Isosceles Right Triangle

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heshamelaziry: Legendary Member; Posts: 869; Joined: Wed Aug 26, 2009 3:49 pmFollowed by: 3 members; Location: California; Thanked: 13 times

by heshamelaziry » Mon Oct 12, 2009 10:49 pm

The perimeter of a certain Isosceles right triangle is 16 + 16âˆš2. What is the length of the hypotenuse of the triangle ?

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xcusemeplz2009: Master | Next Rank: 500 Posts; Posts: 399; Joined: Wed Apr 15, 2009 3:48 am; Location: india; Thanked: 39 times

by xcusemeplz2009 » Tue Oct 13, 2009 12:22 am

h is 16

h^2=2*a^2=>h=asqrt(2)

2a+h=16+16sqrt(2)

2a+sqrt(2)=16+16sqrt(2)

solving a=8*sqrt(2)
h=8*sqrt2*sqrt2=16

It does not matter how many times you get knocked down , but how many times you get up

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NikolayZ: Master | Next Rank: 500 Posts; Posts: 124; Joined: Thu Jun 18, 2009 5:33 am; Thanked: 35 times

by NikolayZ » Tue Oct 13, 2009 12:23 am

Hey there !

Since you have an isosceles triangle, its perimeter = 2x+y. ( 2 sides are x, and the hypotenuse is y)
so the perimeter is
16+16sqrt(2)=2x+y.
Also the hypotenuse equals xsqrt(2), since it is 45 degree isosceles right triangle.

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

x= (16+16*sqrt(2)/(2+sqrt(2))
Then , hypotenuse will be x*sqrt(2)= (16*sqrt(2)+32)/(sqrt(2)+2)=16.

Last edited by NikolayZ on Tue Oct 13, 2009 8:52 am, edited 1 time in total.

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heshamelaziry: Legendary Member; Posts: 869; Joined: Wed Aug 26, 2009 3:49 pmFollowed by: 3 members; Location: California; Thanked: 13 times

by heshamelaziry » Tue Oct 13, 2009 8:45 am

NikolayZ wrote:Hey there !

Since you have an isosceles triangle, its perimeter = 2x+y. ( 2 sides are x, and the hypotenuse is y)
so the perimeter is
16+16sqrt(2)=2x+y.
Also the hypotenuse equals xsqrt(2), since it is 45 degree isosceles right triangle.

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

x= (16+16*sqrt(2)/(2+2*sqrt(2))
Then , hypotenuse will be x*sqrt(2)= (16*sqrt(2)+32)/(sqrt(2)+2)=16.

Last edited by heshamelaziry on Tue Oct 13, 2009 9:05 am, edited 2 times in total.

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NikolayZ: Master | Next Rank: 500 Posts; Posts: 124; Joined: Thu Jun 18, 2009 5:33 am; Thanked: 35 times

by NikolayZ » Tue Oct 13, 2009 8:51 am

Sorry mate. "typo happens" (c) =) corrected.

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heshamelaziry: Legendary Member; Posts: 869; Joined: Wed Aug 26, 2009 3:49 pmFollowed by: 3 members; Location: California; Thanked: 13 times

by heshamelaziry » Tue Oct 13, 2009 8:56 am

NikolayZ wrote:Sorry mate. "typo happens" (c) =) corrected.

I got it, finally. Thanks a lot.

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NikolayZ: Master | Next Rank: 500 Posts; Posts: 124; Joined: Thu Jun 18, 2009 5:33 am; Thanked: 35 times

by NikolayZ » Tue Oct 13, 2009 9:13 am

heshamelaziry, truly there is a flaw in my solvent, because it is not consistent =(

so, right answer to my mind is

2x+x*sqrt(2)=16+16*sqrt(2)
x(2+sqrt(2)=16+16*sqrt(2)
x=(16+16*sqrt(2))/2+sqrt(2)
x*sqrt(2)=(16*sqrt(2)+16*sqrt(2)*sqrt(2))/2+sqrt(2)
==> since hte x*sqrt(2) is the hypotenuse (h)
then
h=(16*sqrt(2)+32)/2+sqrt(2)
factorizing 16 from the numerator we'll get
16(sqrt(2)+2)/(sqrt(2)+2)=16.

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heshamelaziry: Legendary Member; Posts: 869; Joined: Wed Aug 26, 2009 3:49 pmFollowed by: 3 members; Location: California; Thanked: 13 times

by heshamelaziry » Tue Oct 13, 2009 9:25 am

NikolayZ wrote:heshamelaziry, truly there is a flaw in my solvent, because it is not consistent =(

so, right answer to my mind is

2x+x*sqrt(2)=16+16*sqrt(2)
x(2+sqrt(2)=16+16*sqrt(2)
x=(16+16*sqrt(2))/2+sqrt(2)
x*sqrt(2)=(16*sqrt(2)+16*sqrt(2)*sqrt(2))/2+sqrt(2)
==> since hte x*sqrt(2) is the hypotenuse (h)
then
h=(16*sqrt(2)+32)/2+sqrt(2)
factorizing 16 from the numerator we'll get
16(sqrt(2)+2)/(sqrt(2)+2)=16.

Thanks Dude

I posted some problems form a GMATPrep test that I took. Maybe you like to take a shot at them. I posted them yesterday in an attachement.

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kkpatel1: Newbie | Next Rank: 10 Posts; Posts: 7; Joined: Tue Oct 20, 2009 5:59 pm; Location: North Carolina

by kkpatel1 » Fri Nov 20, 2009 2:16 pm

heshamelaziry wrote:
NikolayZ wrote:heshamelaziry, truly there is a flaw in my solvent, because it is not consistent =(

so, right answer to my mind is

2x+x*sqrt(2)=16+16*sqrt(2)
x(2+sqrt(2)=16+16*sqrt(2)
x=(16+16*sqrt(2))/2+sqrt(2)
x*sqrt(2)=(16*sqrt(2)+16*sqrt(2)*sqrt(2))/2+sqrt(2)
==> since hte x*sqrt(2) is the hypotenuse (h)
then
h=(16*sqrt(2)+32)/2+sqrt(2)
factorizing 16 from the numerator we'll get
16(sqrt(2)+2)/(sqrt(2)+2)=16.

Where I am confused is:
When you set:
x*sqrt(2)=[(16*sqrt(2)+16*sqrt(2)*sqrt(2)]/2+sqrt(2)

why dont you multiply the bottom portion by sqrt(2) also, since the rule is, if you multiply one thing by something, you have to do it to everything!??

therefore shouldnt it be: h=(16*sqrt(2)+32)/2+2 ??

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