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The perimeter of a certain isosceles right triangle is 16 +

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

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New post 22 Sep 2005, 19:02
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A
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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
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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

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

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New post 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.
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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New post 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.
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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New post 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)

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

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New post 10 Apr 2006, 05:41
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I may fall into trap but choose 16..... (1 second thinking approach :) )
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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

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

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New post 10 Apr 2006, 10:45
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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
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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New post 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....
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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New post 11 Apr 2006, 00:19
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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
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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New post 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.
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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New post 03 Oct 2006, 19:38
B

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

solve for x

hyp = xsqrt(2)
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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New post 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)
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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New post 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.
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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New post 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)
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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New post 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 :x
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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New post 29 Nov 2006, 05:38
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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...
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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New post 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


x= [16 (1+(sqrt2))] / (2 +(sqrt2))
x = [16 (1+(sqrt2))] / [sqrt2 (1 +(sqrt2))]
x = 16 / sqrt2
x = 8 sqrt2
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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New post 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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