Third level geometry problem

skanger

I'm sitting the leaving cert at the moment and our teacher gave us a problem she said is 'third level standard'. Well if it is she's not finished her degree so we're probably just doing her homework for her. She has offered a monetary reward to anyone who gets it, so I'm hoping someone here can help.

It's a circle of radius 5 bounded by a paralellogram, the task is to find the area of the paralellogram. No other information is given.

Thanks to anyone who offers any help or guidance in advance.

Brussels Sprout

Ta-dah!

TheBody

Do you mean like this?



The area of a parallelogram is the base times the perpendicular height. Can you figure it out from here?

(If this is the problem, it is not a third level question.)

bnt

TheBody wrote: »

Do you mean like this?

If it's like that, then the slope of the parallelogram needs to be known too - it affects the length of the base.

If you draw this with a parallelogram with the angles at 90° - i.e. a square - then the base = height = 2r, so the area is 10x10=100.

If you look at the picture, however, base > 2r. You can see this more clearly if you draw a horizontal line through the centre of the circle: the length of the line is 2r plus some extra on the sides, and that extra increases as the angle increases. So, the standard base x perp. height formula isn't enough on its own, because the base can't be calculated without the angle of the parallelogram slope.

TheBody

bnt wrote: »

If it's like that, then the slope of the parallelogram needs to be known too - it affects the length of the base.

If you draw this with a parallelogram with the angles at 90° - i.e. a square - then the base = height = 2r, so the area is 10x10=100.

If you look at the picture, however, base > 2r. You can see this more clearly if you draw a horizontal line through the centre of the circle: the length of the line is 2r plus some extra on the sides, and that extra increases as the angle increases. So, the standard base x perp. height formula isn't enough on its own, because the base can't be calculated without the angle of the parallelogram slope.

Yes, good point; you would need the angle too as you suggested. Well spotted.

Pherekydes

Maybe the problem is worded incorrectly?

If a circle of radius r bounds a parallelogram, find the area of the parallelogram.

Something like that.

bnt

It is possible to show that the parallelogram must be a rhombus (equal sides) - see here for example. But that doesn't change the problem.

If it's the other way around, with the parallelogram inscribed inside the circle, that's different. The only way you get all four vertices of the parallelogram on the circle is if the parallelogram is a rectangle (see here).

However, the area of that rectangle is not the same in all cases, and it depends on its sides. (You can have a valid rectangle inscribed in a circle with a width of zero, and therefore an area of zero!)