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Bunuel
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Bunuel
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Hello aimtoteach

Not all right angled triangles follow the Pythagorean Triplet theory
In the sense that AB could equal = 2 and BC = 5
still holding statement (1) AB + BC = 7 true

and the hypotenuse AC would equal (2^2 + 5^2)^.5 = 29^.5
= approximately between 5 and 6

& Perimeter would then = 2+5+5 = 15

And hence statement 1 is Not Sufficient

Thanks
ppnimkar
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Excellent question for today, thank you Bunuel and thank you ppnimkar for your explanation as well...I was wrong myself and chose A) with S1 being sufficient but I see the fallacy in my logic now!
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if it is a right triangle does this not mean that 2 sides of the triangle should be equal , by that logic two sides are 3.5,3.5 and the third side can be found by pythagores theoram
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yashrakhiani
if it is a right triangle does this not mean that 2 sides of the triangle should be equal , by that logic two sides are 3.5,3.5 and the third side can be found by pythagores theoram

No, it does not mean that at all. Why should ABC necessarily be an isosceles right triangle?
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Please see the below attached photos for the explanation
Attachments

IMG_20161102_221042_HDR.jpg
IMG_20161102_221042_HDR.jpg [ 3.8 MiB | Viewed 10688 times ]

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My 2 cents:
As in question it is NOT mentioned whether size of sides AB and BC is integer we CANNOT assume, based on AB + BC = 7, that AB = 3 or 4 & BC = 4 or 3 to form Pythagorean triplet or 3-4-5. As CORRECTLY shown above by others, AB = 6.5 and BC = 0.5.
Honestly, i made the same mistake in this problem. :(
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Bunuel could you please explain why statement 1 is not sufficient if I use the Pythagorean triple theory?
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sadikabid27
Bunuel could you please explain why statement 1 is not sufficient if I use the Pythagorean triple theory?

So you are saying that the sides must be 3 and 4 and the hypotenuse comes out to the 5. But why? How do you know that the sides of this particular triangle must be Pythagorean triple? Why cannot they be 1 and 6, for example? Or why they cannot be 1.4 and 5.6? We are neither told that the sides are Pythagorean triples nor that the sides are integers.
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One angle is 90 degree. so I took in triangle abc assuming it's a right triangle, ab^2+bc^2=ac^2 , ab+bc=7 so ab^2+bc^2=7^2 , then you can find ac=7, so ab+bc+ca=7+7=14 (that's perimeter), where I am going wrong? Bunuel
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sadikabid27
One angle is 90 degree. so I took in triangle abc assuming it's a right triangle, ab^2+bc^2=ac^2 , ab+bc=7 so ab^2+bc^2=7^2 , then you can find ac=7, so ab+bc+ca=7+7=14 (that's perimeter), where I am going wrong? Bunuel

I tried to explain it above. Here it is again. Yes, the given triangle is right angled at B and yes, AB^2 + BC^2 = AC^2. But AB + BC = 7 does not mean that AB^2 + BC^2 = 7^2. If you square AB + BC = 7 you'll get AB^2 + 2AB*BC + BC^2 = 7^2.
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