When I took calculus getting the area of an irregular object blew my mind. Its important for engineering/structural shapes though.
EVERYONE should take an advanced math, not that they would actually use the formulas; rather it trains the brain to create new neural connections for analytical/critical thinking.
I've seen arguments that calculus should be introduced earlier, like right after algebra. There's no reason why you need trig or geometry to understand the core ideas of rate of change, cumulative functions, and limits. I think a lot of people never get past the trig/pre-calc barrier and never really learn what higher level maths are like (no equation memorization, much more creativity) and have a completely wrong impression of math.
I think common core attempts to address that failing of math education. I know a lot of people don't like it and I'm not claiming that it's good, but I do think it's good at least that math educators recognize that people not really getting to see the beauty of math unless they make it to calculus is a good thing.
You could teach calculus principles without Trig. You just wont get very far. Ideally teachers early one would be hinting at what the fundamental concepts are of higher level math. Trig is only really required to solve specific problems since they have nice derivatives and identities.
No, it isn't. It's only required for taking derivatives of/integrating trig functions, which you don't care about before trig, and for using the identities to integrate some tricky functions.
trig and geometry are super handy in the trades - building or making things or repairing them is made way easier with those math skillz built on arithmetic that is near instinctual.
calculus is a big shoulder shrug in terms of value in my life, but i viewed it as equivalent to reading the classics. our civilization is built on the stuff.
when i use real world stuff to show the power of math, people love the hell out of it. book learned math is like a cancer that ruins the whole field for most of us normies
To get the corners, calculate the area of a square with the radius of the circle, then subtract the area of a quarter-circle from your square which you're left with the area of the corner. Then that is subtracted from the main shape.
Just think of the two parts cut out as quarter circles. You have a point thats the center of the circle and a point thats at the edge. Using the distance formula for points you can calculate the length that line. Thats the radius of the circle, area of the circle is 2piradius, then you have to divide that by 4 to get the area of the missing part.
Same with the bottom left. You calculate those two quarter circle areas and subtract it from the length times width of the entire rectangle.
Ohhh looking closer at the image they actually give you the corresponding radius length of each circle "R10, and R5" so its actually a little easier. No distance formula needed.
How do you do the top one? D:
Edit: assuming the following (starting top left, clockwise): R4, R10, R6, R5
[area of rectangle] - [area of corner cutouts]
(20 x 30) - {[(8^2)-(pi*4^2)]/4} - [(pi*10^2)/4] - {[(12^2)-(pi*6^2)]/4} - [(pi*5^2)/4]
= 600 - (3.4) - (78.5) - (7.7) - (19.6)
= 490.8
I'm glad I understand this lol
love this idea
i used to paint pixel by pixel when i was a kid on a 386 w software downloaded from a bbs
it was insane, made for summer vacation time
You'd have to be assuming r in all four corners though to work that out.
correct. I updated the comment to show my assumptions.
Yes! I got this too! I'm glad I worked it out (though I got 490.66)
Is it typical to refer to the radius as R[Length]? I've never see the radius referred to this way.
Would you refer the diameters as D6, D20, etc.?
In CAD yea.
In some drafting standards yes, but I was mostly copying the format of the picture.
When I took calculus getting the area of an irregular object blew my mind. Its important for engineering/structural shapes though.
EVERYONE should take an advanced math, not that they would actually use the formulas; rather it trains the brain to create new neural connections for analytical/critical thinking.
I've seen arguments that calculus should be introduced earlier, like right after algebra. There's no reason why you need trig or geometry to understand the core ideas of rate of change, cumulative functions, and limits. I think a lot of people never get past the trig/pre-calc barrier and never really learn what higher level maths are like (no equation memorization, much more creativity) and have a completely wrong impression of math.
I think common core attempts to address that failing of math education. I know a lot of people don't like it and I'm not claiming that it's good, but I do think it's good at least that math educators recognize that people not really getting to see the beauty of math unless they make it to calculus is a good thing.
You could teach calculus principles without Trig. You just wont get very far. Ideally teachers early one would be hinting at what the fundamental concepts are of higher level math. Trig is only really required to solve specific problems since they have nice derivatives and identities.
No, it isn't. It's only required for taking derivatives of/integrating trig functions, which you don't care about before trig, and for using the identities to integrate some tricky functions.
trig and geometry are super handy in the trades - building or making things or repairing them is made way easier with those math skillz built on arithmetic that is near instinctual.
calculus is a big shoulder shrug in terms of value in my life, but i viewed it as equivalent to reading the classics. our civilization is built on the stuff.
when i use real world stuff to show the power of math, people love the hell out of it. book learned math is like a cancer that ruins the whole field for most of us normies
Calculate the surface area of the rectangle. (L x W) [600]
Calculate the area [A] of the top right and bottom left circles. (A = pi x r^2) [78.5] and [314]
Calculate the area of the circles encompassing the rectangle, which appears to be 1/4. [19.63] and [78.5]
Add the quarter circle areas [98.13]
Subreact the area in #4 from the total surface area in #1. [501.87]
The only part I can't get is how to get the rounded edges from the top left and bottom right.
Edit: see my response to the first reply.
top left and bottom right:
(Area of the bounding square - area of the circle) divided by 4
So top left A=50.24
Circumference is 8, making an encompassing square 64, subtract circle and quarter it, giving 2.44.
Repeating same for opposite corner, A=113.04. Square 144, quartered difference is 0.24...
499.19?
bounding square is L x W, or D x D, or 2R x 2R = (2x4) x (2x4) = 64
circle is (Pi x R^2) = 3.141 x (4 x 4) = 50.26
subtract full circle from full square: 64 - 50.26 = 13.74
divide by 4 to get the lone quarter: 13.74 / 4 = 3.43
There appears to be text showing radius? just like for the other two corner cut outs but its too blurry and I can't read them.
I assumed the following (starting top left, clockwise): R4, R10, R6, R5
Looks like 4 and 6
This gives you the area of a weird shape [A]. Note that each corner of the shape could fill the gap created by a rounded corner of the rectangle.
Not sure if my edit went through, but the steps should be numbered 1,2,3,4, not 1,2,4,5.
To get the corners, calculate the area of a square with the radius of the circle, then subtract the area of a quarter-circle from your square which you're left with the area of the corner. Then that is subtracted from the main shape.
You find the area of the rectangle, then subtract the missing area from the four corners.
Rectangle =20x30=600
Top right missing portion is a quarter of a circle with radius 10 =(1/4)×π×10^2=78.5
Bottom left is the same as top right but with radius 5 =(1/4)×π×5^2=19.6
Top left (R4?) is 1/4 of the area of a 8x8 square minus an R4 circle =(1/4)×(8×8-π×4^2)=3.43
Bottom right is the same but with R5? =(1/4)×(10x10-π×5^2)=5.37
Total = 600-78.5-19.6-3.43-5.37 =493.1
Just think of the two parts cut out as quarter circles. You have a point thats the center of the circle and a point thats at the edge. Using the distance formula for points you can calculate the length that line. Thats the radius of the circle, area of the circle is 2piradius, then you have to divide that by 4 to get the area of the missing part.
Same with the bottom left. You calculate those two quarter circle areas and subtract it from the length times width of the entire rectangle.
Ohhh looking closer at the image they actually give you the corresponding radius length of each circle "R10, and R5" so its actually a little easier. No distance formula needed.