Perimeter of a certain isosceles right triangle is 16+16 sqrt2. what is the length of the hypotenuse of the triange?
A) 8
B) 16
C) 4 sqrt 2
D) 8 sqrt 2
E) 16 sqrt 2
Thanks!!
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Consider two of the sides to be x and the hypothenuse y.
p=16+16sqrt2 = 2x+y
y^2=2x^2, since it is a right iscoscele triangle.
So, x=y/sqrt2
16+16sqrt2=(2y/sqrt2)+y
Solve and u will get y=16
So, it is B.
p=16+16sqrt2 = 2x+y
y^2=2x^2, since it is a right iscoscele triangle.
So, x=y/sqrt2
16+16sqrt2=(2y/sqrt2)+y
Solve and u will get y=16
So, it is B.
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This may be a simple question - but how do you simplify that down to 16?
I am having some trouble with this. Thanks in advance!
I am having some trouble with this. Thanks in advance!
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I am not sure that last method works... wouldn't that mean that one side is the same length as c? (a^2+b^2=c^2)
I may be wrong, but I think c must always be the longest?
I know I should know this one.. ack.
I may be wrong, but I think c must always be the longest?
I know I should know this one.. ack.
The hypotenuse of the triange cannot be 16. If it was the other side would be 16/ sq root of 2 and the perimeter would be 16 + 32/ sq root of 2
can you please confirm that the perimeter is indeed 16+16 sq root of 2 ? and not 16 + 8 sq root of 2?
thanks
can you please confirm that the perimeter is indeed 16+16 sq root of 2 ? and not 16 + 8 sq root of 2?
thanks
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I took this test this morning and it is indeed 16 + 16 sqrt. 2
Still pretty stumped by this - any help anyone?
Still pretty stumped by this - any help anyone?
Hi
Just started solving all these prbs and came to this one.
If we take x =8sqrt2 then dont we get the perimeter as 16+16sqrt2?
Am not too good with this so may be wrong but 8 sqrt2 +8sqrt2 + 8 sqrt2 *rt2= 16sqrt2 + 16(as rt2 *rt2 =2 Thus 8 * 2=16)
Hope i am correct
Thanks
Preeti
Just started solving all these prbs and came to this one.
If we take x =8sqrt2 then dont we get the perimeter as 16+16sqrt2?
Am not too good with this so may be wrong but 8 sqrt2 +8sqrt2 + 8 sqrt2 *rt2= 16sqrt2 + 16(as rt2 *rt2 =2 Thus 8 * 2=16)
Hope i am correct
Thanks
Preeti
Ok. think of it this way.
16√ 2 = 8√ 2 + 8√ 2
these two are the two equal sides of the isocelse triangle.
and 16 is the hypotenuse.
you can verify this by using the a^2 + b^2 = c^2 formuale
where (8√ 2)^2 +(8√ 2)^2 = 256
which is also 16^2 ( square of 16 = 256)
think strategically, dont assume that the sides of the triangle have to be 8 and 8. and add up to 16.. thats the joe bloggs generic response.
CRACK THE GMAT NOW
16√ 2 = 8√ 2 + 8√ 2
these two are the two equal sides of the isocelse triangle.
and 16 is the hypotenuse.
you can verify this by using the a^2 + b^2 = c^2 formuale
where (8√ 2)^2 +(8√ 2)^2 = 256
which is also 16^2 ( square of 16 = 256)
think strategically, dont assume that the sides of the triangle have to be 8 and 8. and add up to 16.. thats the joe bloggs generic response.
CRACK THE GMAT NOW
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To reiterate what a few have said.
We know that a right isosceles triangle has sides in the ratio of:
x : x : xroot2
So, if we want to try to write the 3 sides in that ratio.
Here, we know that a + a + hypotenuse = 16 + 16root2
and that the hypotenuse is greater than the length of either side.
Using the "numbers have to work out nicely" principle for the GMAT, let's rewrite our equation as:
2a + h = 16root2 + 16
And assume that a = 8root2 and h = 16.
Now to double check:
(8root2)^2 + (8root2)^2 = 16^2
62*2 + 64*2 = 256
128 + 128 = 256
It works, therefore hyp = 16.
We know that a right isosceles triangle has sides in the ratio of:
x : x : xroot2
So, if we want to try to write the 3 sides in that ratio.
Here, we know that a + a + hypotenuse = 16 + 16root2
and that the hypotenuse is greater than the length of either side.
Using the "numbers have to work out nicely" principle for the GMAT, let's rewrite our equation as:
2a + h = 16root2 + 16
And assume that a = 8root2 and h = 16.
Now to double check:
(8root2)^2 + (8root2)^2 = 16^2
62*2 + 64*2 = 256
128 + 128 = 256
It works, therefore hyp = 16.
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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Lets name the triangle as ABC.
Its an Isoceles triangle with an hypotenuse so AB^2+Bc^2=AC^2(Pythogoras)
or 2AB^2=AC^2
AC is the hypotenuse so when AC=16 then 2AB^2=256 or AB=8 sq.root 2
Now the perimeter would be 8rt2+8rt2+16=16rt2+16
Its an Isoceles triangle with an hypotenuse so AB^2+Bc^2=AC^2(Pythogoras)
or 2AB^2=AC^2
AC is the hypotenuse so when AC=16 then 2AB^2=256 or AB=8 sq.root 2
Now the perimeter would be 8rt2+8rt2+16=16rt2+16
Maxx