Right triangle

[Khush]

Hi Ron,

I just wanted to ask a general query on concepts involving right triangle.

Suppose we are given that in a triangle ABC , AB^2+BC^2 = 6^2+8^2, is it sufficient to conclude that ABC is a right triangle or we need some more information?

As far as my knowledge is concerned, if 2 small legs in a right triangle are 5 and 12 respectively, then the hypotenuse of this right triangle will be 5^2+12^2= 169=13^2

However, i am not sure whether the vice-versa can also be true.

Please help me understand.

[RonPurewal]

Hi Ron,

I just wanted to ask a general query on concepts involving right triangle.

Suppose we are given that in a triangle ABC , AB^2+BC^2 = 6^2+8^2, is it sufficient to conclude that ABC is a right triangle or we need some more information?

no, definitely not. there are at least 2 major holes in this reasoning:

1/
you've just given (AB)^2 + (BC)^2 = 100, but there is no guarantee that the actual values of AB and BC are 6 and 8. (they could be √50 and √50, or 9 and √19, or any of infinitely many other combinations.)

2/
even if we did know that AB and BC were 6 and 8 (which we don't), there are still a whole lot of triangles with sides having lengths 6, 8, and ___.
the unspecified third side could actually be any value between 2 and 14.

[Khush]

understood.

but, there is still some problem with my reasoning.

1) Does it mean that provided we know AB^2+BC^2 = 6^2+8^2 =100 , we still can't say whether AB=6 and BC=8 or AB=8 and BC=6 ?

2) Lets say we also know that BC^2+AC^2=8^2+10^2=164

Now, can we conclude, on the basis of both information, that ABC is a right triangle?

[RonPurewal]

1) Does it mean that provided we know AB^2+BC^2 = 6^2+8^2 =100 , we still can't say whether AB=6 and BC=8 or AB=8 and BC=6 ?

... or AB = BC = √50
... or AB = √19, BC = 9
... or infinitely many other possibilities.

remember, "6^2 + 8^2" is just 100. there is nothing at all to indicate that anything has to be 6 or 8.

[RonPurewal]

2) Lets say we also know that BC^2+AC^2=8^2+10^2=164

Now, can we conclude, on the basis of both information, that ABC is a right triangle?

still nope. in fact, every example i gave above still works here, too.

you could have AB = 6, BC = 8, AC = 10, which is presumably the solution you had in mind.
... but you could also have AB = BC = √50, AC = √114
... or AB = √19, BC = 9, AC = √83
... or infinitely many other possibilities again.

[RonPurewal]

here's another way to think about this:
just let AB^2 = "x", BC^2 = "y", AC^2 = "z".
then, in your #2, you have a system of only two equations: x + y = 100, y + z = 164. this is, very clearly, not enough information to solve for three variables.

[Khush]

got it, Ron!

so, the take away is, unless the question exclusively mentions the measurements of sides, as in the above question, we cannot assume that the sides measure so.

Moreover, even if statement 1 mentions that the lengths of two sides are 6 & 8 respectively, the third side can range from 2 to 14, exclusive.

am i correct?

[RonPurewal]

right.

more generally, you should think about what you DON'T know, just as much as you should think about what you DO know.

[RonPurewal]

Moreover, even if statement 1 mentions that the lengths of two sides are 6 & 8 respectively, the third side can range from 2 to 14, exclusive.

am i correct?

yes.

make sure you know where the "2" and the "14" come from.

imagine two sticks, one of length 6 and the other of length 8, connected by a hinge.

• if you close the hinge as much as possible-- so that the sticks are almost overlapping-- then the third side of the triangle will be just barely more than 8 - 6 = 2.

• if you open the hinge-- so that the two sticks almost (but don't quite) make a straight line-- then the third side of the triangle will be just barely less than 8 + 6 = 14.

[Khush]

sure. makes sense.

Thank You very much Ron for the perfect explanation!

[jnelson0612]

Great to hear!

[RonPurewal]

you're welcome.

[Humstudents]

Great and informative posts.Thanks for sharing.

[RonPurewal]

sure.