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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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07 Jul 2007, 17:19
Thanks KillerSquirrel, when I was trying to solve this, i used the formula i assigned x*sqrt(2) to the hypotenuse, and assigned x to each of the legs. So I had:

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

I had a lot of trouble solving this equation for x and getting it into the form that the answer choices had...

I guess it's better here to use x as the hypotenus and x/sqrt(2) as the legs...

thanks again

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

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07 Jul 2007, 18:17
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?

B for me.

Perimeter of issoceles triangle = 16 + 16 sqrt (2)
x + x + x sqrt (2) = 16 + 16 sqrt (2)
2x + x sqrt (2) = 16 + 16 sqrt (2)

by solving for x, x = 8 sqrt (2)

so the hypoteneous of the issoceles triangle = 16.

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

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08 Jul 2007, 07:52
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ggarr wrote:
Himalayan wrote:
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?

B for me.

Perimeter of issoceles triangle = 16 + 16 sqrt (2)
x + x + x sqrt (2) = 16 + 16 sqrt (2)
2x + x sqrt (2) = 16 + 16 sqrt (2)

by solving for x, x = 8 sqrt (2)

so the hypoteneous of the issoceles triangle = 16.
will you show how you solved for x?

i dunno how i had trouble solving this...heh, i just figured it out.

you end up with:

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

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

x = [16+16sqrt(2)] / [(2+sqrt(2)]

multiply numerator and denominator by 2-sqrt(2) and distribute, and you end up with:

x = [32-16sqrt(2)+32sqrt(2)-32] / (4-2)

x = 16sqrt(2) / 2

x = 8sqrt(2)

therefore x*sqrt(2) is the hypotenuse, which is 8sqrt(2) * sqrt(2) = 16.

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

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08 Jul 2007, 09:05
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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?

Guys, there is a simpler approach. On this board, with all the practice that everyone's doing, we are all so focused on the various nuances of the GMAT, so this should jump out at you.

We know that the triangle has to be x to x to xroot2. But when we try to make it work, it simply doesn't make sense. I mean, if the sides were an integer and the hypotenuse were the same integer times root 2, then the perimeter would have to just be 2x + xroot2. But it's not that. It's x + xroot2. So something's wrong.

You should instantly think - maybe the hypotenuse is the integer. It's the only other way the GMAT has ever really made these things hard.

So now we can eliminate C, D, and E. And since we're left with just 8 or 16, in this case, plugging in isn't so tough, and we get to 16 in about 31 seconds.

This is essentially what Squirrel was saying. Realize it's backwards ahead of time. But then just get the right answer. Remember, the GMAT doesn't award points for slickness of the math, it awards points for right answers in the shortest amount of time.

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

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04 Jun 2009, 21:01
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As mentioned in the subject, this is a 45:45:90 isosceles triangle. As per properties of 45:45:90 triangle, the sides are in the ratio of 1:1: √2 (check out OG, or you can try this with couple of numbers)

So, let the sides be x, x, and x√2 (where x is the length of legs and x√2 is hypotenuse). We are supposed to find length of hypotenuse, i.e. x√2

Perimeter =
2x + x√2 = 16 +16√2
or, x√2(√2 + 1) = 16(1 + √2)
or, x√2 = 16

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

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05 Jun 2009, 01:28
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Perimeter =
2x + x√2 = 16 +16√2
or, x√2(√2 + 1) = 16(1 + √2)
or, x√2 = 16

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

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14 Mar 2010, 09:21
nifoui wrote:
The perimeter of a certain isosceles right triangle is 16 + 16√2. What is the length of the hypotenuse of the triangle?
A: 8
B: 16
C: 4√2
D: 8√2
E: 16√2

(OA =
[Reveal] Spoiler:
B
)

Any idea how to solve this??

Lets say hypotenuse is h and the other sides are a
so h = a*sqrt(2)=>a = h/sqrt(2) by pythagoras theoram
now perimeter is h+a+a = 16+16 sqrt(2)
=>h+2*h/sqrt(2) = 16+16 sqrt(2)
=>h(1+sqrt(2)) = 16(1+sqrt(2))
=>h = 16
so hypotenuse is 16 hence B is the answer
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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14 Mar 2010, 10:27
thanks for the answer, however there's a part that I don't get in your explanation:

how do you go from this

h+2*h/sqrt(2)

to this

h(1+sqrt(2))

?

I understand that you factorize with h, but if we take the result h(1+sqrt(2)) and do it the other way around we would get the following: h+h(sqrt2). isn't it?

And also I was wondering, after we established this:
now perimeter is h+a+a = 16+16 sqrt(2)
=>h+2*h/sqrt(2) = 16+16 sqrt(2)
couldn't we just deduce from here that h = 16?

Thanks again for your help, I didn't translate the problem into an equation like that, which definitely seems like a good approach!
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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14 Mar 2010, 10:46
nifoui wrote:
how do you go from this

h+2*h/sqrt(2)

to this

h(1+sqrt(2))

?

h+2*h/sqrt(2) = h+sqrt(2)*sqrt(2)*h/sqrt(2) = h+h*sqrt(2) = h(1+sqrt(2))
I just divided 2 by sqrt(2) to equate it to the right side
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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14 Mar 2010, 21:05
nifoui wrote:
The perimeter of a certain isosceles right triangle is 16 + 16√2. What is the length of the hypotenuse of the triangle?
A: 8
B: 16
C: 4√2
D: 8√2
E: 16√2
(OA =
[Reveal] Spoiler:
B
)
Any idea how to solve this??

In a isoceles triangle two sides must be same. And its a right triangle as question is asking for hypotenuse so it's a 45:45:90 triangle.
Sides are in ratio s:s:s√2.
Perimeter = 2s + s√2 = 16 + 16√2
solving s√2(1+√2)=16(1+√2)
s=8√2
hypotenuse=16 hence B.

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

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30 Apr 2010, 13:55
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Himalayan wrote:
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

much easier to solve, at least saves few seconds:

if x=8√2, => 2x=2*8√2=16√2, Perimeter-2x= hypotenuse, or 16 + 16√2-16√2=16
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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22 Nov 2010, 18:14
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Hey guys,

Nice solution - just one thing I like to point out on these:

Think like the testmaker!

If I were writing the test and knew that everyone studies the 45-45-90 ratio as: 1, 1, sqrt 2

I would make a living off of making the shorter sides a multiple of sqrt 2 so that the long side is an integer. People aren't looking for that! And they often won't trust themselves enough to calculate correctly...they'll look at the answer choices and see that 3 of them are Integer*sqrt 2, and they'll think they screwed up somehow because the right answer "should" have a sqrt 2 on the end.

So...keep in mind that with the Triangle Ratios:

45-45-90
x: x: x*sqrt 2

30-60-90
x : x*sqrt3 : 2x

One of the easiest tricks up the GMAT author's sleeve is to make x equal to a multiple of the radical so that the radical appears on the side you're not expecting and the integer shows up where you think it shouldn't!

Also, as you go through questions like these, ask yourslf "how could they make that question a little harder" or "how could they test this concept in a way that I wouldn't be looking for it". Training yourself to look out for unique cases, from the testmaker's perspective, helps you to get a real mastery of the GMAT from a high level.
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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24 Apr 2011, 20:07
Hi,

I follow the algebraic solution, but also couldn't do it in less than 2 mins. Somehow can't quite figure out the quick solution

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

I don't follow the above statement. I understand the hypotenuse cannot be larger than the sum of the two sides, but from the perimeter value of 16 +16sqrt2 how can you know that one of those values is the sum of the two sides and one is the hypotenuse?

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

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24 Apr 2011, 23:36
chloeholding wrote:
Hi,

I follow the algebraic solution, but also couldn't do it in less than 2 mins. Somehow can't quite figure out the quick solution

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

I don't follow the above statement. I understand the hypotenuse cannot be larger than the sum of the two sides, but from the perimeter value of 16 +16sqrt2 how can you know that one of those values is the sum of the two sides and one is the hypotenuse?

The perimeter of a certain isosceles right triangle is 16 +16sqrt2. What is the length of the hypotenuse of the triangle?

16 +16sqrt2

8-- this will make the sides as (8+16sqrt2)/2>24/2=12. Not possible. Hypotenuse is always bigger than sides.
16-- this will make the sides as 16sqrt(2)/2=8sqrt(2). Perfect: 8sqrt(2)*sqrt(2)=16 by 45-90-45 right angle triangle rule.
4sqrt2-- 4sqrt2<8(1st option)-- Not possible
8sqrt2- sides = (16+8sqrt(2))/2=13approx; Hypotenuse<side. Not possible.
16sqrt2- hypotenuse>sum of the sides(16)-- Not Possible.
**********************************************************
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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27 Apr 2011, 10:07
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fluke wrote:
chloeholding wrote:
Hi,

I follow the algebraic solution, but also couldn't do it in less than 2 mins. Somehow can't quite figure out the quick solution

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

I don't follow the above statement. I understand the hypotenuse cannot be larger than the sum of the two sides, but from the perimeter value of 16 +16sqrt2 how can you know that one of those values is the sum of the two sides and one is the hypotenuse?

The perimeter of a certain isosceles right triangle is 16 +16sqrt2. What is the length of the hypotenuse of the triangle?

16 +16sqrt2

8-- this will make the sides as (8+16sqrt2)/2>24/2=12. Not possible. Hypotenuse is always bigger than sides.
16-- this will make the sides as 16sqrt(2)/2=8sqrt(2). Perfect: 8sqrt(2)*sqrt(2)=16 by 45-90-45 right angle triangle rule.
4sqrt2-- 4sqrt2<8(1st option)-- Not possible
8sqrt2- sides = (16+8sqrt(2))/2=13approx; Hypotenuse<side. Not possible.
16sqrt2- hypotenuse>sum of the sides(16)-- Not Possible.
**********************************************************

Dear Fluke
for me the easiest approach was that
for a isosceles triangle if the hypotenuse is H then other sides will be$$H/sqrt2$$

A H=8 so S1=S2=8/sqrt2 Perimeter will be 8+16/sqrt2
B H=16 so S1=S2=16/sqrt2 Perimeter will be 16+132/sqrt2=16+16sqrt2 (Answer)
C H=4sqrt2 so S1=S2=4 Perimeter will be 8+4sqrt2
D H=8sqrt2 so S1=S2=8 Perimeter will be 16+8sqrt2
E H=16sqrt2 so S1=S2=16 Perimeter will be 32+16sqrt2

like wise for the others as well

hope it was quite easy
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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28 Apr 2011, 03:43
Let Hypotenuse = x

Sides = x/root(2)

So 2*x/root(2) + x = 16 +16root(2)

x + xroot(2) = 16 + 16root(2)

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

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11 May 2011, 05:19
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Please point out what am I doing wrong:

Let the non-hypotenuse sides be $$x$$. Therefore, hypotenuse= $$\sqrt{2}x$$

$$So,Perimeter= 2x+\sqrt{2}x= 16+16\sqrt{2}$$

$$=> x(2+\sqrt{2})= 8(2+\sqrt{2})$$

Comparing both sides, $$x=8$$, Therefore, $$hypotenuse= 8\sqrt{2}$$

This answer is also included in the answer choices which makes me sure that I'm doing a very silly mistake somewhere, just cant see where! Please guys, please point out that small mistake that is eluding my eyes!
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Re: The perimeter of a certain isosceles right triangle is 16 + [#permalink]

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11 May 2011, 16:43
perimeter of an isosceles right angled triangle = 2x + x *sqrt(2) = 16+16*sqrt(2)

=> x(2+sqrt(2)) = 16(1+sqrt(2))
=> x = 8*sqrt(2)

hypotenuse of an isosceles right angled triangle with x as the identical side = x * sqrt(2)
= 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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02 Aug 2011, 09:12
This problem can be solved using 45:45:90 isosceles triangle rule. As per properties of 45:45:90 triangle, the sides are in the ratio of 1:1: √2

Let the sides be x, x, and x√2 (where x is the length of legs and x√2 is hypotenuse). We have to find out length of hypotenuse, i.e. x√2

Perimeter =
2x + x√2 = 16 +16√2
or, x√2(√2 + 1) = 16(1 + √2)
or, x√2 = 16

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

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03 Aug 2011, 12:12
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

you almost solved it..
x(2+ (sqrt2)) = 16(1 + (sqrt2)) take one more sqrt2 common
x* sqrt2 (sqrt2 + 1) = 16 (1 + sqrt2) => x* sqrt2 = 16 or x=> 8*sqrt2 .. hypotenuse = x*sqrt2 = 16 .. answer B

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Re: The perimeter of a certain isosceles right triangle is 16 +   [#permalink] 03 Aug 2011, 12:12

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