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The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)

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

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New post 11 May 2011, 06:19
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Please point out what am I doing wrong:

Let the non-hypotenuse sides be \(x\). Therefore, hypotenuse= \(\sqrt{2}x\)

\(So,Perimeter= 2x+\sqrt{2}x= 16+16\sqrt{2}\)

\(=> x(2+\sqrt{2})= 8(2+\sqrt{2})\)

Comparing both sides, \(x=8\), Therefore, \(hypotenuse= 8\sqrt{2}\)

This answer is also included in the answer choices which makes me sure that I'm doing a very silly mistake somewhere, just cant see where! Please guys, please point out that small mistake that is eluding my eyes!
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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New post 02 Aug 2011, 10:12
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

Let the sides be x, x, and x√2 (where x is the length of legs and x√2 is hypotenuse). We have to find out length of hypotenuse, i.e. x√2

Perimeter =
2x + x√2 = 16 +16√2
or, x√2(√2 + 1) = 16(1 + √2)
or, x√2 = 16

Ans. 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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New post 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 + 16*2^(1/2)  [#permalink]

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New post 19 Apr 2016, 21:00
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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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New post 18 Aug 2016, 22:57
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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 + 16*2^(1/2)  [#permalink]

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New post 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 + 16*2^(1/2)  [#permalink]

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New post 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




23. Geometry




24. Coordinate Geometry




25. Triangles




26. Polygons




27. Circles




28. Rectangular Solids and Cylinders




29. Graphs and Illustrations




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

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New post 02 Jul 2018, 12:15
gmatpapa wrote:
Please point out what am I doing wrong:

Let the non-hypotenuse sides be \(x\). Therefore, hypotenuse= \(\sqrt{2}x\)

\(So,Perimeter= 2x+\sqrt{2}x= 16+16\sqrt{2}\)

\(=> x(2+\sqrt{2})= 8(2+\sqrt{2})\)

Comparing both sides, \(x=8\), Therefore, \(hypotenuse= 8\sqrt{2}\)

This answer is also included in the answer choices which makes me sure that I'm doing a very silly mistake somewhere, just cant see where! Please guys, please point out that small mistake that is eluding my eyes!


16 + 16√2 = 2a + a√2
16 + 16√2 = a(2 + √2)
so
a = (16 + 16√2) / (2 + √2)
you can rationalize the denominator by multiplying by its conjugate, (2 - √2), making the denominator into a difference of squares: (2 + √2)(2 - √2) = 4 - 2 = 2.
therefore
a = (16 + 16√2)(2 - √2) / 2
multiply out --> = (32 - 16√2 + 32√2 - 32) / 2
= 16√2 / 2
= 8√2

so hypotenuse = 8√2 x √2 = 8 x 2 = 16
so hypotenuse=16
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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New post 11 Aug 2018, 23:27
gmatpapa wrote:
Please point out what am I doing wrong:

Let the non-hypotenuse sides be xx. Therefore, hypotenuse= 2√x2x

So,Perimeter=2x+2√x=16+162√So,Perimeter=2x+2x=16+162

=>x(2+2√)=8(2+2√)=>x(2+2)=8(2+2)

Comparing both sides, x=8x=8, Therefore, hypotenuse=8√2[/quote]

could someone please tell me why cannot we do in this way?
Thanks.
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Re: The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)  [#permalink]

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New post 29 Dec 2018, 04:07
The right answer 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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New post 20 Mar 2019, 10:28
All the explanations look so scary!! I came up with one with the techniques I learnt.

Working backwards from the answer choices.

We know its a 45 - 45 -90 triangle with the sides in the ratio 1x:1x:sqrt 2x

Length of the hypotenuse is asked, so plug in from Answer choices. Starting with A or B because they look simple.

I am choosing B.

sqrt 2x = 16. so, x=8 sqrt 2.

Also, the perimeter is 16 + 16 sqrt 2

Now, we know that the sides are 8 sqrt 2 , 8 sqrt 2 and 16. All all of them you would get the perimeter given.

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

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New post 09 Apr 2019, 03:28
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] 09 Apr 2019, 03:28

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