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The perimeter of a certain isosceles right triangle is 16 +1

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by sudhir3127 » Thu Jul 24, 2008 8:12 am
please search before you post the question... the answer is 16.
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by AleksandrM » Thu Jul 24, 2008 9:48 am
This triangle will have two legs of equal lenght. You can look at 16sqroot2 and assume that this must be the hypotenus, but in that case the other two angles must sum to 32. So, you then realize that 16sqroot2 is actually 8sqroot2 + 8sqroot2, and the hypotenus will be 16.

You can also see it if you set up:

(8sqroot2)^2 + (8sqroot2)^2 = h^2

256 = h^2 [take the sqroot of both sides]

h = 16
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by malolakrupa » Fri Jul 25, 2008 7:56 am
I just had a quick question if the perimeter of isosceles triangle is 16 + 16sqrt(2) , then this can be equated to 2a + a*sqrt(2) right ? So this tells us that 2a = 16 hence a = 8 and hence the length of the hypotenuse is 8 sqrt(2) right , but this invalidates the perimeter of isosceles triangle which is
16 + 16sqrt(2) right?
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by Mclaughlin » Fri Jul 25, 2008 10:23 am
AleksandrM wrote:This triangle will have two legs of equal lenght. You can look at 16sqroot2 and assume that this must be the hypotenus, but in that case the other two angles must sum to 32. So, you then realize that 16sqroot2 is actually 8sqroot2 + 8sqroot2, and the hypotenus will be 16.

You can also see it if you set up:

(8sqroot2)^2 + (8sqroot2)^2 = h^2

256 = h^2 [take the sqroot of both sides]

h = 16
I dont understand how by looking at 16 sqroot 2 you know that the remaining angles woudl be 32. please explain how you got that?
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by AleksandrM » Fri Jul 25, 2008 2:31 pm
Since this triangle is isosceles it has 2 legs of equal length, with the third leg being the hypotenus. Hypotenuse triangles have the following ratio of sides: x : x : xsqroot2, the last one being the hypotenuse.

Now, look at the sum 16 + 16sqroot2.

Based on the ratio I provided you with above, you could assume that 16sqroot2 is the hypotenus. However, that means that the other two sides would have to be 16, each, and add up to 32. The other two sides add up to 16, though. Your only other choice is that 16 is actually the hypotenuse, and the 16sqroot2 is the sum of the two equal sides. This means that the sides are 8sqroot2, each. That is all you need to do. HOWEVER, you can test whether this is true using pythagorean theorem:

(8sqroot2)^2 + (8sqroot2)^2 = h^2

256 = h^2 [take the sqroot of both sides]

h = 16, as I have already shown in my previous post.

Hope this helps!
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by jimmytwoshoes » Mon Aug 11, 2008 8:45 am
I though Pythagorean theorem only applied to triangles with a right angle? That isn't specified in the question stem.
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by sudhir3127 » Mon Aug 11, 2008 9:15 am
if u see the question...its a "isoscles right angle triangle"

hope it helps..
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by jimmytwoshoes » Mon Aug 11, 2008 9:27 am
Doh! I am taking my GMAT in 3 hours. Hope this isn't a sign of things to come!! :wink:
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by sudhir3127 » Mon Aug 11, 2008 9:29 am
All the very best jimmy.. stay cool.. i am sure u will do well..
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by aj5105 » Thu Oct 02, 2008 4:37 am
2x + sqrt2 x = 16 + 16 sqrt2

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

ans --16
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by II » Sun Nov 23, 2008 9:59 am
aj5105 wrote:2x + sqrt2 x = 16 + 16 sqrt2

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

ans --16
Can you explain further ?
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Re: The perimeter of a certain isosceles right triangle is 1

by lunarpower » Mon Nov 24, 2008 5:01 am
airan wrote:The perimeter of a certain isosceles right triangle is 16 +16 sqrt(2). What is the length of the hypotenuse of the triangle

1. 8
2. 16
3. 4 sqrt(2)
4. 8 sqrt(2)
5. 16 sqrt(2)
two ways for you to solve this problem much more easily than with the rather insane algebra:

(1) plug in the answer choices
according to the 45-45-90 template, the hypotenuse of an isosceles right triangle is √2 times each of the legs. therefore, each leg is the hypotenuse divided by √2.
if you plug in choice (b), hypotenuse = 16, you'll find that leg = 16/√2 = 8√2 (post back if you don't know how to simplify this). therefore, perimeter = 2(leg) + hypotenuse = 16√2 + 16.
this is WAY easier than going through with all the algebra.
also, you shouldn't have to do that much plugging. if you plug in (a) first, then you'll get exactly half the desired perimeter, proving that (b) is correct. and if you plug in (c), (d), or (e), then you'll get, respectively, 8 + 4√2, 16 + 8√2, or 32 + 16√2, any of which will allow you to deduce that all three are incorrect.

(2) memorize TWO templates for the 45-45-90 triangle
everyone knows the 1-1-√2 template, which has perimeter 2 + 1√2.
you can also memorize an additional template: √2-√2-2.
this template is just √2 times the original template, but the multiples aren't obvious. this template has perimeter 2 + 2√2.
if you memorize this template, then you'll immediately recognize 16 + 16√2 as exactly eight times its perimeter, so that your triangle is 8√2-8√2-16.
Ron has been teaching various standardized tests for 20 years.

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Re: The perimeter of a certain isosceles right triangle is 1

by Natalllllia » Tue Dec 02, 2008 5:53 pm
if you plug in choice (b), hypotenuse = 16, you'll find that leg = 16/√2 = 8√2 (post back if you don't know how to simplify this).
:roll:
Explain please
Thank you
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Re: The perimeter of a certain isosceles right triangle is 1

by dmateer25 » Tue Dec 02, 2008 6:18 pm
Natalllllia wrote:
if you plug in choice (b), hypotenuse = 16, you'll find that leg = 16/√2 = 8√2 (post back if you don't know how to simplify this).
:roll:
Explain please
Thank you
Given the hypotenuse of an isosceles right triangle, you can find the the leg with the following formula:

Let L = leg
Let H = hypotenuse

L= (1/2)H√2

So when 16 is the hypotenuse it would be:

L=(1/2)16√2
L=8√2
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