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

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Difficulty:   65% (hard)

Question Stats: 60% (01:59) correct 40% (02:11) wrong based on 198 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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Posts: 57025
Re: The perimeter of a right isosceles triangle is 32 + 32*2^(1/2). What  [#permalink]

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7
5
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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GMAT 1: 690 Q47 V38 Re: The perimeter of a right isosceles triangle is 32 + 32*2^(1/2). What  [#permalink]

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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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1 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 Manager  Joined: 24 Sep 2009
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GMAT 1: 710 Q49 V37 Re: The perimeter of a right isosceles triangle is 32 + 32*2^(1/2). What  [#permalink]

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

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1
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!

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

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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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GMAT 1: 200 Q1 V1 Re: The perimeter of a right isosceles triangle is 32 + 32*2^(1/2). What  [#permalink]

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

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_________________ Re: The perimeter of a right isosceles triangle is 32 + 32*2^(1/2). What   [#permalink] 11 Aug 2019, 20:51
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