Possible to solve with geometry or other method?

[eber]

In the graphic posted here: Text

And before anyone asks, no, it's not homework...explaination at the very bottom.

Given: All the distances 396, 402, 411, and 400 are from point X. The angle at 396 is 90 degrees. All of the other angles are unknown (at least as far as I know).

Is is possible to find the lengths of "a" and "b"? And furthermore, is it possible to find the area enclosed by the three points at 402, 411, and 400. I'm not asking you to solve it (but if you could that'd be great), at the least I'd like to know if it's possible and what methods you would use.

I'll probably be able to walk on the field tomorrow (a special event) and take a rough estimate of lengths "a" and "b" by walking along side of them (no I probably won't be all nerdy and breakout the tape measure). I'm just curious.

And for those curious to what this is, it's a mock-up graphic for the new fences for Petco Park, where "X" is home plate. I'm just curious as to how much square footage was enclosed by the new fence.

Edit: Do you think you might need to know distances between each of the points at 396, 402, and 411?

 

[2Xtreme21]

a ~ 145ft
b ~ 94.45ft

You can find the area yourself by looking up tutorials on how to find areas of irregular quadrilaterals. It's essentially going to be (.5*b*h)triangle1 + (.5*b*h)triangle2.

 

[eber]

Damn...nice work. Don't know if it's right or not, but I'll take your word for it. I'm curious in the methods you used, but it was probably a bunch of handwork, so if it was, that's ok then. On second thought, I guess you could draw it up in a CAD program or something similar and dimension it. But is it possible to find the area enclosed by 402, 411, and 400?

Heh...it does sound like homework, but I just set it up that way. I'm just really geeky about ballparks.

 

[2Xtreme21]

It was real easy actually. Start out by drawing lines from home plate to the distances measured. Then compute the missing sides of the triangles using the pythagorean theorem. Rinse and repeat until you have all the missing sides. I'll get into further detail with pics if you want.

 

[eber]

Heh..I've haven't used geometry in a while. I tried figuring this out a while ago with pythagorean theorem and the law of cosines and whatnot but got stuck somewhere and just quit...I guess I didn't try hard enough. It's no problem about drawing everything up.

I guess what I'm more curious about is the area enclosed by 402, 411, and 400. They brought in the new fence for what seems to be a very small area. I guess I should've asked this from the start. To me, I can't seem to figure that part out without knowing any of the angles in that "area".

 

[Ready]

probably possible, but I got too lazy when i forgot how to find the other 2 lengh of a triangle given all 3 angles and one of the lenghs

 

[BigPoppa]

Originally posted by: Ready
probably possible, but I got too lazy when i forgot how to find the other 2 lengh of a triangle given all 3 angles and one of the lenghs

Use some trig functions.

 

[DrPizza]

Can I be the first person to say, "no, it's not possible"??
It isn't, unless I'm missing something in that diagram.

 

[DrPizza]

Yeah, I just read your OP again.. It's not possible. Let me explain.

Let's leave the 396, 402, and 411 points where they're at. Grab a compass. Set your compass to represent 400 feet (to scale). With the point at X, draw an arc (which will include the point where you have the 400.) Clearly, A and B can be drawn to connect to any point on this arc, and in doing so, they will be either longer or shorter.

 

[2Xtreme21]

You can make 4 separate triangles by simply connecting the distance points to X.

 

[DrPizza]

Here. In this edited diagram, I've shown another point that's also 400 feet away from point X. Clearly the lengths a and b are different, as is the area.

edit: I may be off by a pixel or so; just extend or shorten the line segment from x as appropriate. As I pointed out in an earlier post above, there are many such points, all lying on a circle with the center at X and having a radius of 400 feet.

 

[2Xtreme21]

Gotcha. Sorry DrPizza, it's been quite a while.

 

[DrPizza]

Hey, no problem No need to apologize to me! If it *is* a homework problem, the OP hasn't returned since your other answer. Apologize to him!

 

[Ready]

? I didn't solve it but I think it's solvable using the equations
Side Angle Side and Angle Side Angle
I have a picture but now way of showing it. not sure if i'm right or wrong.

The only thing I'm not sure about is if using Side angle side etc..do they have to connect?

 

[eber]

Originally posted by: 2Xtreme21
a ~ 145ft
b ~ 94.45ft

You can find the area yourself by looking up tutorials on how to find areas of irregular quadrilaterals. It's essentially going to be (.5*b*h)triangle1 + (.5*b*h)triangle2.

I'm not doubting your methods to reach your answer, but according to real pictures of the area, side B appears to be way off:
Diagramed Graphic
Actual Before
Actual After

I think what's wrong is that the points 396, 402, 411, and 400 aren't exact and most likely approximations... because of that the angles from home plate to each of those points would be very small and thus there'd be huge margins for error.

And really guys...it's not homework . I know, I know...the more I say it, the less you'll believe me...

 

[eber]

Originally posted by: DrPizza
Here. In this edited diagram, I've shown another point that's also 400 feet away from point X. Clearly the lengths a and b are different, as is the area.

edit: I may be off by a pixel or so; just extend or shorten the line segment from x as appropriate. As I pointed out in an earlier post above, there are many such points, all lying on a circle with the center at X and having a radius of 400 feet.

Hmmm...were you assuming my graphic was to scale? If you were, sorry, it's not and I maybe should've told you beforehand...I just made it up real fast in a paint program.

Edit: And really...no need to stress over this...it's not homework!

 

[DrPizza]

Originally posted by: eber

Originally posted by: DrPizza
Here. In this edited diagram, I've shown another point that's also 400 feet away from point X. Clearly the lengths a and b are different, as is the area.

edit: I may be off by a pixel or so; just extend or shorten the line segment from x as appropriate. As I pointed out in an earlier post above, there are many such points, all lying on a circle with the center at X and having a radius of 400 feet.

Hmmm...were you assuming my graphic was to scale? If you were, sorry, it's not and I maybe should've told you beforehand...I just made it up real fast in a paint program.

Edit: And really...no need to stress over this...it's not homework!

No, I wasn't assuming the graphic was to scale. However, if you look at your picture and my picture, you can see that even if it were to perfect scale, both your point where the 400 foot point and my picture (in red) where the 400 foot point is both meet all of the criteria listed in the problem. i.e. The point 400 feet out isn't uniquely defined. It can be anywhere on an arc. I'll try again to explain this on the picture.

Here.jpg

I would have loved to assume that other angle in your first diagram (at the 400 point) was 90 degrees. However, looking at that original graphic that you provided here, it's pretty apparent that the angle behind home plate is not 90 degrees; thus, I wouldn't assume the other angle is 90 degrees. However, if you really want an answer, and this second graphic is to scale, merely measure the angle at X. Then, the formulas you're looking for are law of sines: SinA/a = SinB/b = SinC/c or the law of cosines: a^2 = b^2 + c^2 -2bcCosA. But, if you're simply looking for the area of the triangle they reduced the size of the field from, after finding 2 of the sides of that triangle, you can use 1/2 the product of those sides times the sin of the angle between those sides.