Using Perimeter to find Hypotenuse of a triangle : PS Archive
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 20 Feb 2017, 22:37

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Using Perimeter to find Hypotenuse of a triangle

Author Message
Manager
Affiliations: CFA L3 Candidate, Grad w/ Highest Honors
Joined: 03 Nov 2007
Posts: 130
Location: USA
Schools: Chicago Booth R2 (WL), Wharton R2 w/ int, Kellogg R2 w/ int
WE 1: Global Operations (Futures & Portfolio Financing) - Hedge Fund ($10bn+ Multi-Strat) WE 2: Investment Analyst (Credit strategies) - Fund of Hedge Fund ($10bn+ Multi-Strat)
Followers: 1

Kudos [?]: 107 [0], given: 9

Using Perimeter to find Hypotenuse of a triangle [#permalink]

### Show Tags

29 Jul 2009, 19:54
00:00

Difficulty:

(N/A)

Question Stats:

50% (01:03) correct 50% (02:03) wrong based on 1 sessions

### HideShow timer Statistics

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.
Manager
Joined: 22 Mar 2008
Posts: 56
Followers: 1

Kudos [?]: 15 [0], given: 1

Re: Using Perimeter to find Hypotenuse of a triangle [#permalink]

### Show Tags

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.
Manager
Joined: 29 Jul 2009
Posts: 123
Location: France
GPA: 3.95
WE: Information Technology (Computer Software)
Followers: 5

Kudos [?]: 102 [0], given: 15

Re: Using Perimeter to find Hypotenuse of a triangle [#permalink]

### Show Tags

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)
_________________
Manager
Joined: 29 Jul 2009
Posts: 123
Location: France
GPA: 3.95
WE: Information Technology (Computer Software)
Followers: 5

Kudos [?]: 102 [0], given: 15

Re: Using Perimeter to find Hypotenuse of a triangle [#permalink]

### Show Tags

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
_________________
Manager
Joined: 29 Jul 2009
Posts: 120
Followers: 3

Kudos [?]: 90 [0], given: 23

Re: Using Perimeter to find Hypotenuse of a triangle [#permalink]

### Show Tags

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
Manager
Affiliations: CFA L3 Candidate, Grad w/ Highest Honors
Joined: 03 Nov 2007
Posts: 130
Location: USA
Schools: Chicago Booth R2 (WL), Wharton R2 w/ int, Kellogg R2 w/ int
WE 1: Global Operations (Futures & Portfolio Financing) - Hedge Fund ($10bn+ Multi-Strat) WE 2: Investment Analyst (Credit strategies) - Fund of Hedge Fund ($10bn+ Multi-Strat)
Followers: 1

Kudos [?]: 107 [0], given: 9

Re: Using Perimeter to find Hypotenuse of a triangle [#permalink]

### Show Tags

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)

SVP
Joined: 30 Apr 2008
Posts: 1887
Location: Oklahoma City
Schools: Hard Knocks
Followers: 40

Kudos [?]: 578 [1] , given: 32

Re: Using Perimeter to find Hypotenuse of a triangle [#permalink]

### Show Tags

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$$ = 2 * 8 = 16 _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

GMAT Club Premium Membership - big benefits and savings

Manager
Joined: 29 Jul 2009
Posts: 123
Location: France
GPA: 3.95
WE: Information Technology (Computer Software)
Followers: 5

Kudos [?]: 102 [0], given: 15

Re: Using Perimeter to find Hypotenuse of a triangle [#permalink]

### Show Tags

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 + $$16\sqrt{2}$$
_________________
SVP
Joined: 30 Apr 2008
Posts: 1887
Location: Oklahoma City
Schools: Hard Knocks
Followers: 40

Kudos [?]: 578 [1] , given: 32

Re: Using Perimeter to find Hypotenuse of a triangle [#permalink]

### Show Tags

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.
_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.

GMAT Club Premium Membership - big benefits and savings

Senior Manager
Status: mba here i come!
Joined: 07 Aug 2011
Posts: 270
Followers: 43

Kudos [?]: 1067 [0], given: 48

Re: Using Perimeter to find Hypotenuse of a triangle [#permalink]

### Show Tags

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
_________________

press +1 Kudos to appreciate posts

Re: Using Perimeter to find Hypotenuse of a triangle   [#permalink] 21 Oct 2011, 06:23
Display posts from previous: Sort by