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The perimeter of a right isosceles triangle is 32 + 32*2^(1/2). What

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The perimeter of a right isosceles triangle is 32 + 32*2^(1/2). What [#permalink]

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New post 04 Feb 2012, 15:50
4
6
00:00
A
B
C
D
E

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Question Stats:

62% (01:12) correct 38% (01:27) wrong based on 273 sessions

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The perimeter of a right isosceles triangle is \(32 + 32\sqrt{2}\). What is the length of the hypotenuse of the triangle?


A. 16

B. \(8 \sqrt{3}\)

C. 32

D. \(8\sqrt{2}\)

E. \(16\sqrt{2}\)


How come answer is E ?

This is how I am trying to solve

Isoscles Right angle triangle mean has x:x:x\(\sqrt{2}\) as its sides. Therefore, x+x+x\(\sqrt{2}\) = 32 + 32\(\sqrt{2}\)
2x+x\(\sqrt{2}\) = 32 + 32\(\sqrt{2}\). I am stuck after this.

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Re: The perimeter of a right isosceles triangle is 32 + 32*2^(1/2). What [#permalink]

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New post 04 Feb 2012, 16:12
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enigma123 wrote:
The perimeter of a right isoscles triangle is 32 + 32 \(\sqrt{2}\). What is the length of the hypotenues of the triangle?

A) 16
B) 8 \(\sqrt{3}\)
C) 32
D) 8 \(\sqrt{2}\)
E) 16\(\sqrt{2}\)

How come answer is E ?

This is how I am trying to solve

Isoscles Right angle triangle mean has x:x:x\(\sqrt{2}\) as its sides. Therefore, x+x+x\(\sqrt{2}\) = 32 + 32\(\sqrt{2}\)
2x+x\(\sqrt{2}\) = 32 + 32\(\sqrt{2}\). I am stuck after this.


In a right triangle where the angles are 45°, 45°, and 90° the sides are always in the ratio \(1 : 1 : \sqrt{2}\). With the \(\sqrt{2}\) being the hypotenuse (longest side). This can be derived from Pythagoras' Theorem. Because the base angles are the same (both 45°) the two legs are equal and so the triangle is also isosceles.

Now, if we proceed the way you suggest:

Let the side of this triangle be \(x\) then, \(P=32+32\sqrt{2}=x+x+\sqrt{2}x\) --> \(32+32\sqrt{2}=x(2+\sqrt{2})\) -->\(x=\frac{32+32\sqrt{2}}{2+\sqrt{2}}\) --> muliptly both nominatro and denominator by \(2-\sqrt{2}\) (to rationalize the denominator): \(x=\frac{(32+32\sqrt{2})(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})}=\frac{64+64\sqrt{2}-32\sqrt{2}-64}{4-2}=\frac{32\sqrt{2}}{2}=16\sqrt{2}\).

So, hypotenuse is \(16\sqrt{2}*\sqrt{2}=32\)

Answer: C.

P.S. Of course you can also solve this problem with approximation method or plug-in method using the answer choices.
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Re: The perimeter of a right isosceles triangle is 32 + 32*2^(1/2). What [#permalink]

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New post 13 Mar 2012, 11:45
2a+c=16+16sqrt2
if c=16, then 16sqrt2=2a or a=8sqrt2

lets check it out-
c^2=(8sqrt2)^2+(8sqrt2)^2=64*4
c=8*2=16

ans is B
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Re: The perimeter of a right isosceles triangle is 32 + 32*2^(1/2). What [#permalink]

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New post 13 Mar 2012, 13:32
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Given isosceles right triangle
So two angles = 45
Let the side opposite to Angle 45 is x
Then hypotenuse2 = x^2 + x^2
Hypotenuse = √2 x


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Re: The perimeter of a right isosceles triangle is 32 + 32*2^(1/2). What [#permalink]

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New post 13 Mar 2012, 15:49
Idiot me found the length of one of the two other sides and chose that as the answer.

SMDH.
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Re: The perimeter of a right isosceles triangle is 32 + 32*2^(1/2). What [#permalink]

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New post 01 May 2013, 05:56
bangalorian2000 wrote:
zz0vlb wrote:
The perimeter of a certain isoceles right triangle is 16 + 16\sqrt{2}. what is the hypotenuse of the triangle?

A. 8
B. 16
C. 4\sqrt{2}
D. 8\sqrt{2}
E. 16\sqrt{2}

Source: GMAT Prep

let one side be x
hypotenuse is x\sqrt{2}
perimeter = 2x + x\sqrt{2} = 16+16\sqrt{2}
solving x = 8\sqrt{2}
hence hypotenuse is 16
hence B.


How is the value of x calculated here ?
perimeter = 2x + x\sqrt{2} = 16+16\sqrt{2}
solving x = 8\sqrt{2}

Thanks
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Re: The perimeter of a right isosceles triangle is 32 + 32*2^(1/2). What [#permalink]

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New post 18 May 2013, 07:21
The perimeter of a certain isosceles right triangle is 16 + 16 * sqrt(2). What is the length of the hypotenuse of the triangle?

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

Please attack my approach the question.

P = 16 + 16 * sqrt(2)

The "isosceles right triangle" must have the angles 90:45:45, and bisecting through the middle of the base at 90 degrees must give two smaller "isosceles right triangles" with sides equal to half the original hypotenuse.

Therefore, the hypotenuse is 2 * P / 4. This gives something approximating 19. E is also approximately 19. Why is the answer exactly 16?

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Re: The perimeter of a right isosceles triangle is 32 + 32*2^(1/2). What [#permalink]

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New post 18 May 2013, 07:43
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stormbind wrote:
The perimeter of a certain isosceles right triangle is 16 + 16 * sqrt(2). What is the length of the hypotenuse of the triangle?

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

Please attack my approach the question.

P = 16 + 16 * sqrt(2)

The "isosceles right triangle" must have the angles 90:45:45, and bisecting through the middle of the base at 90 degrees must give two smaller "isosceles right triangles" with sides equal to half the original hypotenuse.

Therefore, the hypotenuse is 2 * P / 4. This gives something approximating 19. E is also approximately 19. Why is the answer exactly 16?

Thanks,

The answer is [B].

For an "isosceles right triangle", the sides can be {r, r and \(sqrt(2)*r\)}.
Hence, perimeter = \(2r + sqrt(2)*r = 16 + 16 * sqrt(2)\)
and r comes out to be \(16(1+sqrt(2))/(2+sqrt(2))\).

Now the hypotenuse is \(sqrt(2)*r\) from the above. Hence multiplying sqrt(2) we have the value = 16. Answer!

Hope the answer is accurate!

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Re: The perimeter of a right isosceles triangle is 32 + 32*2^(1/2). What [#permalink]

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New post 28 Nov 2015, 10:38
this is a tricky question, yet, we have to remember that right isosceles triangle has the ratio of the sides x:x:x(sqrt2)

clearly, something doesn't add up. if 32(sqrt2) is the hypotenuse, then the other sides should be 32 each, and the total would be 64+32sqrt2
ok, but what if the hypotenuse is 32?
x(sqrt2) = 32
x = 32/sqrt2 or 32*sqrt2/2 = 16(sqrt2)
ok, now that's something better.
we have 2 x's and one hypotenuse.
2x = 32(sqrt2) and hypotenuse 32.
the answer is C.
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Re: The perimeter of a right isosceles triangle is 32 + 32*2^(1/2). What [#permalink]

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New post 01 Apr 2018, 19:09
Bunuel wrote:
enigma123 wrote:
The perimeter of a right isoscles triangle is 32 + 32 \(\sqrt{2}\). What is the length of the hypotenues of the triangle?

A) 16
B) 8 \(\sqrt{3}\)
C) 32
D) 8 \(\sqrt{2}\)
E) 16\(\sqrt{2}\)

How come answer is E ?

This is how I am trying to solve

Isoscles Right angle triangle mean has x:x:x\(\sqrt{2}\) as its sides. Therefore, x+x+x\(\sqrt{2}\) = 32 + 32\(\sqrt{2}\)
2x+x\(\sqrt{2}\) = 32 + 32\(\sqrt{2}\). I am stuck after this.


In a right triangle where the angles are 45°, 45°, and 90° the sides are always in the ratio \(1 : 1 : \sqrt{2}\). With the \(\sqrt{2}\) being the hypotenuse (longest side). This can be derived from Pythagoras' Theorem. Because the base angles are the same (both 45°) the two legs are equal and so the triangle is also isosceles.

Now, if we proceed the way you suggest:

Let the side of this triangle be \(x\) then, \(P=32+32\sqrt{2}=x+x+\sqrt{2}x\) --> \(32+32\sqrt{2}=x(2+\sqrt{2})\) -->\(x=\frac{32+32\sqrt{2}}{2+\sqrt{2}}\) --> muliptly both nominatro and denominator by \(2-\sqrt{2}\) (to rationalize the denominator): \(x=\frac{(32+32\sqrt{2})(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})}=\frac{64+64\sqrt{2}-32\sqrt{2}-64}{4-2}=\frac{32\sqrt{2}}{2}=16\sqrt{2}\).

So, hypotenuse is \(16\sqrt{2}*\sqrt{2}=32\)

Answer: C.

P.S. Of course you can also solve this problem with approximation method or plug-in method using the answer choices.


Good tip. I solved it the long way, but now that i think about it, we can arrive at a much quicker solution by estimating.

Perimeter is \(32+32\sqrt{2}\)
\(\sqrt{2}\) is approx 3/2
so perimeter is approx: 32 + 32(3/2) = 80

If we take the second largest choice, option E, the hypothenuse is approx 16(3/2)=24. The other 2 legs must be less than the hypothenuse, so the max possible perimeter is less than 24+24+24=72. This leaves option C as the only sensible choice.

Or another way...
The sides of the triangle are in the ratio \(x:x:x\sqrt{2}\)
so perimeter, in terms of x, is approx \(x+x+\frac{3}{2}x=\frac{7}{2}x\)
given perimeter is approx 80,
\(\frac{7}{2}x=80\)
since we need to find the value of (3/2)x
\(\frac{7}{2}x*\frac{2}{7}*\frac{3}{2}=80*\frac{2}{7}*\frac{3}{2}\) = approx 30
Re: The perimeter of a right isosceles triangle is 32 + 32*2^(1/2). What   [#permalink] 01 Apr 2018, 19:09
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