Author
Message
TAGS:

Hide Tags
Director

Joined: 05 Feb 2006

Posts: 893

Kudos [? ]:
131 [3 ], given: 0

+1 Kudos

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

Show Tags
29 Nov 2006, 06:04
3

This post received KUDOS

1

This post was BOOKMARKED

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

Kudos [? ]:
131 [3 ], given: 0

+1 Kudos

Senior Manager

Joined: 01 Oct 2006

Posts: 493

Kudos [? ]:
44 [0 ], given: 0

+1 Kudos

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

Show Tags
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

Kudos [? ]:
44 [0 ], given: 0

+1 Kudos

Director

Joined: 28 Dec 2005

Posts: 915

Kudos [? ]:
59 [0 ], given: 0

+1 Kudos

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

Show Tags
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)

Kudos [? ]:
59 [0 ], given: 0

+1 Kudos

Manager

Joined: 24 Jun 2006

Posts: 60

Kudos [? ]:
4 [0 ], given: 0

+1 Kudos

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

Show Tags
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

Kudos [? ]:
4 [0 ], given: 0

+1 Kudos

Senior Manager

Joined: 05 Oct 2006

Posts: 266

Kudos [? ]:
22 [0 ], given: 0

+1 Kudos

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

Show Tags
23 Dec 2006, 21:49

1

This post was BOOKMARKED

x+x+xsqrt2 = 16+16sqrt2
x= 8sqrt2
hyp=xsqrt2 = 16

Kudos [? ]:
22 [0 ], given: 0

+1 Kudos

VP

Joined: 25 Jun 2006

Posts: 1161

Kudos [? ]:
204 [0 ], given: 0

+1 Kudos

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

Show Tags
28 Dec 2006, 21:20

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

Kudos [? ]:
204 [0 ], given: 0

+1 Kudos

Manager

Joined: 19 Aug 2006

Posts: 213

Kudos [? ]:
76 [0 ], given: 0

+1 Kudos

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

Show Tags
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

Kudos [? ]:
76 [0 ], given: 0

+1 Kudos

Manager

Joined: 18 Nov 2006

Posts: 123

Kudos [? ]:
2 [0 ], given: 0

+1 Kudos

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

Show Tags
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)
Kudos [? ]:
2 [0 ], given: 0

+1 Kudos

Senior Manager

Joined: 24 Nov 2006

Posts: 349

Kudos [? ]:
28 [0 ], given: 0

+1 Kudos

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

Show Tags
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

ItÂ´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.

Kudos [? ]:
28 [0 ], given: 0

+1 Kudos

Manager

Joined: 10 Dec 2005

Posts: 112

Kudos [? ]:
5 [0 ], given: 0

+1 Kudos

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

Show Tags
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

_________________

"Live as if you were to die tomorrow. Learn as if you were to live forever." - Mahatma Gandhi

Kudos [? ]:
5 [0 ], given: 0

+1 Kudos

Director

Joined: 13 Dec 2006

Posts: 506

Kudos [? ]:
253 [0 ], given: 0

+1 Kudos

Location: Indonesia

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

Show Tags
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

Kudos [? ]:
253 [0 ], given: 0

+1 Kudos

Intern

Joined: 03 Mar 2007

Posts: 7

Kudos [? ]:
1 [0 ], given: 0

+1 Kudos

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

Show Tags
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.

Kudos [? ]:
1 [0 ], given: 0

+1 Kudos

VP

Joined: 28 Mar 2006

Posts: 1367

Kudos [? ]:
38 [0 ], given: 0

+1 Kudos

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

Show Tags
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

Kudos [? ]:
38 [0 ], given: 0

+1 Kudos

Manager

Joined: 04 Oct 2006

Posts: 76

Kudos [? ]:
38 [1 ], given: 0

+1 Kudos

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

Show Tags
03 May 2007, 02:40
1

This post received KUDOS

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

Kudos [? ]:
38 [1 ], given: 0

+1 Kudos

Director

Joined: 26 Feb 2006

Posts: 900

Kudos [? ]:
165 [0 ], given: 0

+1 Kudos

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

Show Tags
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

Kudos [? ]:
165 [0 ], given: 0

+1 Kudos

Manager

Joined: 22 May 2007

Posts: 121

Kudos [? ]:
6 [0 ], given: 0

+1 Kudos

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

Show Tags
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

Kudos [? ]:
6 [0 ], given: 0

+1 Kudos

Senior Manager

Joined: 29 Nov 2006

Posts: 318

Kudos [? ]:
162 [0 ], given: 0

+1 Kudos

Location: Orange County, CA

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

Show Tags
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.

Kudos [? ]:
162 [0 ], given: 0

+1 Kudos

VP

Joined: 10 Jun 2007

Posts: 1431

Kudos [? ]:
388 [0 ], given: 0

+1 Kudos

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

Show Tags
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

Kudos [? ]:
388 [0 ], given: 0

+1 Kudos

Director

Joined: 14 Jan 2007

Posts: 775

Kudos [? ]:
183 [0 ], given: 0

+1 Kudos

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

Show Tags
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.

Kudos [? ]:
183 [0 ], given: 0

+1 Kudos

VP

Joined: 08 Jun 2005

Posts: 1141

Kudos [? ]:
257 [0 ], given: 0

+1 Kudos

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

Show Tags
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)

Last edited by

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

Kudos [? ]:
257 [0 ], given: 0

+1 Kudos

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

Go to page
Previous
1 2 3 4
Next
[ 75 posts ]