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Iscoceles Triangle ?!

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Iscoceles Triangle ?!

by zico20 » Tue Jun 09, 2009 9:04 am
Hey guys, I wanted some help with this question. I would appreciate your help. Thanks in advance.


The perimeter of a certain isoceles triangle is 16 + 16 (root 2).

What is the length of the hypotenuse of the triangle?

A. 4
B. 16
C. 4 (root2)
D. 8 (root 2)
E. 16 (root 2)

Thanks
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by mike22629 » Tue Jun 09, 2009 9:08 am
Set up equation:

16 + 16*sqrt2 = 2x + x*sqrt2

x = 8*sqrt2

so hyp = 16 (8*sqrt2*sqrt2)

IMO B
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by abhinav85 » Tue Jun 09, 2009 9:15 am
Mike Can you explain in detail the equation you made???
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by zico20 » Tue Jun 09, 2009 9:19 am
Wow mike. you lost me pal !

can you kindly explain in detail .

thanks
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by zico20 » Tue Jun 09, 2009 9:29 am
Got it. Thanks alot!

ok, the isosceles triangle is a 1 1 root 2

therefore

in 16+ 16 root 2

16 represents x + x
and 16 root 2 represents root 2

therefore

16 + 16 root 2 = x + x + x root 2
or 2x + x root 2

so 2 multiplied by what = 16 ? 8

so x is basically 8

so the x root 2 = 8 root 2

so 8 root 2 multiplied by the hypotenuse's ratio root 2 is equal to 8 multiplied by 2!

merci merci
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by abhinav85 » Tue Jun 16, 2009 5:11 am
Hey Zico

can u explain this

"so 2 multiplied by what = 16 ? 8

so x is basically 8

so the x root 2 = 8 root 2

so 8 root 2 multiplied by the hypotenuse's ratio root 2 is equal to 8 multiplied by 2! "
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by niksraza » Tue Jun 16, 2009 5:39 am
Hey guys,

The answer is D. The easiest way to view this question is to remember the golden triangle ratio i.e. "x - x - 2sqrtx" or in other words the perimeter of a right angle triable of 2x + 2sqrtx. Hence using this form you can dedue that x = 8 i.e the two equal sides of an isoseles triangle are equal to 8 which makes the third side (hypotenus) = 8sqrtx.
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by b2eby » Tue Jun 16, 2009 9:25 am
IMO B
The perimeter is equal to 16 + 16(sqrt2) = x + x + x(sqrt2).
The hypotenuse is probably going to be equal to either 16 or 16(sqrt2).

If the hypotenuse is equal to x(sqrt2) = 16(sqrt2) then the perimeter will be:
16(sqrt2) + 16 + 16 = 16(sqrt2) + 32
This is not the correct perimeter.

If 16 is the hypotenuse then 16(sqrt2) must represent the x + x.
8(sqrt2) + 8(sqrt2)

Now the hypotenuse of an isoceles triangle with side x = 8(sqrt2) will be:
x(sqrt2) = 8(sqrt2)(sqrt2) = 8(2) = 16

This works and the perimeter is 16 + 16(sqrt2) with hypotenuse 16.
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by tohellandback » Tue Jun 16, 2009 7:10 pm
I don't know why but I like to solve the questions just by reasoning

now in 90-45-45 triangle the sides are 1,1,sqrt 2
so if sides are two equal sides are 8 each, the hypotenuse will be 8sqrt(2)
perimeter will be 16+8sqrt(2)
now just reject the answer choices, perimeter is 16+16sqrt(2). so Hypotenuse must be greater than 8sqrt(2)
options A,C and D are out
if the hypotenuse is 16sqrt(2), the other sides will be 16 each and perimeter will be 32+16sqrt(2)
so E is out
option B is the answer
The powers of two are bloody impolite!!
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by rah_pandey » Tue Jun 16, 2009 9:41 pm
since the triangle is right angled and isoceles =>

sin45=p/hypotenuse

p=perp
b=base
and p=b

=> hypotenuse=p/sin45
=p*root(2)

=> perimeter=p+p+p*root(2)
16+16*root(2)=p*(2+root(2))

=> p=16*(1+root(2))/root(2)*(1+root(2))=16/root(2)

hypotenuse=p*root(2)
=16
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