time stamps: 1st way 0:43 2nd way 2:17 3rd way 4:15 4th way 7:09 5th way 9:16 I didn't use Heron's formula because all the sides were irrational so it wouldn't be considered "easy". However, here's the proof of the Heron's formula for you guys: th-cam.com/video/9qbpmYqr4so/w-d-xo.html
I thought you would mention the formula that the area of a triangle with vertices (a,b) (c,d) and (e,f) is 1/2 | a(d-f)+c(f-b)+e(b-d) |. I can also give the proof if you want,and if I'm unable to respond,then you can search it up or work it out yourself by constructing trapeziums by drawing lines parallel to the y-axis(pretty simple proof,especially for you).And as always-Nice video!
Wow, this is what I thought of on the spot: 1. Use Pythagoras' theorem to calculate the lengths of the sides. 2. Use the Cosine rule to calculate the angles. 3. Flip the triangle so 1 line is horizontal. 4. Draw a vertical line from the highest point of the triangle to the base. 5. Use the Sine rule to get the length of that vertical line. 6. Base * Height / 2. I have a very convoluted mind...
What I did: calculate distance from (2,1) and (7,5) Distance = sqrt((7-2)^2 + (5-1)^2) = sqrt(41) {this will be the base of our triangle} find the equation of the line connecting (2,1) and (7,5) y - 1 = m(x-2) find the gradient of the line m m = (5-1)/(7-2) = 4/5 y - 1 = 4(x-2)/5 => 5y - 5 = 4(x-2) => 5y - 5 = 4x-8 => 4x - 5y - 3 = 0 Use the perpendicular distance from a line to a point formula: Perpendicular distance formula: |ax1 + by1 + c|/sqrt(a^2+b^2) where (x1,y1) = (3,7), a = 4, b = -5, c = -3 Let the line connecting (2,1) and (7,5) be the base. => Perpendicular height = |4(3) -5(7) -3|/sqrt(4^2 + (-5)^2) = 26/sqrt(41) => Area of triangle = bh/2 = (26/sqrt(41)) * (sqrt(41)/2) units^2 = 13 u^2 Got 13 in the end.
Hardik Bhatia that really isn't an /r/iamverysmart lol. James seems to have forgotten that he'll need to compute the third integral too, but the principle is perfectly reasonable. You can find all of the line equations trivially, and integrating 3 linear equations is about as easy as it gets. Hell, given the diagram you could literally just look at the image and work out the values of the integrals yourself.
People say you're a showoff, but really, if you like integrals and you're currently working with them a lot it might be the first thought This question was also set up a bit like a shaded area problem where it might be a lot more difficult to get the answer in any other way, and integrals get you to the solution right away
I liked Pick's theorem the best. Because it generalizes to more complex polygons, as long as the vertices have integer coordinates (on lattice points).
I went straight for Pick's Theorem, because not enough people know it and it's so cool. I was mostly expecting it to be a sixth easy way, and was pleasantly surprised it was on your list. Plus it's fun when someone asks you the area of some complicated shape to just count the dots and get the right answer.
The shoelace method is great. It can be easily implemented in a spreadsheet to calculate the area of any simple polygon. And similar formulas can be derived to calculate other geometric properties of the polygon, like the centroid and second moment of area.
When you solve for the area by integration and learn you got it right, but then realize you could've solved it in less than 1 minute: *_"I've won, but at what cost?"_*
This video has taught me a lot. Thanks to you, I know many more ways of finding the area of a polygon. Here's how I worked it out: We first need to find the equations corresponding to each segment. To do so, since we are given 2 points that correspond to each equation, we may use the slope formula (RISE÷RUN works just as well). Lastly, to find the constant term, we may plug the points given in each of the equations. Let the function corresponding to the top left side be f(x), the top right one, g(x), and the bottom one h(x). Then, to find the area of the triangle, we will use integrals. Focus on the x-coordinates, which will give you the boundaries of the integrals. The area under the bottom segment (between its boundaries 2-7) subtracted from the area under the 2 top segments (between their boundaries 2-3;3-7) will give us the final area of the triangle. Meaning that the total area will be the integral from 2 to 3 of f(x)dx + the integral from 3 to 7 of g(x) - the integral from 2 to 7 of h(x)dx.
A less elegant and more a brute force approach would be using Heron's formula. Let s be equal to (a+b+c)/2, where a, b and c are the side lengths, then the area of this triangle is: A = sqrt(s*(s-a)*(s-b)*(s-c)) Then find the side lengths with distance formula: a = sqrt(4^2+5^2)=sqrt(41) b = sqrt(4^2+2^2)=sqrt(20) c = sqrt((-1)^2+(-6)^2)=sqrt(37) so s ≈ 8,4790... Pluggin all in(I'm not writing s out, as i have to use the exact value) gives us: sqrt(s*(s-sqrt(41)*(s-sqrt(20)*(s-sqrt(37)) Which is equal to 13
I extended the line that intersects (3,7) and (7,5) until the line reaches the point where y is 1. So now I have a 4th point with coordinates (15, 1) which lies on that extended line. From there I found the area of the big triangle (I’ll call it triangle A) with the points (2, 1), (3, 7) and (15, 1). Base is 15-2 = 13 and height is 7 - 1 which is 6. 13*6/2 = 39. Then I found the area of another triangle (I’ll call it triangle b) with points (2, 1), (7, 5) and (15, 1) and I know that if I do triangle A - Triangle B, I’d get the answer. So for triangle B, Base is 13 as well and height is 5 - 1 which is 4. So area is 13*4/2 = 26 Then I do 39 - 26 = 13 (answer).
1. Find the slope of line with endpoints (2,1) and (7,5) 2. The slope of that line is 4/5. Therefore, the slope of the line perpendicular to it is -5/4. 3. Find the equation of the line segment with end points (2,1) and (7,5) and the perpendicular line to it running through (3,7). 4. Find the point of intersection of the two equations. 5. Find the distance between (2,1) and (7,5) and the distance between (3,7) and the point of intersection. 6. Let line segment with ends points (2,7) and (7,5) be the base and line segment with end points (3,7) and the point of intersection be the height. 7. (Base x Height)/2 = 13 8. The area of the triangle is 13units^2
I found another way. Draw a line at y=5 to split the triangle in two. This will create two new traingles, and each will have the same base. (The bottom triangle will obviously be upside down.) The top triangle will have a height of 2, and the bottom's height is 4. (1/2)bh will give areas of b and 2b respectively. Their sum yields 3b (which is the area of the original triangle. Using the slope of the line between (2,1) and (3,7), m=6; therefore, the side of the triangle on the left crosses the line y=5 at (2.666,5). So the base of our triangles (b) is 4.333. The sum from above was 3b = 3*4.333 = 13.
Considering the length of each side is The square root of a whole number, there are at least three versions of Heron’s formula involving the squares of each side of the triangle, with no need to compute the semi perimeter. Those versions of the formula work like a charm.
I thought you would mention the formula that the area of a triangle with vertices (a,b) (c,d) and (e,f) is 1/2 | a(d-f)+c(f-b)+e(b-d) |. I can also give the proof if you want,and if I'm unable to respond,then you can search it up or work it out yourself by constructing trapeziums by drawing lines parallel to the y-axis(pretty simple proof,especially for you).And as always-Nice video!
Hi, I have another simple solution using Heron's formula. P=sqrt[s*(s-a)*(s-b)*(s-c)] Where "s" is (a+b+c)/2 (half circumference) and a,b,c are the length of the sides of a triangle. Best regards.
line (3,7)(7,5) crosses horizontal line through (2,1) at (15,1) so the base is 13. The wanted area is the big triangle whose top is at (3,7) minus the smaller whose top is at (7,5). 13(6-4)/2=13
Here's how I would do this (paused at the start): 5×6=30=area of rectangle 5×4/2=10=area of lower right corner 4×2/2=4=area of upper right corner 6×1/2=3=area of upper left corner 30-10-4-3=13=area of triangle
I liked the inside the box and the picks theorem But I actually calculated it by the herons formula. That is √s(s-a)(s-b)(s-c) s is half of the perimeter and a,b,c are the lengths of the three sides of the triangle.
We apply a translation of to the triangle to move it to (0,0), (1,6) and (5,4) and then rotate the triangle to (0,0), (x1, y1) and (x2, 0) and then the area = 1/2 * x2 * y1. The triangle translation is from (2,1) to (0,0) and from (3,7) to (1,6) and from (7,5) to (5,4). Then we can calculate the cosine and sine of the angle between the x-axis and the point (5,4), which given cos(theta) = 5/sqrt(41) and sin(theta) = 4/sqrt(41) Now we build the matrix to rotates clockwise by theta, and then we can then find x1, y1 and x2 and calculate the area. So the matrix that rotates clockwise by theta degrees is |cos(theta) sin(theta)| = | 5/sqrt(41) 4/sqrt(41)| |-sin(theta) cos(theta)| |-4/sqrt(41) 5/sqrt(41)| So , let's rotate the point (5,4) clockwise by theta | 5/sqrt(41) 4/sqrt(41)| [5] = [ 41/sqrt(41)] = [x2], so x2 = 41/sqrt(41) = sqrt(41) ... this translates (5,4) onto the x-axis. |-4/sqrt(41) 5/sqrt(41)| [4] [ 0 ] [y2] So , let's rotate the point (1,6) clockwise by theta | 5/sqrt(41) 4/sqrt(41)| [1] = [ 29/sqrt(41)] = [x1], so y1 = 26/sqrt(41) |-4/sqrt(41) 5/sqrt(41)| [6] [ 26/sqrt(41 ] [y1] Now area = 1/2 * x2 * y1 = 1/2 * sqrt(41) * 26/sqrt(41) = 13.
Yes. I like Pick’s theorem a lot too (it also works for any simple polygon with lattice pts) but it requires me to draw a perfect picture with all the dots.
I dropped lines from the points to the x-axis and then calculated the areas of the right-trapeziums made by each side (easy, just width * average height). Add two of them and subtract the third: 1*(7+1)/2 + 4*(7+5)/2 - 5*(1+5)/2 = 24 + 4 - 15 = 13
this is how I did it 1. I found the equation for the line from point A(2,1) to B(7,5) which is y=⅘x-⅗ 2. find the perpendicular line that falls down on line AB from point C.(call the point of running into each other point L) we know that the slope for the perpendicular line is the negative reciprocal of the slope of the AB line ==> m(CL)=-5/4 CL eq==> y= -5/4x+b we know the line passes through point C(3,7): 7=-5/4*3+b b=10¾ ==>CL eq : -5/4x+10¾ 3. we need to know where it runs into line AB -5/4x+10¾=⅘x-⅗ 11.35=2.05x x≈5.537==>L(5.537,3.829) 4. using the distance formula we get that CL≈4.06 5. distance formula again but on line AB we get AB≈6.4 6. Area ABC= ½*4.06*6.4=12.996≈13 if I were to type all the numbers after the decimal it would have said 13. ty for following along :)
Shoelace method can be deduced by calculating the positive oriented closed loop line integral of the vector field (P(x,y), Q(x,y)) = 1/2 * (-y, x) over the edges of the triangle. Since dQ/dx - dP/dy = 1, it follows by Green's Theorem that this integral is equal to the area of the triangle. This result can also be extended for any simple polygon. I didn't know this method and, coincidentally, I stumbled upon this problem a couple days ago. Math is so surprising!
Thank you so much sir. This showed me not one, not two but FIVE ways of finding area of a triangle with the given vertices. This is really cleared all my doubts regarding this. Thank you sooo much sir!!!
Pick's theorem provides a formula for the area of a simple polygon with integer vertex coordinates, in terms of the number of integer points within it and on its boundary. The result was first described by Georg Alexander Pick in 1899.[1] It was popularized in English by Hugo Steinhaus in the 1950 edition of his book Mathematical Snapshots.[2][3] It has multiple proofs, and can be generalized to formulas for certain kinds of non-simple polygons.
An alternative is to use the cross product of two vectors, which was my first thought. You can also use the dot product of two vectors to find the cosine of one angle. With that you can find the height of the triangle, which gives the area as 1/2 * base * height. Another method is to find the linear function of each line segment and use two integrals to find the difference in area between them. Although it's more of a brute force method, it covers other shapes as well. Another calculus method is Green's Theorem.
On our programming courses we use trapezoids to find the area of a polygon. If you have a polygon ABC...Z you find points A', B', C', ..., Z' which are projections of A, B, C, ..., Z on Ox axis. Then you find signed areas of trapezoids ABB'A', BCC'B', CDD'C', ..., YZZ'Y', ZAA'Z' and sum them up. Then you calculate the absolute value of the sum. This algorithm works for any polygon even for concave ones.
U can find the two vectors and do a cross product then find its norm and divide it by 2 because cross as value =ab sin theta and b sin thate actually is the height and divide it by 2 give us the area. I know it’s not a very efficient way but it’s an extra one u asked for it.
Blackpenredpen please make a video saying why the 3rd and 4th method yield the same numerical values while computing before even getting the answer please😫🙏🙏💓
The 4th way is often used in map software for calculating the area of building, lakes, etc (for example, openstreet map, etc). However this method doesn't work for figures where any of its line intersects another line. as well as if there are holes in the figure or isolated islands
Funny that you showed 5 ways, and I solved it correctly using a completely different way. I'd never heard of any of these methods(except the first one, but that didn't occur to me) solving for the side lengths and the height involves more number crunching but I enjoyed it.
There's also a straightforward computation using complex numbers. You find the three complex numbers for which A, B and C are the affixes and call them z1, z2 and z3. Denote by Zi the complex conjugate of zi. Then, the area is given by half of the imaginary part of the sum z1Z2+z2Z3+z3Z1
6th and 7th way Calculate the point of the altitude from any vertex to the base using straight lines equations Then use 1/2*base*height 7th way Use heron's formula By calculating the semiperimeter of the triangle and then use √s(s-a)(s-b)(s-c)
I solved by making (2,1) co-ordinate to (0,0) [ By adding (-2,-1) to all the co-ordinates ] and then used the shoelace theorem....As one point was (0,0) so it was easy to multiply. But the first two methods of the video were really good and worth using.
You can also solve this problem by doing (1/2)IIABII⋅distance between C and AB. IIABII is simply sqrt(4^2+(-2)^2) and the distance between C and AB is IICA projected on n(from AB)II
Thinking with vectors, you have points A, B and C given, which can be used to calculate c, and want to find the height h. The problem can be reduced to finding the minimum distance between a point (here C) and a line (let's call it L). That is, you define a point P which is found on L, defined as OA + t * AB, and take the absolute value of OC - OP. You'll get a radical with a square function. Said radical will always be positive so the minimum of the radical is the same as the square root of the square function's minimum so you don't have to perform square root derivatives to get the desired result. The result will be the height which you can plug in for (c * h)/2. Slightly more complicated than the above solutions (due to radicals and calculus) but hey, it also works. Edit: The minimum distance has to be the height since any vector between a point on the line and C is ultimately constructed by the height vector and a remainding length. If said length is 0 (minimum distance), all you have left is an orthagonal vector.
Or you count the number of unit squares covered (even partially) by the triangle (22) then you add the number of unit squares fully covered by the triangle (5) and you do the mean: (22+5)/2 = 13.5 ≈ 13. It's about the same process that is used for integral approximations with a left sum and a right sum counting rectangles under the curve.
Nice video. I’m really interested to know how why Pick’s Theorem works: could you do something like a proof/intuition of this formula? My method of solving this problem involved mainly some introductory vector geometry. If we translate the shape to move the vertex (2,1) to the origin, we have a triangle spanned by two vectors (1,6) and (5,4). Take the base of the triangle to be the vector (5,4) and project (1,6) onto the base. Then: the area of the triangle works out to be 1/2 * length of (5,4) * length of [(1,6) - projection of (1,6) onto (5,4)] = 1/2 * sqrt(41) * sqrt(27716)/41 = 1/2 * sqrt(676) = 1/2 * 26 = 13 as required.
I did exactly what you said not to do at the beginning and got 12.9989 I found the length of the longest side with Pythagorean theorem and got sqrt(41), then found the slope of that side to be 0.8 and found the y-intercept of the line it lies on to be -0.6. From there, I found the formula of the line with the negative reciprocal of that slope (perpendicular to that side, y=-1.25x + 10.75) that passes through the point (3, 7), then set it equal to the line the bottom side lies on to find their intersection (5.53658 all repeating, 3.82926 all repeating) then found the distance between that point and (3, 7) with Pythagoras again (4.0601940101). This is where the rounding errors start to set in, but I assumed since my answer was so close to 13 that it was probably equal to 13.
I prefer to use determinant to calculate area Double of this area can be useful for convex hull Fun fact Determinant of 2x2 matrix gives us area of parallelogram Determinant of 3x3 matrix gives us volume of parallelepiped Definite integral gives us area under a curve Double integral gives us volume
i think there are some formulas i know that can help with this problem 1st way: find distance between the two points (2,1) (7,5) using Pythagorean which is squareroot41. Then find the equation of the line that passes thru (2,1) (7,5) which is 4x-5y-3=0, now here's the formula: distance from any point to a line=(|Ax+By+C|)/square root of(A^2+B^2) where x and y are the coordinates of the desired point. after substituting the distance turns out to be 26/squareroot41, then we do the distanceXbaseX1/2 which is 13. 2nd way: area of a triangle= squareroot of[s(s-a)(s-b)(s-c)] where S is the semi-perimeter s=(a+b+c)/2, we use the coordinates to find out the 3 lengths between the 3 points and substitute.
Heron's formula actually worked out nicely here since it reduces to 1/4 * sqrt[(b+c+a)(b+c-a)(a+b-c)(a-b+c)], and two differences of squares, the second being 1/4 * sqrt[(16+4 * sqrt 185) * (-16 + 4 * sqrt 185)] = 1/4 * sqrt[16 * (4 + sqrt 185) * (-4 + sqrt 185)] = 1/4 * 4 sqrt(185 - 16) = sqrt(169) = 13
I liked Pick's method. Other methods are 1. Heron's formula using semi-perimeter 2. Finding coordinates of one altitude, it's length and length of the side 3. Finding angle of triangle using dot/cross product, then finding projection (altitude)
you could choose a side and make a linear function to represent it then just find the perpendicular bisector connecting it to the opposite point finding where they intersect and then you can easily just find the lengths using Pythagorean theorem to find base length and height and simply do b*h/2
The only methods that I know, are making a bunch of calculations with trigonometric functions, or simply using Heron's formula, so I went for the second option. So, for the sides, the top one will be a, the right one, b, and left, c. a = 2*sqrt(5), b=sqrt(41), c=sqrt(37). S=sqrt(5)+sqrt(41)/2+sqrt(37)/2. sqrt(S(S-a)(S-b)(S-c)). I'm too lazy to write all that out find what the area is so I used desmos and got a nice answer of 13.
You can also use tangents to find the angle between two lines, and then use sine rule to find the area. I'm guessing you won't like it cos it involves finding length using pythagoras
Area of trapezium/trapezoid is (separation of // sides) × (mean length of // sides). Drop verticals to axis from each vertex to the axis. Add the areas of the trap. below the upper lines, subtract the area of the trap below the lower line. =13. You can also draw horizontals to the y axis and do the same sideways. =13 again :)
Love it! Some of them were surprises; some were vaguely familiar but forgotten (my memory's not what it was - at least I don't think so). Being a Bear of Little Brain, my first thought was the "rectangle minus 3 triangles" one, which reminds me of one of the proofs of the theorem of Pythagoras. If I were an engineer (I'm not, any more than I'm a mathematician), I might have got some paper of known weight (the units are a sort of planar density, typically g/m²); picked¹ a scale; drawn and cut out the triangle; weighed it; and (finally) the area would be the weight of the triangle divided by the paper's weight/area suitably scaled back down (or up, I suppose). The error bars would be huge, but you could mitigate that by weighing a lot of them and dividing by how many you weighed. ¹ Not that Pick, but I was intrigued to see that method. I vaguely think I've seen something like it before but had forgotten it.
I calculated the length of every side with Pythagoras, used the law of cosines to find an angle then did 1/2 a b sin C. This process involved literally drawing a box around the triangle. Yet somehow I was still mad when "it's just a box" was revealed.
Haven't learnt Matrices nor Pick's Theorem, I solved it using only with vectors. I used this function to find one of the angles: cos(x) = (u * v) / (|u| * |v|) And from there just used A = (d1 * d2 * sin(x)) / 2
why does the third way work? the scalar triple product gives us the volume of a parallelotop, but this parallelotop does not have a hight of 1. And it´s base is not the triangle (parallelogram) from the beginning. Sorry for my bad english.
6th method - use Protagoras and calculate each side length. then by Heron formula, calculate the area. :) I saw your comment, but it puts square roots in the mix and makes method extra difficult...
time stamps:
1st way 0:43
2nd way 2:17
3rd way 4:15
4th way 7:09
5th way 9:16
I didn't use Heron's formula because all the sides were irrational so it wouldn't be considered "easy". However, here's the proof of the Heron's formula for you guys: th-cam.com/video/9qbpmYqr4so/w-d-xo.html
You really boss
You make me a great lover of the mighty math
I call math as mighty math
6th way: Heron's formula (en.m.wikipedia.org/wiki/Heron%27s_formula )
Shoelace formula is eloquent and a simple form of Heron’s formula
I thought you would mention the formula that the area of a triangle with vertices (a,b) (c,d) and (e,f) is 1/2 | a(d-f)+c(f-b)+e(b-d) |. I can also give the proof if you want,and if I'm unable to respond,then you can search it up or work it out yourself by constructing trapeziums by drawing lines parallel to the y-axis(pretty simple proof,especially for you).And as always-Nice video!
Is there any theorem u have not proved
The virgin Shoelace Theorem vs. The Chad Heron's Formula
And then the Stacy Pick's theorem
The A S C E N D E D Integral[2-3](6x-11)dx + Integral[3-7](-(1/2)x+(17/2))dx - Integral[2-7]((4/5)x-(3/5))dx
Goes to youtube to take a break from studying.
*accidentaly learns even more*
Sufyan Khan loll nice!!!
lol
I came to this video to get a way of solving a problem.
I liked Pick's method the most because I've never heard of it before!
How does picks method work if the area of a triangle isn't a rational number?
@@jakebrowning2373 If the points are on the lattice, then the triangle always have rational area.
@@thedoublehelix5661 oh I didn't realize that was one of the constraints of the problem, thanks
@@thedoublehelix5661 lattice means points right?
Wow, this is what I thought of on the spot:
1. Use Pythagoras' theorem to calculate the lengths of the sides.
2. Use the Cosine rule to calculate the angles.
3. Flip the triangle so 1 line is horizontal.
4. Draw a vertical line from the highest point of the triangle to the base.
5. Use the Sine rule to get the length of that vertical line.
6. Base * Height / 2.
I have a very convoluted mind...
I thought of the shoelace on the spot but i couldn't remember how it worked so i did the 1st way
With the sides you could just use heron
Once you have the angles you can just do 1/2 ab sinC
You can also use the semiperimeter formula A=√((s)(s-a)(s-b)(s-c)) Where s=Perimeter/2 and a, b and c are the distances between the points
legends you integration for this
What I did:
calculate distance from (2,1) and (7,5)
Distance = sqrt((7-2)^2 + (5-1)^2) = sqrt(41) {this will be the base of our triangle}
find the equation of the line connecting (2,1) and (7,5)
y - 1 = m(x-2)
find the gradient of the line m
m = (5-1)/(7-2) = 4/5
y - 1 = 4(x-2)/5
=> 5y - 5 = 4(x-2)
=> 5y - 5 = 4x-8
=> 4x - 5y - 3 = 0
Use the perpendicular distance from a line to a point formula:
Perpendicular distance formula: |ax1 + by1 + c|/sqrt(a^2+b^2) where (x1,y1) = (3,7), a = 4, b = -5, c = -3
Let the line connecting (2,1) and (7,5) be the base.
=> Perpendicular height = |4(3) -5(7) -3|/sqrt(4^2 + (-5)^2) = 26/sqrt(41)
=> Area of triangle = bh/2 = (26/sqrt(41)) * (sqrt(41)/2) units^2 = 13 u^2
Got 13 in the end.
Very nice!!!
I also thought the exact same way but never calculated😂
I thought the same thing right about until calculating the height, I gave up
Lol
Finally a mathematician with a sense of humor. I love this channel. Keep it up
My first thought was to convert the lines into functions and do two integrals
wtf
Hardik Bhatia that really isn't an /r/iamverysmart lol. James seems to have forgotten that he'll need to compute the third integral too, but the principle is perfectly reasonable. You can find all of the line equations trivially, and integrating 3 linear equations is about as easy as it gets. Hell, given the diagram you could literally just look at the image and work out the values of the integrals yourself.
@@generic8891 3 equations, but 2 integral only, one for X=2 to X=3 and X=3 to X=7
@@generic8891 it is an "/r/iamverysmart tho cause this method is just completely inefficient and the comment's only purpose is to try to sound smart.
People say you're a showoff, but really, if you like integrals and you're currently working with them a lot it might be the first thought
This question was also set up a bit like a shaded area problem where it might be a lot more difficult to get the answer in any other way, and integrals get you to the solution right away
As soon as he started to draw the box in the first method I was like: "oh lol how did i not think of this?
blackpointerredpointer is back :o
AthyXray GD long time no see!
indeed :)
I liked Pick's theorem the best. Because it generalizes to more complex polygons, as long as the vertices have integer coordinates (on lattice points).
I went straight for Pick's Theorem, because not enough people know it and it's so cool. I was mostly expecting it to be a sixth easy way, and was pleasantly surprised it was on your list. Plus it's fun when someone asks you the area of some complicated shape to just count the dots and get the right answer.
3:11 para-LORY-gram :o
3:13 area of a paralorigram? yea man those paralorigrams are poppin' up everywhere these days
The shoelace method is great. It can be easily implemented in a spreadsheet to calculate the area of any simple polygon. And similar formulas can be derived to calculate other geometric properties of the polygon, like the centroid and second moment of area.
Btw, the music in the background is called “The Entertainer.” I forgot the composer but the title should be correct.
Yes! I forgot the composer too. I got it from the YT audio library
blackpenredpen *Scott Joplin.
If you’re wondering, that is Scott Joplin.
Scott Joplin
@@einsteingonzalez4336 Thanks for the information!
When you solve for the area by integration and learn you got it right, but then realize you could've solved it in less than 1 minute:
*_"I've won, but at what cost?"_*
This video has taught me a lot. Thanks to you, I know many more ways of finding the area of a polygon.
Here's how I worked it out:
We first need to find the equations corresponding to each segment. To do so, since we are given 2 points that correspond to each equation, we may use the slope formula (RISE÷RUN works just as well). Lastly, to find the constant term, we may plug the points given in each of the equations. Let the function corresponding to the top left side be f(x), the top right one, g(x), and the bottom one h(x).
Then, to find the area of the triangle, we will use integrals. Focus on the x-coordinates, which will give you the boundaries of the integrals. The area under the bottom segment (between its boundaries 2-7) subtracted from the area under the 2 top segments (between their boundaries 2-3;3-7) will give us the final area of the triangle. Meaning that the total area will be the integral from 2 to 3 of f(x)dx + the integral from 3 to 7 of g(x) - the integral from 2 to 7 of h(x)dx.
This entire video I was screaming “USE PICK’S THEOREM!!!”
Nice!!! And I did!!
blackpenredpen
Saved the best until last 😂
Since I’m currently taking linear algebra, the determinant ones were great
A less elegant and more a brute force approach would be using Heron's formula.
Let s be equal to (a+b+c)/2, where a, b and c are the side lengths, then the area of this triangle is:
A = sqrt(s*(s-a)*(s-b)*(s-c))
Then find the side lengths with distance formula:
a = sqrt(4^2+5^2)=sqrt(41)
b = sqrt(4^2+2^2)=sqrt(20)
c = sqrt((-1)^2+(-6)^2)=sqrt(37)
so s ≈ 8,4790...
Pluggin all in(I'm not writing s out, as i have to use the exact value) gives us:
sqrt(s*(s-sqrt(41)*(s-sqrt(20)*(s-sqrt(37))
Which is equal to 13
I really liked the 2x2 determinant method for how intuitive it is when visualizing and I like Pick's theorem for originality!
I extended the line that intersects (3,7) and (7,5) until the line reaches the point where y is 1. So now I have a 4th point with coordinates (15, 1) which lies on that extended line. From there I found the area of the big triangle (I’ll call it triangle A) with the points (2, 1), (3, 7) and (15, 1). Base is 15-2 = 13 and height is 7 - 1 which is 6. 13*6/2 = 39.
Then I found the area of another triangle (I’ll call it triangle b) with points (2, 1), (7, 5) and (15, 1) and I know that if I do triangle A - Triangle B, I’d get the answer. So for triangle B, Base is 13 as well and height is 5 - 1 which is 4. So area is 13*4/2 = 26
Then I do 39 - 26 = 13 (answer).
1. Find the slope of line with endpoints (2,1) and (7,5)
2. The slope of that line is 4/5. Therefore, the slope of the line perpendicular to it is -5/4.
3. Find the equation of the line segment with end points (2,1) and (7,5) and the perpendicular line to it running through (3,7).
4. Find the point of intersection of the two equations.
5. Find the distance between (2,1) and (7,5) and the distance between (3,7) and the point of intersection.
6. Let line segment with ends points (2,7) and (7,5) be the base and line segment with end points (3,7) and the point of intersection be the height.
7. (Base x Height)/2 = 13
8. The area of the triangle is 13units^2
I found another way.
Draw a line at y=5 to split the triangle in two. This will create two new traingles, and each will have the same base. (The bottom triangle will obviously be upside down.) The top triangle will have a height of 2, and the bottom's height is 4. (1/2)bh will give areas of b and 2b respectively. Their sum yields 3b (which is the area of the original triangle. Using the slope of the line between (2,1) and (3,7), m=6; therefore, the side of the triangle on the left crosses the line y=5 at (2.666,5). So the base of our triangles (b) is 4.333. The sum from above was 3b = 3*4.333 = 13.
Considering the length of each side is The square root of a whole number, there are at least three versions of Heron’s formula involving the squares of each side of the triangle, with no need to compute the semi perimeter. Those versions of the formula work like a charm.
I thought you would mention the formula that the area of a triangle with vertices (a,b) (c,d) and (e,f) is 1/2 | a(d-f)+c(f-b)+e(b-d) |. I can also give the proof if you want,and if I'm unable to respond,then you can search it up or work it out yourself by constructing trapeziums by drawing lines parallel to the y-axis(pretty simple proof,especially for you).And as always-Nice video!
4:02 really helped me with that counterclockwise thing , ty
Hi, I have another simple solution using Heron's formula.
P=sqrt[s*(s-a)*(s-b)*(s-c)]
Where "s" is (a+b+c)/2 (half circumference) and a,b,c are the length of the sides of a triangle. Best regards.
Gonna do much algebra to find a, b and c, though...
The way you say “tell you” makes me dance inside 😊 Along with your great explanations of course ^_^
How about Heron’s formula?:
A = √(s(s-a)(s-b)(s-c))
where s = ½(a+b+c)
Though it takes more time…
line (3,7)(7,5) crosses horizontal line through (2,1) at (15,1) so the base is 13. The wanted area is the big triangle whose top is at (3,7) minus the smaller whose top is at (7,5). 13(6-4)/2=13
Integrate the line cyclically!
A = int(Y_AB dx, A → B) + int(Y_BC dx, B → C) + int(Y_CA dx, C → A)
Take modulus for the unsigned area
I just wanted to say, you are the only reason I am passing math right now. I love the work you do, and what you are literally a public hero.
I love 4th way (Shoelace way), I love it!! Your videos are always interesting!🔝
Glad you like them! Thank you!
The only method I knew was to form three equations using two-point slope form, and then integrate. Nice video mate, very informative!
Here's how I would do this (paused at the start):
5×6=30=area of rectangle
5×4/2=10=area of lower right corner
4×2/2=4=area of upper right corner
6×1/2=3=area of upper left corner
30-10-4-3=13=area of triangle
I just did the same.
I liked the inside the box and the picks theorem
But I actually calculated it by the herons formula. That is
√s(s-a)(s-b)(s-c)
s is half of the perimeter and a,b,c are the lengths of the three sides of the triangle.
The last one was the best. I didn't expected something as ambiguous as litteraly counting the points inside. I'm going to find the prof for that!
We apply a translation of to the triangle to move it to (0,0), (1,6) and (5,4) and
then rotate the triangle to (0,0), (x1, y1) and (x2, 0) and
then the area = 1/2 * x2 * y1.
The triangle translation is from (2,1) to (0,0) and from (3,7) to (1,6) and from (7,5) to (5,4).
Then we can calculate the cosine and sine of the angle between the x-axis and the point (5,4),
which given cos(theta) = 5/sqrt(41) and
sin(theta) = 4/sqrt(41)
Now we build the matrix to rotates clockwise by theta, and then we can then find x1, y1 and x2 and calculate the area.
So the matrix that rotates clockwise by theta degrees is
|cos(theta) sin(theta)| = | 5/sqrt(41) 4/sqrt(41)|
|-sin(theta) cos(theta)| |-4/sqrt(41) 5/sqrt(41)|
So , let's rotate the point (5,4) clockwise by theta
| 5/sqrt(41) 4/sqrt(41)| [5] = [ 41/sqrt(41)] = [x2], so x2 = 41/sqrt(41) = sqrt(41) ... this translates (5,4) onto the x-axis.
|-4/sqrt(41) 5/sqrt(41)| [4] [ 0 ] [y2]
So , let's rotate the point (1,6) clockwise by theta
| 5/sqrt(41) 4/sqrt(41)| [1] = [ 29/sqrt(41)] = [x1], so y1 = 26/sqrt(41)
|-4/sqrt(41) 5/sqrt(41)| [6] [ 26/sqrt(41 ] [y1]
Now area = 1/2 * x2 * y1 = 1/2 * sqrt(41) * 26/sqrt(41) = 13.
very interesting and useful, often need to calculate triangles, centroide ,coordinates ...... Will need to go over Picks theorem again. Thankyou
Shoelace is my favorite because it can be used for any polygone.
Yes. I like Pick’s theorem a lot too (it also works for any simple polygon with lattice pts) but it requires me to draw a perfect picture with all the dots.
I dropped lines from the points to the x-axis and then calculated the areas of the right-trapeziums made by each side (easy, just width * average height). Add two of them and subtract the third: 1*(7+1)/2 + 4*(7+5)/2 - 5*(1+5)/2 = 24 + 4 - 15 = 13
Picks method was so elegant; it was surprising that the area could be determined with just 2 inputs. I love math.
What about using Semiperimeter?
teacher: “think outside the box”
bprp: “there is a box”
*perfect solution*
Pick's theorem is such an elegant way to do it; math never disappoints.
I did it by Pick's theorem but I actually proved which points are inside, which are on the sides and which aren't both.
this is how I did it
1. I found the equation for the line from point A(2,1) to B(7,5) which is y=⅘x-⅗
2. find the perpendicular line that falls down on line AB from point C.(call the point of running into each other point L)
we know that the slope for the perpendicular line is the negative reciprocal of the slope of the AB line ==> m(CL)=-5/4
CL eq==> y= -5/4x+b
we know the line passes through point C(3,7):
7=-5/4*3+b
b=10¾
==>CL eq : -5/4x+10¾
3. we need to know where it runs into line AB
-5/4x+10¾=⅘x-⅗
11.35=2.05x
x≈5.537==>L(5.537,3.829)
4. using the distance formula we get that CL≈4.06
5. distance formula again but on line AB we get AB≈6.4
6. Area ABC= ½*4.06*6.4=12.996≈13
if I were to type all the numbers after the decimal it would have said 13.
ty for following along :)
Shoelace method can be deduced by calculating the positive oriented closed loop line integral of the vector field (P(x,y), Q(x,y)) = 1/2 * (-y, x) over the edges of the triangle. Since dQ/dx - dP/dy = 1, it follows by Green's Theorem that this integral is equal to the area of the triangle. This result can also be extended for any simple polygon. I didn't know this method and, coincidentally, I stumbled upon this problem a couple days ago. Math is so surprising!
Thank you so much sir. This showed me not one, not two but FIVE ways of finding area of a triangle with the given vertices. This is really cleared all my doubts regarding this. Thank you sooo much sir!!!
Man, i love this channel
Pick's theorem provides a formula for the area of a simple polygon with integer vertex coordinates, in terms of the number of integer points within it and on its boundary. The result was first described by Georg Alexander Pick in 1899.[1] It was popularized in English by Hugo Steinhaus in the 1950 edition of his book Mathematical Snapshots.[2][3] It has multiple proofs, and can be generalized to formulas for certain kinds of non-simple polygons.
6:52 is it necessary to put them counterclockwise? I don t remember putting them in a certain order
shoelace method is always the best works for any polygon actually ...really helpful
An alternative is to use the cross product of two vectors, which was my first thought. You can also use the dot product of two vectors to find the cosine of one angle. With that you can find the height of the triangle, which gives the area as 1/2 * base * height. Another method is to find the linear function of each line segment and use two integrals to find the difference in area between them. Although it's more of a brute force method, it covers other shapes as well. Another calculus method is Green's Theorem.
On our programming courses we use trapezoids to find the area of a polygon. If you have a polygon ABC...Z you find points A', B', C', ..., Z' which are projections of A, B, C, ..., Z on Ox axis. Then you find signed areas of trapezoids ABB'A', BCC'B', CDD'C', ..., YZZ'Y', ZAA'Z' and sum them up. Then you calculate the absolute value of the sum. This algorithm works for any polygon even for concave ones.
U can find the two vectors and do a cross product then find its norm and divide it by 2 because cross as value =ab sin theta and b sin thate actually is the height and divide it by 2 give us the area. I know it’s not a very efficient way but it’s an extra one u asked for it.
That's basically what he did on the second one as crossproduct relates to matrix determinant.
Thank you for this interesting video! I’m wondering... what software/hardware did you use as you were drawing on the screen?
Blackpenredpen please make a video saying why the 3rd and 4th method yield the same numerical values while computing before even getting the answer please😫🙏🙏💓
The 4th way is often used in map software for calculating the area of building, lakes, etc (for example, openstreet map, etc). However this method doesn't work for figures where any of its line intersects another line. as well as if there are holes in the figure or isolated islands
Pythagorean theorem + Heron's formula works
And with a lot of detour, you can also use (1/2)*b*c*sin(alpha)
The last one is amazing!!!
Funny that you showed 5 ways, and I solved it correctly using a completely different way. I'd never heard of any of these methods(except the first one, but that didn't occur to me) solving for the side lengths and the height involves more number crunching but I enjoyed it.
There's also a straightforward computation using complex numbers. You find the three complex numbers for which A, B and C are the affixes and call them z1, z2 and z3. Denote by Zi the complex conjugate of zi. Then, the area is given by half of the imaginary part of the sum z1Z2+z2Z3+z3Z1
From (2,1) to (3,7) -> f1(x) = 6x - 11
From (3,7) to (7,5) -> f2(x) = (-x + 17
)/2
From (2,1) to (7,5) -> f3(x) = (4x - 3)/5
A = integral(f1) + integral(f2) - integral(f3)
shoelace method better if have time constrian.
6th and 7th way
Calculate the point of the altitude from any vertex to the base using straight lines equations
Then use 1/2*base*height
7th way
Use heron's formula
By calculating the semiperimeter of the triangle and then use √s(s-a)(s-b)(s-c)
Thank you for working on geometry😀
I solved by making (2,1) co-ordinate to (0,0) [ By adding (-2,-1) to all the co-ordinates ] and then used the shoelace theorem....As one point was (0,0) so it was easy to multiply. But the first two methods of the video were really good and worth using.
You can also solve this problem by doing (1/2)IIABII⋅distance between C and AB. IIABII is simply sqrt(4^2+(-2)^2) and the distance between C and AB is IICA projected on n(from AB)II
Distance formula for the three side lengths, law of cosines for the angle, 1/2 * ab * sin C for the area.
This is a great Math video.
Like the first methods. More methods base times height/2 or by integral
Thinking with vectors, you have points A, B and C given, which can be used to calculate c, and want to find the height h. The problem can be reduced to finding the minimum distance between a point (here C) and a line (let's call it L). That is, you define a point P which is found on L, defined as OA + t * AB, and take the absolute value of OC - OP. You'll get a radical with a square function. Said radical will always be positive so the minimum of the radical is the same as the square root of the square function's minimum so you don't have to perform square root derivatives to get the desired result.
The result will be the height which you can plug in for (c * h)/2.
Slightly more complicated than the above solutions (due to radicals and calculus) but hey, it also works.
Edit: The minimum distance has to be the height since any vector between a point on the line and C is ultimately constructed by the height vector and a remainding length. If said length is 0 (minimum distance), all you have left is an orthagonal vector.
Or you count the number of unit squares covered (even partially) by the triangle (22) then you add the number of unit squares fully covered by the triangle (5) and you do the mean: (22+5)/2 = 13.5 ≈ 13. It's about the same process that is used for integral approximations with a left sum and a right sum counting rectangles under the curve.
Wrong
Hey....u could also have used integration..... its a bit lengthy but still a way....
Nice video. I’m really interested to know how why Pick’s Theorem works: could you do something like a proof/intuition of this formula?
My method of solving this problem involved mainly some introductory vector geometry. If we translate the shape to move the vertex (2,1) to the origin, we have a triangle spanned by two vectors (1,6) and (5,4). Take the base of the triangle to be the vector (5,4) and project (1,6) onto the base. Then: the area of the triangle works out to be
1/2 * length of (5,4) * length of [(1,6) - projection of (1,6) onto (5,4)]
= 1/2 * sqrt(41) * sqrt(27716)/41
= 1/2 * sqrt(676)
= 1/2 * 26
= 13 as required.
I did exactly what you said not to do at the beginning and got 12.9989
I found the length of the longest side with Pythagorean theorem and got sqrt(41), then found the slope of that side to be 0.8 and found the y-intercept of the line it lies on to be -0.6.
From there, I found the formula of the line with the negative reciprocal of that slope (perpendicular to that side, y=-1.25x + 10.75) that passes through the point (3, 7), then set it equal to the line the bottom side lies on to find their intersection (5.53658 all repeating, 3.82926 all repeating) then found the distance between that point and (3, 7) with Pythagoras again (4.0601940101). This is where the rounding errors start to set in, but I assumed since my answer was so close to 13 that it was probably equal to 13.
I prefer to use determinant to calculate area
Double of this area can be useful for convex hull
Fun fact
Determinant of 2x2 matrix gives us area of parallelogram
Determinant of 3x3 matrix gives us volume of parallelepiped
Definite integral gives us area under a curve
Double integral gives us volume
i think there are some formulas i know that can help with this problem
1st way: find distance between the two points (2,1) (7,5) using Pythagorean which is squareroot41. Then find the equation of the line that passes thru (2,1) (7,5) which is 4x-5y-3=0, now here's the formula: distance from any point to a line=(|Ax+By+C|)/square root of(A^2+B^2) where x and y are the coordinates of the desired point. after substituting the distance turns out to be 26/squareroot41, then we do the distanceXbaseX1/2 which is 13.
2nd way: area of a triangle= squareroot of[s(s-a)(s-b)(s-c)] where S is the semi-perimeter s=(a+b+c)/2, we use the coordinates to find out the 3 lengths between the 3 points and substitute.
People are so creative, so many different solutions of a single problem!
Heron's formula actually worked out nicely here since it reduces to 1/4 * sqrt[(b+c+a)(b+c-a)(a+b-c)(a-b+c)], and two differences of squares, the second being 1/4 * sqrt[(16+4 * sqrt 185) * (-16 + 4 * sqrt 185)] = 1/4 * sqrt[16 * (4 + sqrt 185) * (-4 + sqrt 185)] = 1/4 * 4 sqrt(185 - 16) = sqrt(169) = 13
I liked Pick's method. Other methods are 1. Heron's formula using semi-perimeter 2. Finding coordinates of one altitude, it's length and length of the side 3. Finding angle of triangle using dot/cross product, then finding projection (altitude)
you could choose a side and make a linear function to represent it then just find the perpendicular bisector connecting it to the opposite point finding where they intersect and then you can easily just find the lengths using Pythagorean theorem to find base length and height and simply do b*h/2
The only methods that I know, are making a bunch of calculations with trigonometric functions, or simply using Heron's formula, so I went for the second option. So, for the sides, the top one will be a, the right one, b, and left, c. a = 2*sqrt(5), b=sqrt(41), c=sqrt(37). S=sqrt(5)+sqrt(41)/2+sqrt(37)/2. sqrt(S(S-a)(S-b)(S-c)). I'm too lazy to write all that out find what the area is so I used desmos and got a nice answer of 13.
You can also divide the triangle into a right-angled triangle and two other triangles
connect 3 vertices to (3,5)
there you split 1 into 3
You can also use tangents to find the angle between two lines, and then use sine rule to find the area. I'm guessing you won't like it cos it involves finding length using pythagoras
Area of trapezium/trapezoid is (separation of // sides) × (mean length of // sides).
Drop verticals to axis from each vertex to the axis. Add the areas of the trap. below the upper lines, subtract the area of the trap below the lower line. =13.
You can also draw horizontals to the y axis and do the same sideways. =13 again :)
Love it! Some of them were surprises; some were vaguely familiar but forgotten (my memory's not what it was - at least I don't think so).
Being a Bear of Little Brain, my first thought was the "rectangle minus 3 triangles" one, which reminds me of one of the proofs of the theorem of Pythagoras.
If I were an engineer (I'm not, any more than I'm a mathematician), I might have got some paper of known weight (the units are a sort of planar density, typically g/m²); picked¹ a scale; drawn and cut out the triangle; weighed it; and (finally) the area would be the weight of the triangle divided by the paper's weight/area suitably scaled back down (or up, I suppose). The error bars would be huge, but you could mitigate that by weighing a lot of them and dividing by how many you weighed.
¹ Not that Pick, but I was intrigued to see that method. I vaguely think I've seen something like it before but had forgotten it.
Which app and which device are you using?
does the pick Pick's theorem work for any shape on a grid
I calculated the length of every side with Pythagoras, used the law of cosines to find an angle then did 1/2 a b sin C. This process involved literally drawing a box around the triangle. Yet somehow I was still mad when "it's just a box" was revealed.
Either use cosine rule or take cross product of the 3d version of two vectors and divide it by two
Haven't learnt Matrices nor Pick's Theorem, I solved it using only with vectors.
I used this function to find one of the angles:
cos(x) = (u * v) / (|u| * |v|)
And from there just used
A = (d1 * d2 * sin(x)) / 2
why does the third way work? the scalar triple product gives us the volume of a parallelotop, but this parallelotop does not have a hight of 1. And it´s base is not the triangle (parallelogram) from the beginning. Sorry for my bad english.
4:10, couldn’t you also do 5 x 4 + 1 x 6?
Your a genius bro. Keep up the good work. This really helped me.
May I know the solutionif the given is not a triangle but any polygon, given only points-not showing in draw. Thanks @bprp
2nd is best for me
6th method - use Protagoras and calculate each side length. then by Heron formula, calculate the area. :)
I saw your comment, but it puts square roots in the mix and makes method extra difficult...