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

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Senior Manager
Joined: 27 Aug 2005
Posts: 331
The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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22 Sep 2005, 19:02
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The perimeter of a certain isosceles right triangle is $$16 + 16 \sqrt{2}$$. What is the length of the hypotenuse of the triangle?

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

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

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

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

$$(E) \hspace{5} 16\sqrt{2}$$
[Reveal] Spoiler: OA
Intern
Joined: 22 Sep 2005
Posts: 3
Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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23 Sep 2005, 00:44
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It's B

16+16(rt2)=2a+a(rt2) (where a is the lenght of each leg)

16+16(rt2)=a(2+rt2)

(16+16(rt2))/(2+rt2)=a

Hypoteneuse of isosceles rigth triangle is a(rt2)

H=a(rt2)=((16+16(rt2)(rt2))/(2+rt2)

H=(16(rt2)+16x2))/(2+(rt2))

H=16(rt2+2)/(rt2+2)=16
Manager
Joined: 06 Aug 2005
Posts: 197
Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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23 Sep 2005, 01:03
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Let the length of the opposite and adjacent = a
Length of hypotenuse = h = sqrt(a^2+a^2) = sqrt(2) * a

So a = h/sqrt(2) = (sqrt(2)/2) * h

Perimeter = 2a + h
= 2 * (sqrt(2)/2) * h + h
= h (sqrt (2) + 1 )

Also given
Perimeter = 16 + 16 sqrt (2) = 16 (sqrt (2) + 1 )

So h = 16
SVP
Joined: 05 Apr 2005
Posts: 1705
Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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23 Sep 2005, 06:02
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Macedon wrote:
Hima...could you please put down how you got to that result? shouldn't be the proportion 1:1:Rt2? Thanks

ooooppppppppppsssssssssssssssssss....................

the ratio of the sides of issoceles triagles is: a:a:h=1:1:sqrt(2)

the perimeter of that triangle= a+a+h=2a+a [sqrt(2)]
= 16 + 16(rt2) = 8(rt2)+8(rt2)+16

So h= 16 and a= 8(rt2).

mistakenly, i posted the value of one side of the issoceles sides.
Manager
Joined: 17 Aug 2005
Posts: 165
Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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23 Sep 2005, 06:44
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I also get B

If the length of one of the two equal sides is x, then the other side is also x and the hypotenuse is xrt2.

we know that P= 16+16rt2

to get the perimeter you add the lengths of the three sides so either the equal sides are 16/2 = 8 and the other side is 8rt2 (which it isn't)

or the equal sides are 16rt2/2 = 8rt2 and the hypotenus is 16.

2(8rt2) = 16rt2 so it's pretty obvious that the hypotenuse is 16.
Manager
Joined: 28 Dec 2005
Posts: 117
Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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29 Jan 2006, 11:55
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The perimeter of a certain isosceles right triangle is 16+16 sqrt(2). What is the length of the hypontenuse of the triangle

A. 8
B. 16
C. 4 sqrt (2)
D. 8 sqrt (2)
E. 16 sqrt (2)

Director
Joined: 05 Feb 2006
Posts: 893
Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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10 Apr 2006, 05:41
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I may fall into trap but choose 16..... (1 second thinking approach )
Senior Manager
Joined: 08 Sep 2004
Posts: 257
Location: New York City, USA
Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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10 Apr 2006, 10:09
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The Hypotenous is 16. The side of the triangle is 8*sqrt(2).

- Vipin
Senior Manager
Joined: 24 Jan 2006
Posts: 251
Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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10 Apr 2006, 10:45
2
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y^2 = x^2+x^2 => x = y/sqrt(2)

2x+y=16+16sqrt(2)
ysrrt(2)+y=16+16sqrt(2)
y=16
Director
Joined: 05 Feb 2006
Posts: 893
Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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11 Apr 2006, 00:10
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gmatacer wrote:
y^2 = x^2+x^2 => x = y/sqrt(2)

2x+y=16+16sqrt(2)
ysrrt(2)+y=16+16sqrt(2)
y=16

I looked at it in a more simple manner....

Simply kept in mind that two sides must be equal and that the sum of the two sides must be more than the length of the remaining side....
VP
Joined: 29 Apr 2003
Posts: 1402
Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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11 Apr 2006, 00:19
1
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Put each sides to be X
Then hyp = xsqrt(2)

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

Solving for x, we get x = 8Sqrt(2)

Hence, Hyp = 16
Manager
Joined: 25 May 2006
Posts: 226
Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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03 Oct 2006, 18:00
Using the special triangles.

A isosceles right triangle is a 1:1:SQRT(2)

So by using reverse engineering:

If h=8, then the = sides are 4*SQRT(2) --> and the perimeter is not the given one.
If h=16, then the = sides are 8*SQRT(2) --> and 16+16SQRT(2) is the given perimeter. BINGO.

B it is.
_________________

Who is John Galt?

Senior Manager
Joined: 11 Jul 2006
Posts: 377
Location: TX
Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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03 Oct 2006, 19:38
B

side side hyp
x + x + xsqrt(2) = 16 + 16sqrt(2)

solve for x

hyp = xsqrt(2)
Director
Joined: 06 Sep 2006
Posts: 734
Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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03 Oct 2006, 21:39
the perimeter of a certain isosceles right triangle is 16 +16*sqrt. of 2. What is the length of the hypotenuse of the triangle?

8
16
4*sqrt of 2
8*sqrt of 2
16*sqrt of 2

since h^2 = a^2 + b^2
a = b (isosceles triangle) h^2 = 2a^2 or
h = sqrt(2) * a --A

a + b + h = 16 + 16 * sqrt(2)
2a + h = 16 ( 1 + sqrt(2))
from equation A
2a + sqrt(2)*a = 16(1+sqrt(2))
sqrt(2) * a (1 + sqrt(2)) = 16* (1 + sqrt(2))

a * sqrt(2) = 16
a = 8 * sqrt(2)
D.
Director
Joined: 06 Sep 2006
Posts: 734
Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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04 Oct 2006, 06:15
asaf wrote:
the perimeter of a certain isosceles right triangle is 16 +16*sqrt. of 2. What is the length of the hypotenuse of the triangle?

8
16
4*sqrt of 2
8*sqrt of 2
16*sqrt of 2

since h^2 = a^2 + b^2
a = b (isosceles triangle) h^2 = 2a^2 or
h = sqrt(2) * a --A

a + b + h = 16 + 16 * sqrt(2)
2a + h = 16 ( 1 + sqrt(2))
from equation A
2a + sqrt(2)*a = 16(1+sqrt(2))
sqrt(2) * a (1 + sqrt(2)) = 16* (1 + sqrt(2))

a * sqrt(2) = 16
a = 8 * sqrt(2)
D.

Ok my bad, length of hypotnuse is required
From equation A above
h = sqrt(2) * a = sqrt(2) * 8 * sqrt(2) = 16
I correct myself. B is the answer.
Current Student
Joined: 29 Jan 2005
Posts: 5201
Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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04 Oct 2006, 06:24
Clearly it must be a 45-45-90 triangle. Intuitively, both legs must be shorter than the hypotnuse, so it can't be 16sqrt2. This leaves 16. (D)
Senior Manager
Joined: 23 May 2005
Posts: 263
Location: Sing/ HK
Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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29 Nov 2006, 05:14
The perimeter of a certain isosceles right triangle is 16 +16sqrt2. What is the length of the hypotenuse of the triangle?

8
16
4sqrt2
8sqrt2
16sqrt2

I don't know why I can't seem to arrive at the correct answer for this...

x+x+x(sqrt2) = 16+16sqrt2
x(2+ (sqrt2)) = 16(1 + (sqrt2))

x= [16 (1+(sqrt2))] / (2 +(sqrt2))

... I'm stumped. Don't know what to do after this. Help! I'm so annoyed
_________________

Impossible is nothing

Director
Joined: 05 Feb 2006
Posts: 893
Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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29 Nov 2006, 05:38
2
KUDOS
This is actually more simple than you think....

Remember the rule that the sum of any two sides of the triangular must not be less than the length of the remaining side....

Thus isosceles triangular has two sides equal in length... so the sum of those two sides must be more than the remaining side...

You are given a perimeter, you can trace down the answer quickly.... and without calculations...
Senior Manager
Joined: 08 Jun 2006
Posts: 335
Location: Washington DC
Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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29 Nov 2006, 05:44
Hermione wrote:
The perimeter of a certain isosceles right triangle is 16 +16sqrt2. What is the length of the hypotenuse of the triangle?

8
16
4sqrt2
8sqrt2
16sqrt2

I don't know why I can't seem to arrive at the correct answer for this...

x+x+x(sqrt2) = 16+16sqrt2
x(2+ (sqrt2)) = 16(1 + (sqrt2))

x= [16 (1+(sqrt2))] / (2 +(sqrt2))

... I'm stumped. Don't know what to do after this. Help! I'm so annoyed

x= [16 (1+(sqrt2))] / (2 +(sqrt2))
x = [16 (1+(sqrt2))] / [sqrt2 (1 +(sqrt2))]
x = 16 / sqrt2
x = 8 sqrt2
Senior Manager
Joined: 20 Feb 2006
Posts: 372
Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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29 Nov 2006, 05:46
Perimeter of isoceles = x+x+xsqrt2)

Hypotenuse = x*sqrt2

Therefore:

x+x+x(sqrt2) = 16+16(sqrt2)

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

x(2+sqrt2) = 16(1+sqrt2)

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

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

x(sqrt2) = 16 = Hypotenuse

I didn't get this in the time and am keen to know SimaQs faster method.
Re: The perimeter of a certain isosceles right triangle is 16 +   [#permalink] 29 Nov 2006, 05:46

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