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# Using Perimeter to find Hypotenuse of a triangle

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Using Perimeter to find Hypotenuse of a triangle [#permalink]  29 Jul 2009, 19:54
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(N/A)

Question Stats:

50% (01:03) correct 50% (02:03) wrong based on 1 sessions
If the perimeter of a isosceles right triangle is 16 + 16*sqrt(2), what is the length of its hypotenuse?

a) 8
b) 16
c) 4*sqrt(2)
d) 8*sqrt(2)
e) 16*sqrt(2)

Last edited by robertrdzak on 30 Jul 2009, 19:46, edited 1 time in total.
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Re: Using Perimeter to find Hypotenuse of a triangle [#permalink]  29 Jul 2009, 20:42
let each equal side = d
therefore hypoteneous = sqrt(2) * d
perimeter = 2d + sqrt(2) * d = 16 + [sqrt(2) * 16]
from this d = 16
and sqrt(2) d = 16 * sqrt(2)

hence e.
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Re: Using Perimeter to find Hypotenuse of a triangle [#permalink]  29 Jul 2009, 22:59
robertrdzak wrote:
If the perimeter of a isosceles right triangle is 16 + 16(2)^2, what is the length of its hypotenuse?

a) 8
b) 16
c) 4(2)^2
d) 8(2)^2
e) 16(2)^2

Is this question == > 16 + 16*sqrt(2)
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Re: Using Perimeter to find Hypotenuse of a triangle [#permalink]  29 Jul 2009, 23:02
Spoilt wrote:
robertrdzak wrote:
If the perimeter of a isosceles right triangle is 16 + 16(2)^2, what is the length of its hypotenuse?

a) 8
b) 16
c) 4(2)^2
d) 8(2)^2
e) 16(2)^2

Is this question == > 16 + $$16\sqrt{2}$$

I have faced the same question == during preparation

The ration of sides of the a isosceles right triangle is 1: 1 : $$\sqrt{2}$$

So it must me $$16/\sqrt{2}$$ : $$16/\sqrt{2}$$ : 16

So OA : B
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Re: Using Perimeter to find Hypotenuse of a triangle [#permalink]  30 Jul 2009, 07:44
robertrdzak wrote:
If the perimeter of a isosceles right triangle is 16 + 16(2)^2, what is the length of its hypotenuse?

a) 8
b) 16
c) 4(2)^2
d) 8(2)^2
e) 16(2)^2

LET a be the side of the the triangle

16 + 16 $$\sqrt{2}$$= 2a + a $$\sqrt{2}$$
16 + 16 $$\sqrt{2}$$ = a(2 + $$\sqrt{2}$$)
so
a = (16 + 16 $$\sqrt{2}$$) / (2 + $$\sqrt{2}$$)

After rationalizing by (2-$$\sqrt{2}$$)

we get a = 8$$\sqrt{2}$$

FROM THIS WE CAN FIND HYPOTENUSE = 8$$\sqrt{2}$$ * $$\sqrt{2}$$ = 16

HTH
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Re: Using Perimeter to find Hypotenuse of a triangle [#permalink]  30 Jul 2009, 19:47
Sorry everyone, I messed up the Q, i put (2)^2 instead of *sqrt(2)....I reposted below and fixed the Q above.

If the perimeter of a isosceles right triangle is 16 + 16*sqrt(2), what is the length of its hypotenuse?

a) 8
b) 16
c) 4*sqrt(2)
d) 8*sqrt(2)
e) 16*sqrt(2)

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Re: Using Perimeter to find Hypotenuse of a triangle [#permalink]  30 Jul 2009, 20:23
1
KUDOS
I don't disagree with any of the posts that got 16 as the answer, but some of them were not the easiest to follow. Here is how I did it.

formula for a right isosceles triangle is $$1:1:\sqrt{2}$$ so the perimeter for the most basic (easiest to figure out) triangle of this type would be $$1+1+\sqrt{2}$$. If we let d = one side (NOT THE HYPOTNUSE!!) we get the formula d + d + d$$\sqrt{2}$$ With the other equation, we don't put the 1 in front of the square root of 2, but it's "implied" so the d is required in the second forumla/equation.

We know the perimeter is 16 + 16$$\sqrt{2}$$, so

d + d +d$$\sqrt{2}$$ = 16 + 16$$\sqrt{2}$$

2d +d$$\sqrt{2}$$ = 16 + 16$$\sqrt{2}$$
FACTOR OUT d

d(2+ $$\sqrt{2}$$) = 16 + 16$$\sqrt{2}$$

divide both sides by (2 + $$\sqrt{2}$$) and we get

d = $$\frac{16 + 16\sqrt{2}}{2 + \sqrt{2}}$$

Now, if we have a radical in the bottom, how do we get rid of it? We cant just muliply by $$\sqrt{2}$$ because when we do 2 * $$\sqrt{2}$$ we would still be creating a radical in the denominator.

Remember this rule: (x + y)(x - y) = $$x^2 - y^2$$? This is how we get rid of complex radicals (not sure if that is the proper word for it, but i think they're complex because they're a pain in the a$$to get rid of. so $$(2 + \sqrt{2})(2 - \sqrt{2}) = 2^2 - \sqrt{2}^2$$ which = 4 - 2 = 2 Now multiple the numerator by the (2 - $$\sqrt{2}$$ also, because it has to be even or it changes the problem/answer. $$16 + 16\sqrt{2}) ( 2 - \sqrt{2})$$ $$32 - 16\sqrt{2} + 32\sqrt{2} - 16\sqrt{2}^2$$ becomes $$32 - 16\sqrt{2} + 32\sqrt{2} - 32$$ becomes $$-16\sqrt{2} + 32\sqrt{2}$$ becomes $$16\sqrt{2}$$ Don't forget this is all over 2 in the denominator. 16 / 2 = 8 so we now have d = $$8\sqrt{2}$$ Remember that we have the ratio $$1:1:\sqrt{2}$$ and d = the side, not the hypotnuse, so we must take the value of d and multiply it by $$\sqrt{2}$$ to get the final value for the hypotnuse. $$\sqrt{2}$$ * $$\sqrt{2} * 8$$ = 2 * 8 = 16 _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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Re: Using Perimeter to find Hypotenuse of a triangle [#permalink]  31 Jul 2009, 04:03
Spoilt wrote:
Spoilt wrote:
robertrdzak wrote:
If the perimeter of a isosceles right triangle is 16 + 16(2)^2, what is the length of its hypotenuse?

a) 8
b) 16
c) 4(2)^2
d) 8(2)^2
e) 16(2)^2

Is this question == > 16 + $$16\sqrt{2}$$

I have faced the same question == during preparation

The ration of sides of the a isosceles right triangle is 1: 1 : $$\sqrt{2}$$

So it must me $$16/\sqrt{2}$$ : $$16/\sqrt{2}$$ : 16

So OA : B

I am sorry to say : But I still dont understand, what is wrong with this method ?

1 : 1 : $$\sqrt{2}$$

i.e. $$1/\sqrt{2}$$ : $$1/\sqrt{2}$$ : 1 (hypo)

Perimeter ==> $$16/\sqrt{2}$$ + $$16/\sqrt{2}$$ + 16 (hypo)

Then perimeter is 2 * $$16/\sqrt{2}$$(sides) + 16 (Hypo)
= 16 + $$16\sqrt{2}$$
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Re: Using Perimeter to find Hypotenuse of a triangle [#permalink]  31 Jul 2009, 06:23
1
KUDOS
Spoilt,

There is not necessarily anything wrong with it, but the start of your method actually skips some steps that most people won't realize are necessary.

You see that if you have 8$$\sqrt{2}$$ + 8$$\sqrt{2}$$ as the base and height of the triangle, then that would give you 16 for your hypotnuse and a perimeter of 16 + 16$$\sqrt{2}$$. Most people are not going to realize that. We are going to see 16$$\sqrt{2}$$ and think...ok, the ratio is 1:1:$$\sqrt{2}$$so the one with the $$\sqrt{2}$$ must be part of the hypotnuse. You were able to look at the problem and in seconds see the correct answer. That's impressive, but not necessarily a "method" for the rest of us to use.
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**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.

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Re: Using Perimeter to find Hypotenuse of a triangle [#permalink]  21 Oct 2011, 06:23
hypotenuse if different than the legs in isosceles. H could be either $$16$$ or $$16\sqrt{2}$$

if H = $$16\sqrt{2}$$, then each side must be 16 and P must be 32 + $$16\sqrt{2}$$ ---> wrong
if H = 16, then each side must be $$8\sqrt{2}$$ and P must be 16 + $$16\sqrt{2}$$ ---> looks good

let's confirm this by pythagoras.
$$2x^2 = H^2$$ ---> in an isosceles
$$2*64*2 = 16^2$$ ----> proved 4*64 is basically 16*16
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Re: Using Perimeter to find Hypotenuse of a triangle   [#permalink] 21 Oct 2011, 06:23
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