500 million points for anyone who can solve this maths problem

[RlCKY]

I drew it in AutoCAD...

Area = 13.09915, Perimeter = 13.16775


Without CAD... Figure out the radius of the arc...2.31745

Circumscribe the circle about a square (draw square inside of circle)...figure out the area of the square in the circle and divide by four to give you the area of the 4 arc sections...

Multiply that by 2 for your area of the arcs, and then add the area of the rectangular part of the object.

[da_beast]

You can't use that ellipse method because you don't know that the segments are half of an ellipse. For that to be true the gradient at the top needs to be zero.

Btw what is that test on?

I though the ellipse one was out tbh. It's on mathematics for com*****g. It's a past paper. I managed all other 12 questions with relative easy apart from this one. It's beyond me lol

[da_beast]

I drew it in AutoCAD...

Area = 13.09915, Perimeter = 13.16775

I've never used autocad before. Is it reliable for this sort of thing?

[RlCKY]

I've never used autocad before. Is it reliable for this sort of thing?

If I drew the part right.

[RlCKY]

The image looks like ****, but there ya go.

[fighter1234]

That looks right^. The question did say that it was only an approximation because we assumed that the two segments form a perfect circle around the rectangle. 12.85 is reasonably close to 13.1, so the segment method is the way to go.

[RlCKY]

That looks right^. The question did say that it was only an approximation because we assumed that the two segments form a perfect circle around the rectangle. 12.85 is reasonably close to 13.1, so the segment method is the way to go.

You were close but the center point of the arc isn't directly in the center of the rectangular part. The radius is slightly larger because of that.

[Sane Man]

I tried to read through this thread, got two pages in and then then my head exploded.

I thought I was pretty good at Math, but damn, you some smart motherfuckers.

[Citizen Koba]

I've never used autocad before. Is it reliable for this sort of thing?

No - you need to provide a mathematical solution - and having given myself a quick refresher in basic trig I realised that it's quite soluable.

(sorry, the ancient laptop I'm using overheated last pm while I was trying to post - at which point I realised it was like 2pm UK time - and I had to be up early)

You're given the dimensions to make 2 sides of a triangle describing half of the inside of each curved section 3.55 /2 = 1.775 by 0.7975. This gives you your opp and adj side so inv. tan to get the angle. The radius line can be found by subtracting this angle from 90 and then a second triangle made using this angle (90 - theta) and the original adj 1.775. use the cosine to find a value for the hyponteneuse - this is the radius of your circle - it's also worth finding the angle formed at the centre of the circle at this point ( easily done once you have (90- theta) and a right angle. By doubling this angle you have the angular arc described and the radius from which you can easily calculate the area of the arc (pi r squared/ (angle/360)) subtract from this the triangular area covered using 0.5xba***height and you have the area of one of your curved sections. multiply by 2 to include the other and then add in the area of the sqare (or rectangular middle section and you're done).

Sorry - couldn't provide diagrams or even pause for breath, but I'm worried this piece of junk might overheat again. I'm kinda busy this pm but get back if you need a more detailed work through an I'll see what I can do.

K

[da_beast]

No - you need to provide a mathematical solution - and having given myself a quick refresher in basic trig I realised that it's quite soluable.

(sorry, the ancient laptop I'm using overheated last pm while I was trying to post - at which point I realised it was like 2pm UK time - and I had to be up early)

You're given the dimensions to make 2 sides of a triangle describing half of the inside of each curved section 3.55 /2 = 1.775 by 0.7975. This gives you your opp and adj side so inv. tan to get the angle. The radius line can be found by subtracting this angle from 90 and then a second triangle made using this angle (90 - theta) and the original adj 1.775. use the cosine to find a value for the hyponteneuse - this is the radius of your circle - it's also worth finding the angle formed at the centre of the circle at this point ( easily done once you have (90- theta) and a right angle. By doubling this angle you have the angular arc described and the radius from which you can easily calculate the area of the arc (pi r squared/ (angle/360)) subtract from this the triangular area covered using 0.5xba***height and you have the area of one of your curved sections. multiply by 2 to include the other and then add in the area of the sqare (or rectangular middle section and you're done).

Sorry - couldn't provide diagrams or even pause for breath, but I'm worried this piece of junk might overheat again. I'm kinda busy this pm but get back if you need a more detailed work through an I'll see what I can do.

K

That looks correct. However where did you get the figure for the adjacent side? Should it not be 0.8275? Instead of 0.7975?

4.195 - 2.54 = 1.655. That is the depth of the 2 side segments. Half that and it gives us the adjacent side (0.8275?). Half of 3.55 gives us the opposite side