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# The perimeter of an isosceles right triangle is 16 + 16 sq

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Intern
Joined: 25 Jun 2007
Posts: 23
The perimeter of an isosceles right triangle is 16 + 16 sq [#permalink]

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04 Sep 2007, 21:32
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The perimeter of an isosceles right triangle is 16 + 16 sq rt 2. What is the length of the hypotenuse of the triangle?

a) 8
b) 16
c) 4 sq rt 2
d) 8 sq rt 2
e) 16 sq rt 2

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-perimeter-of-a-right-isoscles-triangle-is-127049.html
[Reveal] Spoiler: OA
Director
Joined: 03 May 2007
Posts: 872
Schools: University of Chicago, Wharton School
Re: The perimeter of an isosceles right triangle is 16 + 16 sq [#permalink]

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04 Sep 2007, 22:24
mexicanhoney wrote:
The perimeter of an isosceles right triangle is 16 + 16 sq rt 2. What is the length of the hypotenuse of the triangle?

a) 8
b) 16
c) 4 sq rt 2
d) 8 sq rt 2
e) 16 sq rt 2

Can someone please explain the answer using the 1:1: sq rt 2 ratio? Thanks.

x + x + x (sqrt2) = 16 + 16 sqrt2
2x + x (sqrt 2) = 16 + 2 (8 sqrt2)
2x + x (sqrt 2) = (8 sqrt2) (sqrt 2) + 2 (8 sqrt2)

now we can say it is 16.
Senior Manager
Joined: 19 Feb 2007
Posts: 325
Re: The perimeter of an isosceles right triangle is 16 + 16 sq [#permalink]

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04 Sep 2007, 22:59
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In an isosceles right triangle the ratio of the sides is

1:1:sqrt(2)

Explanation

Let the side be x,

So the other side is also x since it is an isosceles triangle

Using Pythagoras theorem

x^2 +x^2 = H^2

=> H^2 = 2(x^2)

=> H = x*sqrt(2)

so ratio is x:x:x*sqrt(2), or 1:1:sqrt^2
Manager
Joined: 10 Aug 2007
Posts: 63
Re: The perimeter of an isosceles right triangle is 16 + 16 sq [#permalink]

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05 Sep 2007, 05:28
Let's call x the equal sides of the triangle and y the hypotenuse.

From the parameter we have: x+x+y=16sqrt(2) hence 2x+y=16sqrt(2) (1)

From definition of Pythagoras, x^2 +x^2 = y^2 hence y=xsqrt(2) (2)

Substitute (2) in (1):

2x+sqrt(2)x = 16+16sqrt(2) or sqrt(2)x[1+sqrt(2)] = 16[1+sqrt(2)] therefore

sqrt(2)x=16 or x=8sqrt(2)
VP
Joined: 10 Jun 2007
Posts: 1438
Re: The perimeter of an isosceles right triangle is 16 + 16 sq [#permalink]

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05 Sep 2007, 07:01
mexicanhoney wrote:
The perimeter of an isosceles right triangle is 16 + 16 sq rt 2. What is the length of the hypotenuse of the triangle?

a) 8
b) 16
c) 4 sq rt 2
d) 8 sq rt 2
e) 16 sq rt 2

Can someone please explain the answer using the 1:1: sq rt 2 ratio? Thanks.

Set hyp of the triangle to be x
Using 45-45-90 relationship, other sides will be x/sqrt(2)
so x+2*x/sqrt(2) = 16+16sqrt(2)
x+x*sqrt(2) = 16+16sqrt(2)
x= 16
Intern
Joined: 25 Aug 2007
Posts: 14
Re: The perimeter of an isosceles right triangle is 16 + 16 sq [#permalink]

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07 Sep 2007, 09:30
It should be D.

2a + a*sqrt(2) = 16+16*sqrt(2)

sqrt(2)*a{sqrt(2) + 1} =16{1+sqrt(2)}

Therefore, sqrt(2) * a = 16

Implying a = 8*sqrt(2)
Senior Manager
Joined: 04 Jun 2007
Posts: 366
Re: The perimeter of an isosceles right triangle is 16 + 16 sq [#permalink]

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10 Sep 2007, 21:11
side= X, hypotenuse= XSQRT2

16+16SQRT2= X( 2+2SQRT2)
X= 16+16SQRT2/2+SQRT2
SIMPLIFY TO,
X= (32-16SQRT2+32SQRT2-32)/ = 8SQRT2

Hence, hypotenuse= XSQRT2= 8SQRT2*SQRT2= 8*2= 16
Director
Joined: 17 Sep 2005
Posts: 903
Re: The perimeter of an isosceles right triangle is 16 + 16 sq [#permalink]

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11 Sep 2007, 06:39
mexicanhoney wrote:
The perimeter of an isosceles right triangle is 16 + 16 sq rt 2. What is the length of the hypotenuse of the triangle?

a) 8
b) 16
c) 4 sq rt 2
d) 8 sq rt 2
e) 16 sq rt 2

Can someone please explain the answer using the 1:1: sq rt 2 ratio? Thanks.

In this case, three sides of the triangle will be in ratio 1:1:sqrt(2)
Now let's multiply these ratio with x and add to get the perimeter of the triangle:

x+x+xsqrt(2) = 16 + 16sqrt(2)
=> 2x + x*sqrt(2) = 16 + 16*sqrt(2)

One way is to solve the equation and get the value of x.

x = 16(1+sqrt(2))/(2+sqrt(2))
= 8sqrt(2)

[Note : We can solve above equation by rationalizing the denominator]

Other way is to make Left Hand Side and Right Hand Side equations in equivalent form as shown below:

2x + x * sqrt(2) =

2x + x * sqrt (2) = 2 (8 sqrt(2)) + (8 sqrt(2)) * sqrt(2)

Clearly x = 8sqrt(2) [By comparing LHS and RHS equations]

Now Hypotenuse = xsqrt(2)
= 8sqrt(2) * sqrt(2) [substituting the value of x]
= 8 * 2
= 16

- Brajesh
Senior Manager
Joined: 27 Aug 2007
Posts: 253
Re: The perimeter of an isosceles right triangle is 16 + 16 sq [#permalink]

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11 Sep 2007, 08:31
Agree with 16

Ans: B
Manager
Status: Post MBA, working in the area of Development Finance
Joined: 09 Oct 2006
Posts: 169
Location: Africa
Re: The perimeter of an isosceles right triangle is 16 + 16 sq [#permalink]

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11 Sep 2007, 08:48
B. Caught with the side/hypotenuse!
mexicanhoney wrote:
The perimeter of an isosceles right triangle is 16 + 16 sq rt 2. What is the length of the hypotenuse of the triangle?

a) 8
b) 16
c) 4 sq rt 2
d) 8 sq rt 2
e) 16 sq rt 2

Can someone please explain the answer using the 1:1: sq rt 2 ratio? Thanks.
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1857
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: The perimeter of an isosceles right triangle is 16 + 16 sq [#permalink]

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08 Aug 2014, 00:59
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Refer diagram below

Given that perimeter $$= 16+16\sqrt{2}$$

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

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

$$x = \frac{16+16\sqrt{2}}{2 + \sqrt{2}} = \frac{16(1+\sqrt{2})}{\sqrt{2}(1+\sqrt{2})}$$

$$x = \frac{16}{\sqrt{2}}$$

$$\sqrt{2}x = 16 = Hypotenuse$$

Bunuel: Kindly update OA for this question. Thanks
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Math Expert
Joined: 02 Sep 2009
Posts: 39607
Re: The perimeter of an isosceles right triangle is 16 + 16 sq [#permalink]

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12 Aug 2014, 01:28
OPEN DISCUSSION OF THIS QUESTION IS HERE: the-perimeter-of-a-right-isoscles-triangle-is-127049.html
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Re: The perimeter of an isosceles right triangle is 16 + 16 sq   [#permalink] 12 Aug 2014, 01:28
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