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The perimeter of a certain isosceles right triangle 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 Nov 2006, 06:04
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My method for this is simple....

two equal sides
hypotenuse

Perimeter=16+16sqrt2

the sum of the two equal sides cannot be less than the hypotenuse.

Thus hypotenuse must be less than the sum of the other two equal sides:

Perimeter=16+16sqrt2.... the lower value in this equation is the hypotenuse, thus answer is 16

B

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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, 13:38
B it is

Perimeter of right isoceles trainagle =x+x+xsqrt(2)
= 2x+xsqrt(2)
= sqrt(2)x(1+sqrt(2))
Thus
16+16sqrt(2)=sqrt(2)x(1+sqrt(2))
16(1+sqrt(2))=sqrt(2)x(1+sqrt(2))
thus
16 = sqrt(2)x
16 = hypotenuese of isoceles rt triangle

Thus 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 23 Dec 2006, 13:39
mitul 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


Ans: 8sqrt(2)

This is testing knowledge of a special triangle - right, isoseles. that is the angles which are equal are both 45 degrees, and the sides are of the ratio: 1:1:sqrt(2)

Thus, given: x + x + x*sqrt(2) = 16 + 16 sqrt(2)
=> 2x+xsqrt(2) = 16(1+sqrt(2))
=> xsqrt(2) ( sqrt(2) +1) = 16(1+sqrt(2))

Cancelling common term on both sides,

x = 16/sqrt(2)

= 16*sqrt(2)/2 = 8sqrt(2)

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

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New post 23 Dec 2006, 13:44
Since they have asked the length of hypoteneuse, shouldn't we not multiply x by sqrt 2 to get hypotenuese

= 8sqrt2(sqrt 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 Dec 2006, 21:49
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x+x+xsqrt2 = 16+16sqrt2

x= 8sqrt2

hyp=xsqrt2 = 16

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

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New post 28 Dec 2006, 21:20
x + x/root(2) + x/root(2) = 16(1+root(2))

x = 16.

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

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New post 29 Dec 2006, 03:11
mitul 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


x+x+x root 2= 16 + 16 root 2

2x+x root 2= 16 + 16 root 2

root 2. x ( 1 + root 2) = 16 (1 + root 2)

x= 16/root 2

Hypotenus = (16/ root 2) * root 2

= 16

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

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New post 29 Dec 2006, 12:49
mitul 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

-------------------------
a(2+sqrt2) = 16(1+sqrt2) -> sqrt2.a=16 which is what is asked.

TF, B)

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

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New post 31 Dec 2006, 09:11
mitul 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


It´s 16. Just remember that, for a 45-45-90 triangle, the sides are proportional to sqrt(2), sqrt(2), and 2, respectively => basic perimeter is 2*sqrt(2) + 2, which times 8 yields the expression given in the stem. As the integer part of such expression corresponds to the hypotenuse, it must be 16.

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

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New post 08 Feb 2007, 22:35
above720 wrote:
How would you solve?

The perimeter of a certain isosceles right triangle is 16 + 16√2. What is the length of the hypotenuse of the triangle?



If it is isosceles right angle triangle then

x+x+h = 16 + 16√2 and we know x^2 + x^2 = h^2==> 2x^2 = h^2
Therefore, h = √2*x

Replacing the value of h in previous equation we get
2x + √2*x = 16 + 16√2
√2*x(√2 + 1) = 16(1+√2) ==> x=16/√2 ==> h =16
_________________

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

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New post 17 Mar 2007, 23:24
Answer is B

For isosceles triangle two of its sides should have equal length, and since its a right angle triangle, perpendicular and base should of equal lenght.

lets say perpendicula = base = a
and hypotenuse = b

we know that perimeter = 16+16sqrt 2

= 2a + b = 16 + 16sqrt2.... eq 1

also by using pythagoras them we will have b^2 = 2.a^2

or a = bsqrt2 ( since a and b are positive)

putting this in equation one we will have b + bsqrt2 = 16 + 16sqrt2

hence b = 16=hypotenus

regards,

Amardeep

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

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New post 18 Mar 2007, 01:36
Thanks Amardeep.

The crux to the solution is to derive the two solutions

2x + y = 16 + 16sqrt2 --> equation 1

2x^2 = y^2
x = y/(sqrt2) --> equation 2

Where x = base and perpendicular lengths, y = hypotenuse

Put equation 2 into equation 1

2y/sqrt2 + y = 16 + 16sqrt2
--> 2y + ysqrt2 = 16sqrt2 + 16*2 (times sqrt2 to both sides)
--> 2y + ysqrt2 = 2*16 + 16sqrt2 (rearranging)

By visual inspection, we can clearly see that y = 16. Therefore the hypotenuse is 16. And the answer is therefore B.

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

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New post 18 Mar 2007, 05:48
let the side be 'a'

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

so

a(2 + sqrt(2) ) = (2 + sqrt(2))*some number [= 16 +16*(sqrt(2))]

So that some number should be 8*sqrt(2)

so a = 8*sqrt(2)

So hypotenuse = sqrt(2)*a = 8*sqrt(2) * sqrt(2) =16

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

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New post 03 May 2007, 02:40
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We know that an isocele right triangle has the following sides dimension: x, x and xsqrt(2) and we are looking for xsqrt(2).

Lets calculate x:

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

Lets caculte xsqrt(2):

xsqrt(2) = 8 (2+2sqrt(2)) sqrt(2) / (2+sqrt(2))
xsqrt(2) = 8 (2sqrt(2)+4) / (2+sqrt(2)) = 16

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

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New post 03 May 2007, 08:56
above720 wrote:
How would you solve?

The perimeter of a certain isosceles right triangle is 16 + 16√2. What is the length of the hypotenuse of the triangle?


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

h^2 = x^2 + x^2
h^2 = 2x^2
h^2 = 2 (8√2)^2
h^2 = 2x2x8x8
h = 2 x 8 = 16

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

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New post 11 Jun 2007, 14:25
As long as you know that a right isoc. triangle has the ratio of bothe sides to the hypotenuse in the form 1 to 1 to root of 2 - u should be fine.

Now if the sides are x, the hypot. will be xsqrt(2)

so x + x +xsqrt(2) = 16 +16sqrt(2)

solve for x --->>> x=( 16+16root2)/(2+root2)

We need the hypotenuse of the triangle, which is xroot2

We already have x=( 16+16root2)/(2+root2)
multiply both sides by root2 to get xroot2

hypotenuse=16(2+root2)/(2+root2) or 16

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

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New post 11 Jun 2007, 15:07
WIth an Isosceles Right Triangle, you know 2 sides are the same (X) and the hyp is X (sqr root) 2.

Hence: X + X + X(sqr root)2 = 16 + 16(sqr root)2
simplify: 2X = X(sqr root)2 = 16 + 16(sqr root)2

2X is greater than X(sqr root)2

2X = the largest number on the other side = 16(sqr root)2; therefore, X = 8(sqr root)2 and the hyp is 16.

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

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New post 11 Jun 2007, 18:07
let x be hyp of 45-45-90 triangle.
Its parameter will be x+[x/sqrt(2)]+[x/sqrt(2)] = x+x*sqrt(2)
Therefore, x=16

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

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New post 12 Jun 2007, 13:01
I think such type of questions can get easily solved by the back tracing rather than by solving the time consuming equations.

Let the each of the two equal sides of the triangle is x and hypotenuse is y

so 2x^2 = y^2
try each of the y values from answer choices and find x. Check the perimeter. If perimeter matches, that's the answer.

answer is 16.

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

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New post 07 Jul 2007, 16:56
plaguerabbit wrote:
The perimeter of a certain isosceles right triangle is 16 + 16sqrt(2). What is the length of the hypotenuse of the triangle?

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

how do you solve this without backsolving?


this is a problem that repeats every so often in this forum:

since its a isosceles right triangle, then the ratio of its angels is 45:45:90 and the ratio of its sides is 1:1:sqrt(2).

let x be the longer side (i.e sqrt(2)) then the shorter side will be x/sqrt(2).

16+16*sqrt(2) = x/sqrt(2)+x/sqrt(2)+x

16+16*sqrt(2) = 2x/sqrt(2) + x ---> multply by sqrt(2)

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

x = 16

this is the hypotenuse !

16 is the answer (B)

:-D

Last edited by KillerSquirrel on 07 Jul 2007, 21:02, edited 2 times in total.

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Re: The perimeter of a certain isosceles right triangle is 16 +   [#permalink] 07 Jul 2007, 16:56

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