Ok, heres some math

The area of Triangle ABC + A circle with a circumference of D= The Base of the bottom of a rectangular prism. The height of this prism is 5. Find the volume of the rectanguler when...

a=7.8, b=10.4, c= (you have to figure c out) and d= b*c]

EASY!

The area of Triangle ABC...

...a=7.8, b=10.4, c= (you have to figure c out)...

But wait! With just two of the sides given, and no other assumptions, we can't find out what c is Maybe we have the wrong triangle.

...unless--

Is ABC a right triangle?

You do have the wrong triangle (we know by the values given we have a scalene triangle).  It is not necessary to know, or to have a right triangle as part of the given information. It makes solving a bit more complicated, but not by much.  It is possible to solve for C.  Even if you only know for every right triangle A squared plus B squared = C squared you can still solve with two equations and two unknowns.

After calculating, the height of the given triangle H = 6.24



Max

It can be a right triangle, with the length of the hypotenuse being 13, assuming side a and b are the lengths of the legs... I really don't know what they're supposed to be (angle measurements?). And the values of the third leg of the triangle can be shown like so: 18.2>c>2.6.  And If it's not a right triangle, you'd need at least one angle value as well to use the law of sines or the law of cosines to find the rest of the measurements. If it is a right triangle, and c is the hypotnuse, then you already have the height and base of the triangle, which means you can find the area, and c is only necessary for finding the circumference.

If this was a right triangle here's the solution as far as I can go:
c=13
So we can find D to be 135.2 (So the diameter was 43.03549661...? Seems odd, but I'll keep going)
The area doesn't require c since c is the hypotenuse, b*a*1/2 = 40.56
The base of the rectangular prism is 175.76 square units.
Multiply that by 5.
878.8 cubic unites
But I completely assumed half of the problem.
If that triangle is not a right triangle I'd need another angle to find out the rest of the measurements, from that I could make a right triangle within that triangle to help me find the height of the triangle. Otherwise, to my knowledge, you cannot find the area of that triangle.

You are not given any angles. How do you know what degree to bisect the angle at? You don't know how bringing an altitude down will effect the angles, not only that, since you aren't given any angles, now you have two right triangles with one measurement each. OR you could bring an altitude down from A and have 7.8 under it,(or the same from b, with 10.4) but since the altitude can be anywhere on that leg, you don't know how to cut up 7.8, it could be half, or maybe not. But you don't have any angles so you can't possibly use the law of sines or cosines to find the measure of a side or angle to help you out from one of those right triangles you made.
Could you please show me your work to find the the measurements of this triangle?

The only way to make two right triangles is bringing down an altitude from one of the angles.
Heres are all 6 of your right triangle possibilities.

From angle B (opposite of side b):
The side being "hit" with an altitude is 10.4, since it's an altitude you don't know how it's altered, and I believe it's also impossible that creates a midpoint, because the only triangle I know that creates a midpoint with its altitude is a 60-60-60 aka an equilateral triangle.
You'd get one triangle:
Sides: Altitude, Side c, and a portion of side b. All lengths unknown.
The other:
Sides: Altitude, Side a, a portion of of side b. A is known to be 7.8.

From angle A:
One triangle:
Sides: Altitude, side c, a portion of side a. All lengths unknown.
The other:
Side: Altitude, side b, a portion of side c. b is known to be 10.4

From angle C:
One triangle:
Sides: Altitude, a potion of unknown c, side a. A is known to be 7.8
The other:
Sides: Altitude, a potion of unknown c, side b. Side b is known to be 10.4

You only have a max of one side measurement and a right angle. The law of cosines or sines cannot help you with a 90 degree angle and one side length.

No, you and Fred are correct.  I went back and looked at the assumptions which i made.  You are absolutely right, there isn't enough information to solve.  Too bad, I love math too.  Nevertheless, even if an angle were given, it would have been too easy to solve.  We need something more challenging.



Max

There was the earth and belt thingy challenge Fred posted, But I already answered it, you can try and just not read my answer if you want to do it on your own, but here is the question:
Imagine a belt snug around a spherical earth's equator so that there's no space between it and the earth, make the belt 6 inches longer, how far does the belt raise above the earth's equator?

I'm not sure but I think you'd solve it like this:
The person's line of sight it tangent to a circle... So that tangent line will be what we are trying to find, well at least until it "touches" earth.
The point from the center of the circle to the point the tangent "hits" the earth creates a right angle, That line segment is equal to the radius of the earth. Then you can make the next leg of the right triangle the radius to the person on the tower. I'm going to assume we're not calculating the height of the person into the equation as well.
So now we have one leg that is 6,380,000 meters (a) and the other that is 6,380,050 meters(c). So c squared - a squared = b squared, how far we can see! Do some calculations... And we find it's about 25,258.71137 meters, I'll round that to 25259 meters. So that's about 25 kilometers. But that would be the bottom of the tower...

So If we're not looking at the bottom the tower, rather the top, I'd be left with an isosceles triangle... Two measures of 6380,050 from the center to the height of the two towers and x, the distance between the tops of the two towers. Well, I know 0<x<12760100. Other than that I can't find that distance using anything I know. Even if I drop an altitude from the earth's center, I couldn't do much considering I'd be left with the radius, which I know, the altitude, which is unknown, and half of x, which is also unknown at that point. So I'll go with my first answer.

all this sounds like mumbo jumbo to me.

You da man Fred.  Answer is ~50 km (50.49).  See the attachment for the calulations.



Max

Only thanks to Mac's starting calculations! Even just reading other people's ideas on a certain problem seems like accomplishing half the work already. Exploring possibilities, eliminating things that won't work.

all this sounds like mumbo jumbo to me.

Hehe that's why I try to slip in a few supposed jokes/funny references here and there; make it a little bit more interesting to most of everyone. Like this Vi Hart video when she's bored in math class. (Dedicated to CloudSong mayor MasterSquiggle)
http://www.youtube.com/watch?v=ik2CZqsAw28#