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Hypoteneuse of the triangle

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Hypoteneuse of the triangle

by Baldini » Fri Mar 06, 2009 3:21 am
The perimeter of a certain isosceles right triangle is 16 + 16 sq root 2. What is the length of the hypoteneuse?
a. 8
b. 16
c. 4 sq rt 2
d. 8 sq rt 2
e. 16 sq rt 2

OA is b.
Could someone please explain this answer?
thanks
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by pakaskwa » Fri Mar 06, 2009 3:59 am
For any isosceles right triangle, the ratio of its 3 sides is 1:1:sqrt2
(sqrt = square root)

Let x is one side, the perimeter will be:
x+x+(sqrt2)x = 16+sqrt16
--> (2+sqrt2)x = 16+sqrt16
--> x = (16+sqrt16)/(2+sqrt2)
--> x = [(16+sqrt16)(2-sqrt2)]/[(2+sqrt2)(2-sqrt2)]
note: numerator and denominator each multiplies by (2-sqrt2), to get rid of sqrt
--> x = 8sqrt2

Hypotenuse = sqrt2 x 8sqrt2 = 16. (remember the ratio of all sides is 1:1:sqrt2)
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by Baldini » Fri Mar 06, 2009 5:34 am
thanks a lot - just wondering if a quicker way (shortcut) exists to find this answer?
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Re: Hypoteneuse of the triangle

by sanju09 » Fri Mar 06, 2009 5:58 am
Baldini wrote:The perimeter of a certain isosceles right triangle is 16 + 16 sq root 2. What is the length of the hypoteneuse?
a. 8
b. 16
c. 4 sq rt 2
d. 8 sq rt 2
e. 16 sq rt 2

OA is b.
Could someone please explain this answer?
thanks
The perimeter of an isosceles right triangle, with its hypotenues as x, is always in the form x + x sqrt 2. So if perimeter is 16 + 16 sqrt 2, hypotenues is 16.
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Another similar way

by Ossa » Fri Mar 06, 2009 6:04 am
Hi,

Here is another way that is very similar.
If x is the hypotenuse, then the 2 sides will be equal to x/SQRT 2
then, x/(SQRT 2)+x/(SQRT 2)+x=16(1+SQRT 2)
then, x(2/(SQRT 2) +1)=16(1+SQRT 2)
and accordingly, x((SQRT 2 +1)=16 (1+SQRT 2), and so x=16
Baldini wrote:thanks a lot - just wondering if a quicker way (shortcut) exists to find this answer?
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by arora007 » Sun Aug 01, 2010 11:26 pm
45-90-45

has sides x, xroot2, x

2x+xroot2 = 16+16root2

taking common terms out from both sides....

xroot2( root2 + 1) = 16( 1 + root2)

now hypotenuse is xroot2 = 16
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by RCV » Tue Sep 07, 2010 11:10 pm
sanju09 wrote:
Baldini wrote:The perimeter of a certain isosceles right triangle is 16 + 16 sq root 2. What is the length of the hypoteneuse?
a. 8
b. 16
c. 4 sq rt 2
d. 8 sq rt 2
e. 16 sq rt 2

OA is b.
Could someone please explain this answer?
thanks
The perimeter of an isosceles right triangle, with its hypotenues as x, is always in the form x + x sqrt 2. So if perimeter is 16 + 16 sqrt 2, hypotenues is 16.
very very quick, thanks a lot
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by GMATGuruNY » Wed Sep 08, 2010 4:17 am
Baldini wrote:The perimeter of a certain isosceles right triangle is 16 + 16 sq root 2. What is the length of the hypoteneuse?
a. 8
b. 16
c. 4 sq rt 2
d. 8 sq rt 2
e. 16 sq rt 2

OA is b.
Could someone please explain this answer?
thanks
The sides of an isosceles right triangle are proportioned: s - s - sRoot2.
So if s=side and h=hypotenuse, h = sRoot2 and s = h/Root2.

Let's plug in the answers -- which represent the hypotenuse -- and determine the length of each leg and the perimeter of the whole triangle. We need the perimeter to be 16 + 16Root2.

We can see quickly that C, D and E won't work:

C:
h = 4Root2
s = 4
p = 4+4+4Root2 = 8 + 4Root2. No.

D:
h = 8Root2
s = 8
p = 8+8+8Root2 = 16 + 8Root2. No.

E:
h = 16Root2
s = 16
p = 16+16+16Root2 = 32 + 16Root2. No.

Let's try B:
h = 16
s = 16/Root2 = [(16*Root2)/(Root2*Root2)]= (16Root2)/2 = 8Root2
p = 8Root2 + 8Root2 + 16 = 16 + 16Root2. Success!

The correct answer is B.
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