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alimad
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alimad
An isoceles triangle has a perimeter of 16 + 16Root(2). What is the length of the hypotenuse.


If this is DS, we will have to find whether A or B will give us sufficient information to find the hypotenuse. However since you did not mention what A and B are, I went ahead and solved for the Hypotenuse. !!!

Let X be the two sides of the isoceles triangle that are equal.
Let H be the hypotenuse of the triangle.

Perimeter = 2X+H=16+16root(2)
Also X^2 + X^2 = H^2 ==> 2X^2 = H^2
so, H = Xroot(2)

Perimeter = 2X + X root(2) = 16+16root(2)
Solving for X you get X = 8root(2)
So H = 16.

Length of Hypotenuse = 16!!
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Manny Calavera
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sonibubu
If it's a DS problem, you need to provide 2 pieces of information.

So far, if we make the two equal sides "x" and the last side "y", then all we have right now is:

2x + y = 16 + 16sq(2)


Following the original data, this is 90:45:45 triangle. :) Therefore, its sides have ratios 1:1:root(2).
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Babin wrote :

Let X be the two sides of the isoceles triangle that are equal.
Let H be the hypotenuse of the triangle.

Perimeter = 2X+H=16+16root(2)
Also X^2 + X^2 = H^2 ==> 2X^2 = H^2
so, H = Xroot(2)

Perimeter = 2X + X root(2) = 16+16root(2)
Solving for X you get X = 8root(2)
So H = 16.

Length of Hypotenuse = 16!!


so xRoot(2) = 16 correct
x = 16/root(2) would be one of the sides.

16/root(2) + 16/root(2) + 16 = Perimeter ?

Not sure where this is heading?
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Manny Calavera
sonibubu
If it's a DS problem, you need to provide 2 pieces of information.

So far, if we make the two equal sides "x" and the last side "y", then all we have right now is:

2x + y = 16 + 16sq(2)

Following the original data, this is 90:45:45 triangle. :) Therefore, its sides have ratios 1:1:root(2).


The original data said it is an isoceles triangle. How did you conclude that it is 90-45-45?
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Babnin
alimad
An isoceles triangle has a perimeter of 16 + 16Root(2). What is the length of the hypotenuse.

If this is DS, we will have to find whether A or B will give us sufficient information to find the hypotenuse. However since you did not mention what A and B are, I went ahead and solved for the Hypotenuse. !!!

Let X be the two sides of the isoceles triangle that are equal.
Let H be the hypotenuse of the triangle.

Perimeter = 2X+H=16+16root(2)
Also X^2 + X^2 = H^2 ==> 2X^2 = H^2
so, H = Xroot(2)

Perimeter = 2X + X root(2) = 16+16root(2)
Solving for X you get X = 8root(2)
So H = 16.

Length of Hypotenuse = 16!!


You are making an assumption that the isoceles triangle is a right triangle. All we know is that 2X + H = 16 + 16root(2). In fact, I don't think the GMAT requires you to know the Pythagorean Theorem. Even if it does, you cannot necessarily apply it here.
We need the two pieces of information here to solve for this DS. Why would a DS problem be provided where you can solve it WITHOUT a single piece of information?



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