ATTN: People who are good at math

[CapitalistClock]

How do I figure out the area of this?

[CANNABISCLOCK]

63

[GearBoxClock]

You don't

[FluxCapacitorClock]

63

[SoBe Clock]

63

[CapitalistClock]

Well, I need proof of how it's done too.

[PirateClock]

you build an exact replica.
fill it with water
measure how much water you put in
profit.

[FluxCapacitorClock]

The answer is clearly 63

[SouvlakiClock]

use the formula you were taught in school

[PropagandaClock]

Turn the tilted corners to Triangles and calculate the area for the whole square then remove the triangles

[Wind-up Clock]

If ya wanna high-falutin' all complicated way, then do this.

Turn the quadrilateral into two triangles, We'll call 'em ABC and  ADC.

First to find the area of ABC, use the area rule (1/2)ABsinC,

(1/2)(31)(68.2)sin(79) = 1037.67 m^2

To find the area of ADC, use the cosine rule to find AC, which is
c^2 = (31)^2 + (68.2)^2 - 2(31)(68.2)cos(79), which has AC turn out to be roughly 69.3382237

Then find the angle CAD using the sin rule

12.4/sinCAD = 69.3/sin94, where CAD becomes 10.28 degrees.

Then find angle ACD by 180--94-10.28 = 75.72 degrees.

Then using the sine rule again, find AD

AD\sin75.72 = 69.3/sin94 = 67.322

Then using the area formula again

(1/2)(67.322)(12.4)(sin94) = 416.379

Then add the areas

1037.67 + 416.37 = 1454.04 m2

That's how i solved it.

[FluxCapacitorClock]

If ya wanna high-falutin' all complicated way, then do this.

Turn the quadrilateral into two triangles, We'll call 'em ABC and  ADC.

First to find the area of ABC, use the area rule (1/2)ABsinC,

(1/2)(31)(68.2)sin(79) = 1037.67 m^2

To find the area of ADC, use the cosine rule to find AC, which is
c^2 = (31)^2 + (68.2)^2 - 2(31)(68.2)cos(79), which has AC turn out to be roughly 69.3382237

Then find the angle CAD using the sin rule

12.4/sinCAD = 69.3/sin94, where CAD becomes 10.28 degrees.

Then find angle ACD by 180--94-10.28 = 75.72 degrees.

Then using the sine rule again, find AD

AD\sin75.72 = 69.3/sin94 = 67.322

Then using the area formula again

(1/2)(67.322)(12.4)(sin94) = 416.379

Then add the areas

1037.67 + 416.37 = 1454.04 m2

That's how i solved it.

There is a real flaw with your math in that the work does not result in 63.

[GreyClock]

If ya wanna high-falutin' all complicated way, then do this.

Turn the quadrilateral into two triangles, We'll call 'em ABC and ADC.

First to find the area of ABC, use the area rule (1/2)ABsinC,

(1/2)(31)(68.2)sin(79) = 1037.67 m^2

To find the area of ADC, use the cosine rule to find AC, which is
c^2 = (31)^2 + (68.2)^2 - 2(31)(68.2)cos(79), which has AC turn out to be roughly 69.3382237

Then find the angle CAD using the sin rule

12.4/sinCAD = 69.3/sin94, where CAD becomes 10.28 degrees.

Then find angle ACD by 180--94-10.28 = 75.72 degrees.

Then using the sine rule again, find AD

AD\sin75.72 = 69.3/sin94 = 67.322

Then using the area formula again

(1/2)(67.322)(12.4)(sin94) = 416.379

Then add the areas

1037.67 + 416.37 = 1454.04 m2

That's how i solved it.

[Wind-up Clock]

If ya wanna high-falutin' all complicated way, then do this.

Turn the quadrilateral into two triangles, We'll call 'em ABC and ADC.

First to find the area of ABC, use the area rule (1/2)ABsinC,

(1/2)(31)(68.2)sin(79) = 1037.67 m^2

To find the area of ADC, use the cosine rule to find AC, which is
c^2 = (31)^2 + (68.2)^2 - 2(31)(68.2)cos(79), which has AC turn out to be roughly 69.3382237

Then find the angle CAD using the sin rule

12.4/sinCAD = 69.3/sin94, where CAD becomes 10.28 degrees.

Then find angle ACD by 180--94-10.28 = 75.72 degrees.

Then using the sine rule again, find AD

AD\sin75.72 = 69.3/sin94 = 67.322

Then using the area formula again

(1/2)(67.322)(12.4)(sin94) = 416.379

Then add the areas

1037.67 + 416.37 = 1454.04 m2

That's how i solved it.

The math checks out. I say this is correct.

Well played Greyclock

[CapitalistClock]

use the formula you were taught in school

Woah man, why didn't I think of that?!

If ya wanna high-falutin' all complicated way, then do this.

Turn the quadrilateral into two triangles, We'll call 'em ABC and ADC.

First to find the area of ABC, use the area rule (1/2)ABsinC,

(1/2)(31)(68.2)sin(79) = 1037.67 m^2

To find the area of ADC, use the cosine rule to find AC, which is
c^2 = (31)^2 + (68.2)^2 - 2(31)(68.2)cos(79), which has AC turn out to be roughly 69.3382237

Then find the angle CAD using the sin rule

12.4/sinCAD = 69.3/sin94, where CAD becomes 10.28 degrees.

Then find angle ACD by 180--94-10.28 = 75.72 degrees.

Then using the sine rule again, find AD

AD\sin75.72 = 69.3/sin94 = 67.322

Then using the area formula again

(1/2)(67.322)(12.4)(sin94) = 416.379

Then add the areas

1037.67 + 416.37 = 1454.04 m2

That's how i solved it.

Damn, man! Thanks.

[Marlin Clock]

Man I only did this stuff a year ago and I've already forgotten how to do it.

Thank god I'm going into Biology, not a lot of trig or calc there.

[Wind-up Clock]

I had a test on this, and i didn't study for it.

Thank god these equations were on the formula sheet.

[MonsterMunch]

Rape someone

[TwistClock]

look at awht i did
 
[flash]http://72.20.18.21/clam/math.swf width=550 height=400[/flash]

63