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Re: 5 or z [#permalink]
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Can you please post an explanation? Thank you
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Re: 5 or z [#permalink]
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Answer=A
I used a^2+b^2=C^2=100
if we use z=5 z^2=25 the rest is 75 there is nothing square smth gives u 75, therefore, it could be less than Z or greater than Z not equal to 5. i tried Z=4 but it is not working then i tried z=6 and it works thereafter, it should be greater than Z which is Z=6 Z^2=36
b=8 , b^2=64 , Z^2+b^2= 36+64=100
Answer A
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Re: 5 or z [#permalink]
settahri wrote:
Answer=A
I used a^2+b^2=C^2=100
if we use z=5 z^2=25 the rest is 75 there is nothing square smth gives u 75, therefore, it could be less than Z or greater than Z not equal to 5. i tried Z=4 but it is not working then i tried z=6 and it works thereafter, it should be greater than Z which is Z=6 Z^2=36
b=8 , b^2=64 , Z^2+b^2= 36+64=100
Answer A
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Re: 5 or z [#permalink]
settahri wrote:
Answer=A
I used a^2+b^2=C^2=100
if we use z=5 z^2=25 the rest is 75 there is nothing square smth gives u 75, therefore, it could be less than Z or greater than Z not equal to 5. I tried Z=4 but it is not working then I tried z=6 and it works thereafter, it should be greater than Z which is Z=6 Z^2=36
b=8 , b^2=64 , Z^2+b^2= 36+64=100
Answer A



why can't we use the triplet property to solve this question? for example 3x,4y,5z!
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5 or z [#permalink]
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mrk9414 wrote:
settahri wrote:
Answer=A
I used a^2+b^2=C^2=100
if we use z=5 z^2=25 the rest is 75 there is nothing square smth gives u 75, therefore, it could be less than Z or greater than Z not equal to 5. I tried Z=4 but it is not working then I tried z=6 and it works thereafter, it should be greater than Z which is Z=6 Z^2=36
b=8 , b^2=64 , Z^2+b^2= 36+64=100
Answer A



why can't we use the triplet property to solve this question? for example 3x,4y,5z!


Hi there!

Just to clarify x and y are the measure of the angles i.e the degrees and not the length of the side. Although I agree that we can use pythagoras theorem, but that will deviate us from the information given y>2z this information is given for some purpose. This is the only information that derives us to the correct answer for
eg. If I use pythagoras the sides can be 6, 8 & 10 or \(5\sqrt{2}, 5\sqrt{2}\), 10

In both the cases z is greater than 5. But we do not surely know that y>2x. When we use this information, we already know that one angle is 90 the other two will sum upto 90 and as y>2x, y>60 and x<30 and z<5 using the 30-60-90 rule.


And yes good to know: Whenever we use pythogaras theorem the constant are same and not distinct i.e 3x, 4x & 5x and not 3x, 4y & 5z!!!!

Let me know if I can help further!

Hope this helps!
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Re: 5 or z [#permalink]
Ks1859 wrote:
mrk9414 wrote:
settahri wrote:
Answer=A
I used a^2+b^2=C^2=100
if we use z=5 z^2=25 the rest is 75 there is nothing square smth gives u 75, therefore, it could be less than Z or greater than Z not equal to 5. I tried Z=4 but it is not working then I tried z=6 and it works thereafter, it should be greater than Z which is Z=6 Z^2=36
b=8 , b^2=64 , Z^2+b^2= 36+64=100
Answer A



why can't we use the triplet property to solve this question? for example 3x,4y,5z!


Hi there!

Just to clarify x and y are the measure of the angles i.e the degrees and not the length of the side. Although I agree that we can use pythagoras theorem, but that will deviate us from the information given y>2z this information is given for some purpose. This is the only information that derives us to the correct answer for
eg. If I use pythagoras the sides can be 6, 8 & 10 or \(5\sqrt{2}, 5\sqrt{2}\), 10

thanks for your quick response and for clearing my doubt.

yes, you're right it should be in the whole x by mistake I wrote x,y,z.



In both the cases z is greater than 5. But we do not surely know that y>2x. When we use this information, we already know that one angle is 90 the other two will sum upto 90 and as y>2x, y>60 and x<30 and z<5.


And yes good to know: Whenever we use pythogaras theorem the constant are same and not distinct i.e 3x, 4x & 5x and not 3x, 4y & 5z!!!!

Let me know if I can help further!

Hope this helps!
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Re: 5 or z [#permalink]
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solving using angles perspective
y+x=90
y=90-x>2x
30>x
max value of sinx<0.5
therefore z<5.
A
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Re: 5 or z [#permalink]
If this were a 30-60-90 triangle, then y=2x (e.g. y=60, x=30) and z=5.

However, we are told that y > 2x, meaning that y > 60 and x <30.

Since z is the side across from the 30-degree angle, if the angle gets smaller, the side length will also get smaller. Thus, z<5.

So the answer is A.
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Re: 5 or z [#permalink]
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