Triangle with 0<a<1<b<2<c<3

**james_bond** · January 9th 2010, 09:09 AM

In a triangle with sides $a$ , $b$ , $c$ the following holds: $0<a\le 1\le b\le 2\le c\le 3$ . What can be the maximum area?

**Henryt999** · January 9th 2010, 11:17 AM

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**Grandad** · January 9th 2010, 12:19 PM

Hello everyone

Originally Posted by Henryt999

Maximum area is when the sides are maximum so a =1 b = 2 c = 3

Do you know how to go from here?
Law of cosines?

C^2 = a^2 + b^2 -2ab*cos(c) this gives you angle c = arccos(((3^2)-(1^2)-(2^2))/(-2*1*2)) that gives you the angle.
Now to find the area use: A = (ab*sin(x))/2 where x is the angel between the sides a;b.
Hope this was not too confusing, I can follow you through if you want.

Sorry, but this is certainly not correct. With $a=1, b=2, c=3$ , the 'triangle' is a straight line. (Note: $1+2=3$ .) So it has zero area.

I'm still thinking about how to answer the question correctly...

Grandad

**Henryt999** · January 9th 2010, 12:23 PM

Oh lord, You are correct Im sorry that was very very wrong. Sorry Hope he didn´t wright that down.

**Henryt999** · January 9th 2010, 02:09 PM

Originally Posted by Grandad

Hello everyoneSorry, but this is certainly not correct. With $a=1, b=2, c=3$ , the 'triangle' is a straight line. (Note: $1+2=3$ .) So it has zero area.
Grandad

You are correct the third side can not be 3 But has to be less.

If for example the angle is $(pi)/2$ then it has area: $(1^2+2^2)^(0.5) = sqrt(5)$

And Area is given by: $A = (a*b*sin(x))/2 = (1*2*sin(x))/2 = sin(x);$ That would give me a maximum area of 1??

If sides are: $1;2$ the last side would be $sqrt:5$ ??
If this is the correct solution its not pretty, hope it was somewhat helpful....

**Grandad** · January 9th 2010, 11:19 PM

Hello everyone

Originally Posted by Henryt999

You are correct the third side can not be 3 But has to be less.

If for example the angle is $(pi)/2$ then it has area: $(1^2+2^2)^(0.5) = sqrt(5)$

And Area is given by: $A = (a*b*sin(x))/2 = (1*2*sin(x))/2 = sin(x);$ That would give me a maximum area of 1??

If sides are: $1;2$ the last side would be $sqrt:5$ ??
If this is the correct solution its not pretty, hope it was somewhat helpful....

Yes, I think this is correct. I didn't have any more time to think about this yesterday (it's now 7:15 am UK time), but I had come to the same conclusion overnight. With $b=2$ as base, the maximum height possible is 1, where $a = 1$ and the angle between these two sides is $90^o$ . So $c = \sqrt5$ , and the area is $1$ sq unit.

Grandad

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Triangle with 0<a<1<b<2<c<3

Maximum area

Thanks

Acctually I might be wrong again but....

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