Math Help!

[Wheatmaster]

triangle has sides 5, 6, and x. The Area is also x. Find the greatest possible variable of x. (the triangle can be any kind). i'm stuck!

edit: i think it's x<11?

 

[mb]

Make that 7.8102496759066543941297227357591

 

[Vortex22]

Originally posted by: Wheatmaster
triangle has sides 5, 6, and x. The Area is also x. Find the greatest possible variable of x. (the triangle can be any kind). i'm stuck!

edit: i think it's x<11?

No, they are looking for one value.

 

[Wheatmaster]

Originally posted by: supafly
Make that 7.8102496759066543941297227357591

well i thought it was that easy, but how does that equal it's area?

 

[mb]

It may not, I didn't work that out, but to find the third side of the triangle its always a squared + b squared = c squared.

 

[Wheatmaster]

Originally posted by: supafly
It may not, I didn't work that out, but to find the third side of the triangle its always a squared + b squared = c squared.

that's only if its a right triangle

 

[mugs]

area = sqrt(s(s-a)(s-b)(s-c)) where s = (a+b+c)/2

 

[mb]

Originally posted by: Wheatmaster

Originally posted by: supafly
It may not, I didn't work that out, but to find the third side of the triangle its always a squared + b squared = c squared.

that's only if its a right triangle

Really? Damn, it's been too long since I've done that sorta stuff. Gimme an accounting problem next time

With the information given I think it's safe to assume its a right triangle, since we don't know any angles..

 

[Bound_Vortex]

area = (1/2)base*height
area = x
base = x

x = (1/2)x*h
h = 2

theta1 and theta2 are the angles made with the base and each respective side

sin(theta1) = 2/6
theta1 = 19.47 degrees
sin(theta2) = 2/5
theta2 = 23.58 degrees

x = 6cos(theta1)+5cos(theta2)
x = 10.239

:beer:

 

[Muzzan]

With the information given I think it's safe to assume its a right triangle, since we don't know any angles..

Eh, the first post says:

(the triangle can be any kind)

 

[mb]

Originally posted by: Muzzan

With the information given I think it's safe to assume its a right triangle, since we don't know any angles..

Eh, the first post says:

(the triangle can be any kind)

Oh god, I should just stop drinking so much, or take more time to read the next morning

 

[Kyteland]

MathHelp.jpg

This should help you greatly in solving the problem.

And yes, x<11

 

[Tremulant]

42

 

[Kyteland]

Originally posted by: Bound_Vortex
area = (1/2)base*height
area = x
base = x

x = (1/2)x*h
h = 2

theta1 and theta2 are the angles made with the base and each respective side

sin(theta1) = 2/6
theta1 = 19.47 degrees
sin(theta2) = 2/5
theta2 = 23.58 degrees

x = 6cos(theta1)+5cos(theta2)
x = 10.239

:beer:

*sigh* I shouldn't have taken the time to draw the picture.....

Also, once you know the height and two of the sides you can use the pythagoean theorem to get an exact answer.

x = sqrt(32) + sqrt(21)

 

[Wheatmaster]

how'd u get height of 2?

 

[MichaelD]

Originally posted by: Bound_Vortex
area = (1/2)base*height
area = x
base = x

x = (1/2)x*h
h = 2

theta1 and theta2 are the angles made with the base and each respective side

sin(theta1) = 2/6
theta1 = 19.47 degrees
sin(theta2) = 2/5
theta2 = 23.58 degrees

x = 6cos(theta1)+5cos(theta2)
x = 10.239

:beer:

Holy crap. That must be the DNA sequence to all living things...or something meaningful like that. Never in a million years could I solve such an equation. It all boils down to that left brain/right brain thing. Whatever side of the brain does math, well, that's the side of my brain that does not work.

 

[TuxDave]

Originally posted by: Wheatmaster
how'd u get height of 2?

area = (1/2)base*height
area = x
base = x

x = (1/2)x*h
1 = 1/2*h
h = 2

 

[ActuaryTm]

If nothing else, threads such as this serve as excellent comic relief (most especially when the CPAs chime in).

 

[Chronoshock]

Originally posted by: mugs
area = sqrt(s(s-a)(s-b)(s-c)) where s = (a+b+c)/2

Hero's thm = :thumbsup: