How do you calculate the area of an equilateral triangle with three 90 degree angles, and sides of length 1?

Question

(may require some "thinking outside the box")

Way to go, Dr.Bob!

Answer 1

OK here's how you do it.

For the three angles of a triangle to be 90 degrees it has to be drawn on something like the surface of a sphere rather than a flat plane.

Now each edge of the triangle becomes one quarter of the circumference of the sphere, the triangle covers one eighth of the surface of the sphere.

The circumference of said sphere is 4 (units)

The radius is then 4/(2(pi))
2/(pi)

The surface area of the whole sphere =4(pi)r^2
= 4(pi)x 2x2/((pi)^2)
=16/(pi)

One eigth of this surface area is (16/(pi))/8
= 2/(pi) units squared

Answer 2

A triangle cannot have 3 ninety degree angles; it would not be a triangle but rather a rectangle. Why? All three angles of a triangle add up to 180 degrees, and 90 * 3 = 270 which is greater than 180. I like dr bob's answer; however if you draw a triangle on a sphere it will no longer be called a triangle; a triangle is a 2-d object, and that would make it 3-d.

Answer 3

This would have to be a triangle on a sphere. One vertex would be at the north pole and the other two vertices on the equator 90° of longitude apart. This triangle is 1/8 the surface area of the sphere. So if the sphere had radius r, the area A, of the triangle would be:

A = (1/8)4πr² = πr²/2

For a sphere with radius r the length of one side s, of this triangle would be:

s = (1/4)2πr = πr/2 = 1
r = 2/π

So the area of the triangle is:

A = πr²/2 = (π/2)(r²) = (π/2)(2/π)² = 2/π

Answer 4

Dr. Bob has it right.

To those who disagree with his answer, you have to look at a sphere. If you were to draw a triangle on a sphere, the angles would be >180. If it were a pseudosphere, the sum of angles < 180.

split the sphere into great circles and you get the 360 angle.

90/360= 1/4 of the circle.

Answer 5

hmmmm...an interesting question. let's see... i don't see an answer directly. however, an additional constraint of the problem would be the the three inner circles do not overlap each other. i have two scenarios from your description: 1) the four circles are all tangent at the same point or 2) it looks like the top view of three bowling balls in a barrel. sorry i don't have an answer (yet) but i will star it and keep an eye on the developments. good luck.

Answer 6

OK - I'm now "outside the box" - and it's lonely out here.

Your question cannot be based on a Euclidean "plain".
The sides cannot be "straight" lines in the conventional meaning.

Since the lines are curved (again, in the conventional sense) it is most probably 1/8th of the surface area of a sphere with a circumference of 4.

Answer 7

Thinking of the triangle as lying on a sphere gives one possible answer for the area, but I don't think this is the only answer. For example, what if you replace the sphere by a surface which has a huge bulge near the center of this triangle, but which looks like a sphere near the edges of the triangle?

Answer 8

there's no such triangle. THe angles of a triangle add up to 180 degrees. What you've descibed are three sides of a square

EDIT

Ha ha okay good one - annoying as I did exactly this kind of this in Cosmology. But there you go, I thought you were another of those werird random question askers. Curse my Euclidean Geometric Mind!

Answer 9

Think of the Earth, 2 meridians and a paralell

Ana

Answer 10

think of earth
north pole so and so forth..
now that area is a slice
as u see I can solve it
but since u also know the solution I say :)
long live minkowski