2017, 20:33
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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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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05 Nov 2017, 22:55
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 + [#permalink]

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08 Mar 2018, 17:46
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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Re: The perimeter of a certain isosceles right triangle is 16 +   [#permalink] 08 Mar 2018, 17:46

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