GMATPREP question #2

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colors: Newbie | Next Rank: 10 Posts; Posts: 4; Joined: Fri Nov 10, 2006 11:20 am

by colors » Sun Dec 24, 2006 10:04 am

Perimeter of a certain isosceles right triangle is 16+16 sqrt2. what is the length of the hypotenuse of the triange?

A) 8
B) 16
C) 4 sqrt 2
D) 8 sqrt 2
E) 16 sqrt 2

Thanks!!

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Source: Beat The GMAT — Problem Solving | Sun Dec 24, 2006 10:04 am

Chrystelle: Junior | Next Rank: 30 Posts; Posts: 19; Joined: Thu Dec 07, 2006 3:20 am

hypothenuse

by Chrystelle » Sat Dec 30, 2006 7:15 am

Consider two of the sides to be x and the hypothenuse y.

p=16+16sqrt2 = 2x+y
y^2=2x^2, since it is a right iscoscele triangle.
So, x=y/sqrt2
16+16sqrt2=(2y/sqrt2)+y

Solve and u will get y=16

So, it is B.

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optimisticsam: Senior | Next Rank: 100 Posts; Posts: 65; Joined: Wed Aug 22, 2007 8:29 pm; Location: Atlanta, GA

by optimisticsam » Wed Sep 05, 2007 8:28 am

This may be a simple question - but how do you simplify that down to 16?

I am having some trouble with this. Thanks in advance!

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myornob: Junior | Next Rank: 30 Posts; Posts: 13; Joined: Tue Aug 21, 2007 9:00 pm

by myornob » Wed Sep 05, 2007 10:04 am

we know one of the ratios of the right triangle is (1:1: root 2); when angle are 45, 45, 90

Here, two values are given 16, 16 root 2.

So, the third value must be 16

I think this is the easiest method

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optimisticsam: Senior | Next Rank: 100 Posts; Posts: 65; Joined: Wed Aug 22, 2007 8:29 pm; Location: Atlanta, GA

by optimisticsam » Wed Sep 05, 2007 11:38 am

I am not sure that last method works... wouldn't that mean that one side is the same length as c? (a^2+b^2=c^2)

I may be wrong, but I think c must always be the longest?

I know I should know this one.. ack.

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ri2007: Master | Next Rank: 500 Posts; Posts: 258; Joined: Mon Aug 27, 2007 12:43 pm; Thanked: 15 times

by ri2007 » Wed Sep 05, 2007 2:00 pm

The hypotenuse of the triange cannot be 16. If it was the other side would be 16/ sq root of 2 and the perimeter would be 16 + 32/ sq root of 2

can you please confirm that the perimeter is indeed 16+16 sq root of 2 ? and not 16 + 8 sq root of 2?

thanks

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optimisticsam: Senior | Next Rank: 100 Posts; Posts: 65; Joined: Wed Aug 22, 2007 8:29 pm; Location: Atlanta, GA

by optimisticsam » Wed Sep 05, 2007 4:38 pm

I took this test this morning and it is indeed 16 + 16 sqrt. 2

Still pretty stumped by this - any help anyone?

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camitava: Legendary Member; Posts: 645; Joined: Wed Sep 05, 2007 4:37 amFollowed by: 5 members; Location: India; Thanked: 34 times

by camitava » Wed Sep 05, 2007 10:03 pm

I believe the qs itself is having some problem! It cannot be 16! But I am not being able to find out any option specified here!

Correct me If I am wrong

Regards,

Amitava

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Preetics: Junior | Next Rank: 30 Posts; Posts: 15; Joined: Sat Aug 25, 2007 11:02 pm; Location: India

by Preetics » Tue Sep 11, 2007 6:29 am

Hi
Just started solving all these prbs and came to this one.

If we take x =8sqrt2 then dont we get the perimeter as 16+16sqrt2?
Am not too good with this so may be wrong but 8 sqrt2 +8sqrt2 + 8 sqrt2 *rt2= 16sqrt2 + 16(as rt2 *rt2 =2 Thus 8 * 2=16)
Hope i am correct

Thanks
Preeti

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stock2007: Senior | Next Rank: 100 Posts; Posts: 65; Joined: Thu Sep 20, 2007 7:54 am; Thanked: 6 times

re

by stock2007 » Tue Apr 22, 2008 5:43 pm

Ok. think of it this way.

16√ 2 = 8√ 2 + 8√ 2

these two are the two equal sides of the isocelse triangle.

and 16 is the hypotenuse.

you can verify this by using the a^2 + b^2 = c^2 formuale

where (8√ 2)^2 +(8√ 2)^2 = 256
which is also 16^2 ( square of 16 = 256)

think strategically, dont assume that the sides of the triangle have to be 8 and 8. and add up to 16.. thats the joe bloggs generic response.

CRACK THE GMAT NOW

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Tue Apr 22, 2008 6:57 pm

To reiterate what a few have said.

We know that a right isosceles triangle has sides in the ratio of:

x : x : xroot2

So, if we want to try to write the 3 sides in that ratio.

Here, we know that a + a + hypotenuse = 16 + 16root2

and that the hypotenuse is greater than the length of either side.

Using the "numbers have to work out nicely" principle for the GMAT, let's rewrite our equation as:

2a + h = 16root2 + 16

And assume that a = 8root2 and h = 16.

Now to double check:

(8root2)^2 + (8root2)^2 = 16^2

62*2 + 64*2 = 256
128 + 128 = 256

It works, therefore hyp = 16.

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
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moneyman: Master | Next Rank: 500 Posts; Posts: 468; Joined: Sat Mar 03, 2007 10:17 pm; Thanked: 5 times

by moneyman » Wed Apr 23, 2008 1:49 am

Lets name the triangle as ABC.

Its an Isoceles triangle with an hypotenuse so AB^2+Bc^2=AC^2(Pythogoras)

or 2AB^2=AC^2

AC is the hypotenuse so when AC=16 then 2AB^2=256 or AB=8 sq.root 2

Now the perimeter would be 8rt2+8rt2+16=16rt2+16

Maxx

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