Print Page - ATTN: People who are good at math

Title: ATTN: People who are good at math
Post by: CapitalistClock on October 21, 2008, 02:49:55 PM

(https://clockcrew.net/talk/proxy.php?request=http%3A%2F%2Fbay01.imagebay.com%2F_upload%2Fimg%2F63%2Fmath.png&hash=21567ca2c33ca02c0d666219c0cf5d464f625e27)

How do I figure out the area of this?

Title: ATTN: People who are good at math
Post by: CANNABISCLOCK on October 21, 2008, 02:51:46 PM

Title: ATTN: People who are good at math
Post by: GearBoxClock on October 21, 2008, 02:54:12 PM

You don't

Title: ATTN: People who are good at math
Post by: FluxCapacitorClock on October 21, 2008, 02:54:14 PM

Title: ATTN: People who are good at math
Post by: SoBe Clock on October 21, 2008, 02:54:44 PM

Title: ATTN: People who are good at math
Post by: CapitalistClock on October 21, 2008, 02:55:58 PM

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

Title: ATTN: People who are good at math
Post by: PirateClock on October 21, 2008, 03:04:07 PM

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

Title: ATTN: People who are good at math
Post by: FluxCapacitorClock on October 21, 2008, 03:05:33 PM

The answer is clearly 63

Title: ATTN: People who are good at math
Post by: SouvlakiClock on October 21, 2008, 03:06:56 PM

use the formula you were taught in school

Title: ATTN: People who are good at math
Post by: PropagandaClock on October 21, 2008, 03:09:00 PM

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

Title: ATTN: People who are good at math
Post by: Wind-up Clock on October 21, 2008, 03:24:35 PM

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.

Title: ATTN: People who are good at math
Post by: FluxCapacitorClock on October 21, 2008, 03:32:25 PM

Quote from: Wind-up Clock;1431295If 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.

Title: ATTN: People who are good at math
Post by: GreyClock on October 21, 2008, 03:33:07 PM

Title: ATTN: People who are good at math
Post by: Wind-up Clock on October 21, 2008, 03:36:53 PM

Quote from: GreyClock;1431303If 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 :golfclap:

Title: ATTN: People who are good at math
Post by: CapitalistClock on October 21, 2008, 04:14:19 PM

Quote from: SouvlakiClock;1431280use the formula you were taught in school

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

Quote from: GreyClock;1431303If 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. :D

Title: ATTN: People who are good at math
Post by: Marlin Clock on October 21, 2008, 04:14:38 PM

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.

Title: ATTN: People who are good at math
Post by: Wind-up Clock on October 21, 2008, 04:16:13 PM

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

Thank god these equations were on the formula sheet.

Title: ATTN: People who are good at math
Post by: MonsterMunch on October 21, 2008, 05:12:16 PM

Rape someone

Title: ATTN: People who are good at math
Post by: TwistClock on October 22, 2008, 12:43:32 PM

Quote from: Franklin G. Hamilton;1431353look at awht i did

[flash]http://72.20.18.21/clam/math.swf width=550 height=400[/flash]

Clock Crew

Community => General Discussion => Topic farted by: CapitalistClock on October 21, 2008, 02:49:55 PM