Here's a request. I searched your extensive videos for cylinder cut by a plane at an angle creating an oval. Trying to characterize the oval created with interest in a chord across said oval. There is a real world interest on my part. I was attempting a home use of a sweep shape to the ends of rafters to extend to the jack rafters and hip rafter then turn 90 degrees short distance back to the wall. Small project on my part and not really needed but I wanted to jazz it up.
@@prbmax I don't know what the structure is that you're describing, and what it exactly is that you're asking for. But the intersection of a cylinder and a plane (at an angle with the cylinder's axis) is not an oval, but an ellipse. An ellipse has two symmetry axes: both principal axes are an axis of symmetry. An oval has only one symmetry axis; only one of its principal axes (usually the longer one) is an axis of symmetry. If the cylinder has diameter D , and the cutting angle between cylinder and plane is 45°, then the length of the short axis of the intersection ellipse is D , and the length of the long axis of the ellipse is D√2 . The circumference of the cylinder is πD = 2πR, where R = D/2 is the radius of the cylinder. Suppose the cylinder is then held vertical (with the cut side upwards), and we draw a horizontal line on the cylinder's mantle near the bottom, around the circumference, with markings t starting from t=0 right under the top/highest point of the ellipse. Then the marking at t = πR (= half the circumference) will be right under the bottom/lowest point of the ellipse, and the marking for t = 2πR (= full circumference) coincides with t = 0 (= starting point). The height of the outline (edge) of the ellipse is then characterized by the formula h(t) = h₀ + R*cos(t/R) where h₀ is the height of the center of the ellipse. In other words, the outline of the ellipse along the circumference of the cylinder is just a scaled (co)sine-function (equally scaled in horizontal and vertical direction), wrapped around the cylinder. If the plane cuts the cylinder at a different angle θ (with θ=0 meaning perpendicular; the normal vector of the plane coincides with the cylinder's axis), then the short axis of the ellipse remains D, the long axis of the ellipse is D/cos(θ) , and the formula for the outline height becomes h(t) = h₀ + R*tan(θ)*cos(t/R) which also has the shape of a (co)sine-function, but a stretched/compressed one (vertically stretched when θ>45° , vertically compressed when 0°
@@yurenchu Thanks. Is there a naming difference in the shape of a cylinder cut by a plane at an angle and a cone cut by a plane at an angle? I guess I always thought the cone shape cut was an ellipse and could be created with a string and two nails and the cylinder cut at angle was called an oval.
@@prbmax You're welcome. And thanks for your reply. There's no difference, they are both an ellipse. Which is just a stretched circle (where the stretch factor in one direction (say, horizontal) is different from the stretch factor in the perpendicular direction (say, vertical)). In the plane, the equation of the ellipse is (x/a)² + (y/b)² = 1 where a>0 and b>0 are half the lengths of the principal axes (in other words, the principal axes of the ellipse have lengths 2a and 2b); a and b can also be seen as respectively the horizontal stretch factor and the vertical stretch factor. And yes, the ellipse can also be created by fixing two pins at its foci, and tracing a path using a string loop around the foci. However, I'm not very familiar with the relation between the parameter set (c, d) , where 2c is the distance between the two pins and d is the length of the string loop (and here, e = 2c/(d - 2c) is defined as the ellipse's _eccentricity_ ), and the parameter set (a,b) as the "stretch factors"/"principal axes' half-lengths" in the planar model formula above. Anyway, I hope this helps.
@@prbmax Oh, one way to look at the naming is that an ellipse is a special type of oval (just as a circle is a special type of ellipse, or a rectangle is a special type of trapezium). So _technically_ these cross-sections are also ovals.
You can also notice that area of top rectangle 3 times less than area of compound rectangle below it. And since both rectangles share same width, bottom rectangle has 3 time height of top rectangle: 2*3=6. So side of big square is 2+6=8.
It is not assuming. Orange is 1, and yellow, green and pink are 3, all statet the same size in the question. Thus orange is 1/4 of the hight, because all 4 is orange wide.
You don't even need the yellow, green and pink rectangles. Take x = orange width and a = square width. Orange is 1/5 of the square, so the total square a² is therefore 5 * 2x = 10x. Blue is 1/5 of the square and a(a - x) or a² - ax. So you end up with 2x = a² - ax, in which you can replace a² with 10x and then you can divide by x, so the orange width is irrelevant. The final equation is 2 = 10 - a |+a -2 a = 8
@@Mrtheunnameable it's not an assumption as they have the same width and it is a given fact thst they all have the same are, given that the area of a rectangle is witdth time height we know that the height has to be 3 times that of the top rectangle. Therefore 3x2=6. And 6+2=8 qed
Thats too complicated bruh. This is my way. Let side length of the whole square be *x* and length of orange be *y* . Therefore; 3( area of orange ) = ( total area of yellow green and pink ) 3(2y) = (x-2)y 6y = (x-2)y 6 = x-2 x = 8 So, Area of the large square = 8.8 = 64 sq units.
When algebra complicates everything. Green, pink and yellow have a combined area of 3 times the orange one, and same width, so height of yellow must be 3 times 2 and the square has sides of 2+3*2=8.
Exactly what I figured out. I'd argue this is an example of someone being so good at technical knowledge, they circle back to being impractical. Alas, it was fun watching him make a big deal of a rather simple problem, and a bit interesting too
@@Sam_on_RU-vid I might have found myself there too, had I had pen and paper in hand. Instead, I just kept looking for possible relations between the proportions, until it clicked.
@@dojelnotmyrealname4018 Using the formula for the area of a rectangle. We have R1, whose area is Base × 2 We have R2, whose area is Base × X, (with X = L - 2, L being the side of the overall square) sharing a base with R1, and being three times its area, since it contains in its area an area equal to three R1 rectangles. For this to be congruent, X must be equal to 3(2) = 6 Therefore, the total length of the sides of the square must be L = 2 + X = 2 + 3(2) = 8. Is this argument insufficient proof?
To those saying "notice how big" the lines are, you arent answering the question. Firstly, these types of examples dont give measurements on purpose, and often clearly state the picture is not proportionally accurate, that you must only use information available to you. Its like being asked to check someones temperature but you just stick your fingernin their mouth and say "warm". Secondly, nobody asked for a guess, they asked you to use data and logic to find an answer. Stop looking for an easy way out, its blocking you from understanding the assignment.
This was so easy. You first blocked off 1/5 of the total area, leaving 4/5 blank. Then next you block off 1/4 of the remaining area(1/4 of 4/5 = 1/5, so the area is still the same as the first piece), and that's as far as we need to go. Yellow, pink, and green are entirely unnecessary to think about. Since that means you'd be able to lay 4 of those pieces side by side in the remaining area after blocking off the first piece, 2 x 4 = 8. That's one side. And the whole thing is a square, so 8 x 8 = 64. ETA: additional context for clarity.
when you had 8x = x^2 there was no reason to move x, just cancel the x and note that x = 0 is a solution, and that leaves 8 = x. Much faster that way, and you can still discout x = 0 as a solution.
I did it such a long way: I set up some equations. The left length is partially given as 2. Let the remaining length be z. So the left side of the square is (2+z) The area is (2+z)^2 let the top left rectangle have side lengths 2 and x. This rectangle has area 2x. Let the middle rectangle horizontal length be b. Notice all the rectangles have same area. This means the height of the middle rectangle and the height of the middle bottom rectangle are the same. Let this height be h. Since we already have z and h, we can compare them. that is, z = 2h. (**) If the whole length (horizontal) of the top left rectangle is x, and the length of the 2 middle rectangles is b, the horizontal width of the bottom left rectangle is x - b. Area of bottom left rectangle = Area of middle rectangle z(x-b) = bh 2h(x-b) = bh (from **) 2(x-b) = b (since h can't be 0) 2x - 2b = b 3b = 2x b = 2x/3 This leaves the width of the bottom left rectangle to be x/3 (since x - b = x - 2x/3 = x/3) Area of bottom left rectangle = area of top left rectangle zx/3 = 2x z/3 = 2 (since x can't be 0) z = 6 z + 2 = 8 (z + 2)^2 = 64 therefore the area of the square is 64 square units.
This one is extremely easy ... you only need orange and blue rectangles ... square side x area x² area of each rect x²/5 Orange : 2a = x²/5 => a = x²/10 Blue : x(x-a) = x²/5 => a = 4x/5 And so x²=8x, x=8 Area 64
After realizing each rectangle is just x²/5 you can solve this with the blue one and the orange one, since the blue has x on one side, the other side will be x/5, while for the same reasoning on the orange one we have x²/10 on one side. Now we have x = x²/10+x/5, or 5x=x²/2+x, so 4x=x²/2, 8x=x², x=8. So the area is 64
The way I did this is as follows: 1: Side length is s 2: The middle two rectangles share their length 3: They must, therefore, share their height as well, as their areas are equal 4: If their heights are equal, and they add up to the left rectangle's height, that must mean they each have half its height 5: To maintain equal areas, their length must be double the left rectangle's length 6: The top rectangle's length is o 7: If the lengths of the bottom rectangles are at a 2 to 1 ratio, and add to create o, the left rectangle's length must be a third of o 8: The left rectangle's height is s-2, as we already know the top rectangle takes 2 away from the side length 9: The left rectangle has the area (s-2)(o/3), and the top rectangle has area 2o 10: The equal areas mean that these quantities wre equal, so (s-2)(o/3) = 2o 11: Simplify, giving you os/3 - 2o/3 = 2o 12: Add the 2o/3 to the other side, giving you os/3 = 8o/3 13: Multiply both sides by 3, giving you os = 8o 14: Divide both sides by o, giving you s = 8. This is a valid step because o is a side length, and if a side length like o is 0, that kinda breaks the whole problem. 15: Square s to give you the total area, which is 64.
That's what I did. Just a lot of tedious algebra and logic. I like the problems when I can use algebra and I don't have to remember identities, geometric oddities (e.g., area of a cone), or trigonometric functions.
A=area of 1 small rectangle X=sides of big square Y=top side of blue rectangle 1) Yellow+green+pink = 3A = (x-2)(x-y) 2) Orange = A = 2 (x-y) Divide 1) by 2) 3 = (x-2)/2 so x=8 0:00
Using first letter of the color lowercase = horizontal-axis uppercase = vertical-axis G+P=Y gG+pP = 2yY = 2y(G+P) = 2Gy+2Py g = p = 2y o = y + g = y + p = 3y 2o = Yy Therefore: Y = 6 and the Area = (2+Y)(2+Y) = 64
My solution: The yellow, green and pink rectangle form a greater rectangle with 3 times the area of a single rectangle. This combined triangle has the same width as the orange rectangle but 3 times the area, so height as to be 3*2 = 6. The whole square has a side length of 6+2 = 8. So the size of the rectangle is 8² = 64.
I took a different approach: 1) I looked at the green and pink rectangles. They had the same width, and since they also had the same area, their height should've also been the same. 2) Then I looked at the yellow rectangle. It had the same height as green + pink, but half of their area. So A (the area of the rectangles) = (x - 2) * y, where y is the width of the yellow shape. 3) Then looking at the orange rectangle, we can also say that A = 2 * 3y, or A = 6y 4) From step (2), we can say that y = A / (x - 2) 5) We can replace y in step (3) with its equivalent from step (4): A = 6A / (x - 2) => 1 = 6 / (x - 2) => x - 2 = 6 => x = 8 6) The area of the square is x^2, so 8^2 = 64
You can actually extend the question further by determining the dimension of every colored rectangle using the fact that the side of the big square is 8 and all 5 rectangles' areas are equal and 1/5 of the area of the big square, which is 64 sq. units. So each of the 5 rectangles has an area of 12.8 sq. units. Solving, you get: Orange: w = 6.4, h = 2 Yellow: w = 32/15, h = 6 Green and Purple: w = 64/15, h = 3 Blue: w = 1.6, h = 8
The only step i did different was find the width of the side rectangle and subtract that from x to get the width of the top rectangle but i cant beleive i got that one right
You did an amazing job! That was fun to watch, especially after trying to work it out in my head. What’s up with these people who think that if a length or area looks like a certain portion of another, that they can simply assume it is -- *on a math question* ?
For the rectangle on the right we know one side is x, so the other one must be x/5. And for the rectangle with the given side of 2, the other side must be x²/10. So now we have x²/10 + x/5 = x. Multiply by 10, rearrange, and solve. x² - 8x = 0. One solution (x=0) we can disregard, the other one (x=8) is the valid one. Area is therfore 64 square units.
That’s one way to do it. The way I did it was that I realized the only way to design this problem would be by making the first rectangle divide the square in fifths like you said. But then take that a step further and see the second rectangle divides the remaining rectangle by 4. So x would have to be 2 times 4, which is 8.
Let x = width of yellow rect. Thus width of green = 2x Thus width of orange = 3x. Thus area of orange = 6x Thus area of green = 6x Thus height of green = 6x / 2x = 3 Thus height of whole square = 2 + 3 + 3 = 8 Thus area = 64
The area of the top rectangle is x^2/5 divide that by 2 for its other side in terms of X. Divide the 4x^2/5 by x^2/10 to cancel out X and get sidelength as 8. 8^2 is 64.
For this mathematical problem, dividing by x is fine, but generally you would avoid it because you lose potential answers. In this case you would lose x=0 as an answer. Since you are calculating a surface which you know sides of the object are bigger than 0 you can drop x=0 as an answer, or faster divide with x as a shortcut.
solution: 5 rectangols have same area so 5A is the surface of the square, so if square has side equal to l => 5A = l². So it's sufficient that we find l or A to resolve the problem. we notice that: - orange rectangle has h = 2 and b = 1/2 A, from that we can recover that - blue rectangle has b = l - 1/2A and h = l We can use relation of blue square to find A: l(l-1/2A) = A l² - 1/2Al = A => A(2+l) = 2l² => A = 2l²/(2+l) now, since we know A: l² = 5A => l² =10l²/(2+l) =>(since l≠0) 1 = 10/(2+l) => l = 8 => 5A = l² = 64
My way of solving: the orange rectangle's area is 1/4 of the total area of orange pink yellow and green, since they each have equal areas. Because the big rectangle (orange yellow green pink) has the same width as the orange rectangle but 4 times the area, than its length has to be 4 times bigger than orange's length: 4*2 is 8, and the total area is just 8²=64.
my way : take ractangle (yellow, green, pink) name the yellow side x and the other side y no x*y/3 = 2y x/3 = 2 x = 6 the side of the big square is (2+x) so 2+6 = 8 then 8^2 = 64 unit square
Okay, here is what I have. Assign x² as the area for each section. The orange is 2×a=x² so a=x²/2 The yellow, green, pink area is 3x² with one side x²/2, meaning the remaining side is: 3x²=x²/2 × b 6x²=x² × b 6=b 2+6= 8 Since the shape was defined as a square, 8×8=64
Here's my attempt; which is... vastly different, much longer and more complicated XD Let 'X' be the the full length of the right side. As such, X can be defined by the following unknown variables For the top side, X = Y + Z For the left side, X = 2 + A For the lower side, X = B + C + Z Also given that all rectangles are equal in volume, then X * Z = 2 * Y = A * B = (1/2 * A) * C [Blue, Orange, Yellow, and Green/Pink respectively] Now work backwards with algebraic logic; If A * B = (1/2 * A) * C, then C = 2 * B Then if X = Y + Z, and X = B + C + Z, then Y = B + C, then Y = B + 2 * B, therefore Y = 3 * B. Then, if 2 * Y = A * B, then 2 * (3 * B) = A * B, then 6 * B = A * B, therefore A = 6 Finally, since X = 2 + A, then X = 8. X ^ 2 gives the final answer of 64.
My attempt: Let x equal the side length of the square, α equal the area of each rectangle, and A equal the area of the square. By definition, we know that A=x^2. And by construction, we can see that A=5α, and that one side of the blue rectangle is x. Let y equal the other side of the blue rectangle. α=xy A/5=xy (x^2)/5=xy. Assuming that x, as a measurement, is greater than 0, we can divide both sides by x. x/5=y Let o equal the width of the orange rectangle. By definition, we know that α=2o. By construction, we can see that o=x-y. o=x-(x/5) o=(4/5)x. α/2=(4/5)x α=(8/5)x 5α=8x A=8x (x^2)=8x Again, if x is greater than 0, we can divide both sides. x=8 A=(8^2)=64 Wow, that took longer to type than it did for me to do mentally. But I didn't want to miss any rigor, so...
My solution, before seeimg video: ->Let Unknown side of Orange be x -> Therefore area of orange be 2x -> Since all rectangles be equal, total area be 10x ->Let Long side of yellow be y ->Since all Rectangles be equal area, removing blue rectangle leaves us with 4/5 of Square total area -> Therefore (y+2)×x is 4/5 of total area ->We know total area is 10x => (y+2)×x = (⅘×10)x => y+2 = 8 => y = 6 -> As Y is long side of yellow, and 2 is short side of yellow, Together they make one side of a square, which is 8. Hence Are of square is 64. QED
Set each rectangle's area =a Also set the side length of square = x By the orange one, we can get the other side length of the orange = a/2 When we look at the left side of the square without the blue one, we can note the area of the left : x(a/2) Also, we knew that the left area contains 4 pieces of rectangles with the same area=a Then we can get the following: x(a/2) = 4a Now we know x is 8, so the area of the square is 64.
A bal felső téglalap alatti 3 téglalap területe 3-szorosa neki, ez pedig csak úgy lehet, ha a bal oldala a 2 háromszorosa, ami 6. 2+6=8. Megvan a négyzet oldala. 8*8=64. So: Left side of square: 2 + 3*2=8. 8*8=64.
Here is my reasoning: The orange rectangle is as wide as the rectangle formed by the green, pink and yellow rectangles. But the 3 color rectangle has trice the area therefore it is trice as long. Since the length of the blue rectangle is 2 the length of the 3 color rectangle is 6. So the length of one side of the big square is 8 and 8^2 = 64.
I took a slightly different route: x^2 = area of square *(SOLVE FOR THIS)* (x^2)/5 = Area of each quadrilateral y = short length of rightmost quadrilateral x = side of square & long side of rightmost quadrilateral xy = (x^2/5) y = x/5 (divide both sides by x)(can rule out x=0 since this is a geometry puzzle) x-y = Long length of upper-left quadrilateral with side length of 2 2(x-y) = (x^2)/5 2(x-(x/5)) = (x^2)/5 2(4x/5) = (x^2)/5 8x/5 = (x^2)/5 8x = x^2 x = 8 (divide both sides by x) x^2 = 64 *(SOLUTION)*
ATTEMPT: Green and pink have the same area. They have the same length too, so they must also have the same height. Let's call the length L and the height H. The area is thus LH. The yellow rectangle has the same area as the green and pink ones. Its height is the height of green and pink combined, so 2H. Its length must therefore be L/2. Orange has the same width as yellow + green, so L/2 + L or 3L/2. Its area is the same as all the other rectangles as before, so it should have area LH. But we know one of the side lengths of orange: its height is 2. So if 2 * 3L/2 = LH, H = 3. The length of the square is the height of orange + green + pink. That's 2 + 2H or 2 + 2 * 3 = 8. The area of the square is 64.
I did it like that but it's easier to just combine green pink and yellow into a rectangle you know is 3 times the area of orange with the same width. It must therefore have a height of 6 so the height of the square is 8.
Solution: x = side of total square a = orange width orange = 2a blue = (x - a)x = x² - ax because all 5 rectangles have the same area, we can say that x² = 5 * 2a = 10a since orange = blue and x² = 10a, we can say that 2a = 10a - ax |:a 2 = 10 - x |+x -2 x = 8 so x² = 64, which is the area of the total square.
My attempt: I used multiple variables to solve each rectangles Orange: (x+y)*2 Yellow: x*a Green & pink: 0.5a*y Blue: c*(a+2) Area of a square: z Starting with green and pink, since they share the same length, they must share the same width. Fusing green, pink, and yellow together, I got (x+y)*a=3z. Seeing how (x+y)*2=z and (x+y)*a=3z, I solved that a=6. I then focused on the fact that x*6=z and, comparing green and pink, that y*6=2z. Seeing this, I concluded that 2x=y, and replaced all y values with 2x. Because 6*x=z, and 3*2x=z, I solved 6*x=3*2x, eventually discovering that x=2, and because y=2x, I knew that y=4. With every variables, I solved each and found that z=12, for each ended up equaling to 12: Orange: (2+4)*2=12 Yellow: 2*6=12 Green and pink: 0.5(6)*4=12 Knowing the area is 12^2, I find that c=1.5, since 1.5*(6+2)=12. The entire square is 5z, so I did 5(12) and found 60 as the final answer of the area of this one big square with 5 smaller squares.
Solution to the problem from the thumbnail (writing this before watching the video) Every rectangle has he same area. Green and pink both share their sides on the y axis with yellow and blue, and since they're all rectangles we can say they have equal sides, and can call the side on the x axis is x, and on the y axis is y. With that knowledge we can tell the side on the y axis of the yellow is equal to 2y, so for it to have an equal area it's x axis side needs to be equal to 0.5x. The unknown side of orange is equal to 0.5x + x = 1.5x, telling us the area of the orange is 3x (and since the areas are equal, 3x=xy, therefore y=3) Knowing y=3, the blue's y axis side must be equal to 3+3+2=8, and since the blue square shares that side with the entirety of the side of the square, the area of the square is 8*8=64.
There is much faster & easier way! The rectangle with the 2 dimension & the 3 rectangles below it form a larger rectangle. The area of the 3 rectangles below have 3 times the area of the top one. Since both have the same width, the height is proportional to the area. So the height of the lower 3 is 2 x 3 = 6. This makes the square's side = 8 & the area = 64. Who needs all those equations when a simple proportion will do?
my solution: let the other side of the orange rectangle equal y, the side of the square x. since all rectangles are equal, it follows that (1) x^2 = 10y (that is, five times the area of the orange one equals the area of the square) and (2) 2y = x(x - y) (the area of the orange one equals the blue one, whose one side is x, and the other side is x - y) expand the right hand side of (2) then rearrange to get (2) x^2 = 2y +xy from combining (1) and (2) follows that 10y = 2y + xy take away 2y, then divide by y on both sides to get x = 8 and thus the area of the square is A = 8^2 = 64 q.e.d.
Answer (from the thumbnail): 64 Calculation: height × width of each rectangle area: green = a × b pink = c × b = a × b ==> c = a yellow = 2a × d = a×b ==> d = b/2 orange = 2 × (b/2 + b) = 2×(1.5b) = a×b ==> a = 3 blue = (2+2a)×e = (2+2*3)×e = 8×e ==> height of blue is 8 square area = (blue height)² = 8² = *64* By the way, blue = 8×e = a×b = 3×b ==> e = 3b/8 . Width of square = (b/2 + b + 3b/8) = (15/8)b = 8 ==> b = 64/15 a×b = 3×(64/15) = 64/5
The way I did this: Green and Pink have to be equivalent, so we can use their height and width as base reference points h and w. Yellow is the combined height of green and yellow, so has a height of 2h, and to balance that, must have a width of w/2. Orange is the combined width of p/g and yellow, so has a width of 3w/2, and therefore must have a height of 2h/3. 2h+2h/3 6h/3+2h/3=8h/3 The height of Orange is defined to be 2. 2h/3=2 2h=6 h=3 8(3)/3=8 8×8=64 Bonus: Blue has the combined height of orange and yellow, 8h/3, the height of the square, and therefore must have a width of 3w/8 3w/2+3w/8 12w/8+3w/8=15w/8 15w/8=8 15w=64 w=64/15 w=4.26666 repeating 4.2666 rep ×3=12.8 5×12.8=64
Similar reasoning here, noticed that the rectangle on the right was area x^2/5 so the right part of the top edge must be x^2/5x = x/5, then to the left of that is x - x/5 = 4x/5. Etc.
call the middle 2 rectangles bottom amd top sides y, and their left and right sides x. This will make the side length of the entire square 2 + 2x and the area of each rectangle xy. It will also make the top left rectangle have a top and bottom of xy/2 which means the bottom left rectangle has a top and bottom of xy/2 - y with the other 2 sides being 2x. So (xy - y)2x = xy x - 2 = 1 x - 3 Area of entire square = (2 + 2x)^2 = (2 + 2(6))^2 = 8^2 = 64.
If we put the length of the side of the square is s and the width of the blue rectangle is w The area of blue rectangle =sw, and the area of orange rectangle =2×(s-w) =2(s-w), and the area of square =s²=5sw=10(s-w) 5sw=10(s-w) 5sw=10s-10w 5sw-10s-10w=0 10w+5sw-10s=0 5w(2+s)-10s-20=-20 5w(2+s)-10(s+2)=-20 (5w-10)(2+s)=-20 s>0》2+s>0, so 5w-102 we can't take number (4). From the equation s²=5sw s²-5sw=0 s(s-5w)=0 s-5w=0》s=5w So we can take only ..(2)when w=8/5, s=8 Area=s²=8²=64
2 месяца назад
Or, the easier way: The combined area of the green (G) and pink (P) is the twice of the yellow (Y) area. I call the Y's width (w) and its height (h). Because the GP area's height equal to the Y's (h), therfore their width is the twice of the Y (2w). It means the orange (O) area width is w + 2w = 3w. The O area is 2 * 3w = 6w. And because every area is equal, every area is 6w. I know the Y area's width (w) and the area (6w). So, the equation is wh = 6w => h = 6. And because the Y's h is 6, the big square side is 6+2, and its area is 8*8 = 64.
let's call orange long side "a" then orange area is "2a" if all the rectangles have the same area then orange + yellow + green + pink = 4 * 2a = 8a callinge "x" the side of the square, the big rectangle (so everything except blue) has an area of ax since we took the same area in both cases then ax=8a cancel the a, x=8
I first noticed that the central and the bottom rectangle have the same length and since they have the same area they therefore also must have the same height. The left rectangle has twice their height. If I call the height of the central rectangle y and its length 2x its area is 2x*y or 2xy. The bottom rectangle is identical to it. Since the left rectangle has twice their height (2y) and its area is equal to the other rectangles (2xy) its length must be x, half of the length of the central and bottom rectangles. The height of the top rectangle is 2 and since its area is 2xy as well, its length is xy. But at the same time its length is the sum of the lengths of the left rectangle (x) and central rectangle (2x), which is 3x. This means that the area of the top rectangle can be described as 2*xy or 2*3x, so 2*xy = 2*3x. If we divide by 2x we get y=3. Since the height of the left rectangle is 2y and the height of the top rectangle is 2 we can calculate the side of the entire square as 2y+2 or 2*3+2, which is 8. Thus the area of the square is 8² = 64.
I am wondering your thoughts on collatz conjecture. Recently I have tried to develop a formula for it, but could not go to the depth of it. All I know about it is, it cannot be a power of 2, it cannot make a loop in a way where numbers come odd-even-odd-even... in turn, or it cannot make a loop as this way: odd-even-even-even... (ax+b)/c is the best I could get, where a,b,c are all positive integers. But we can equal this to x since it is a loop and we get: b/(c-a) = x. (and here b>c>a>1) We know that "a" must be a power of 3 and it is always odd, "c" must be a power 2 and it is always even. "b" is also always odd but I could not make a formula for finding "b". What I know about "b" is it must not be a prime number.
Of course I didn't notice the obvious fact that the area of the 3 rectangles below the orange one together is 3*that, I started with my target area being X, and figured that the other side of the orange one must therefore be X/10. The long side of the blue one has to be sqrt(X), and the short side - sqrt(X)/5. So the long side of orange + short side of blue together must be sqrt(X) too, and solved that, for sqrt(X) = 8 so X=64
your approach for finding the 0.8X distance was not too straightforward. Might as well computed the side of the right rectangle: a.X=X2/5 so a=X/5. Now the top left one is X-X/5=4X/5
Without watching. Here's how I did it. The rectangle on the right has to be a ratio of 1:5 so the ratio of the rectangle with a value has to be (and I'm sorry for purists) 4:1.25 So how many 1.25's go into 2 2÷1.25=1.6 Now I skip all ahead as all original measurements need to be multiplied by 1.6 and originally my side was 5... 1.6*5=8 8*8=64
And here I was ending up with a system of equations that solves for the side lengths, because I didn't pay close enough attention that it was a square and not a rectangle.
Once you figured x=8, i just left and came back with an area of 64 (it was at the end, though, so it was most of the legwork lol) Edit: i got the area of the rectangles as 12.8 on the way because i didnt think very efficient lol
A Redditor friend of mine sent me this one in email--it's cute. The original diagram fools you a little by not being entirely to scale (which is actually helpful at keeping you from making unwarranted assumptions). I figured it out without (explicitly) doing any algebra, by just thinking about the dimensions of the three rectangles at the lower left. The two vertically stacked ones must together have twice the area of the tall one, so the tall one must be one third of the width of that whole block. That means it's 1/3 the width of the rectangle at upper left, so if it has the same area, it must be 3 times the height. So its height is 6, and everything else follows.
That's WAY too much math. Extend the line between green and pink through yellow. Extended green and extended pink are now both 50% larger than they were since they each added half of yellow. Since they have the same width as orange, they must be 50% taller. 50% more than 2 is 3. 2+3+3=8. 8^2=64. Use geometry not algebra.
That seems more complicated to me. You know the blue one has width x/5, and the orange one has width x^2 /10 x /5 + x^2 /10 = x 2x +x^2 = 10x x^2 - 8x = 0 x = (0 or) 8 Area = 64 No need to extend any rectangles or work with ratios
@@vortexlegend101 i don't know those things. Also, geometry questions are all about drawing new lines. That's how you're supposed to solve them. and uhh... 3:2 isn't exactly the craziest ratio to have to work with.
It’s fine Define length of orange as a Define height of green as b Define volume of orange as x You’d have { 2a=x, ab=3x/2 } Solve for b to get 3 (2+2b)^2=A 8^2=A 64=A
@@zachansen8293it's math, there isn't a "correct" way to do anything. It's just preference, and the algebra done here is minimal and pretty easy to derive.
Is there anything in the actual post that says the colored bits have to be rectangles? Is it actually 100% required assuming all the lines are in fact straight lines but not necessarily all 90 degrees? The outer area is specified to be a square. But could pink and green just be parallelograms and not rectangles?
For this question it’s ok, but if it’s not a geometry question, then you would also need the x=0 solution. So, if you divided by x, then you would lose the x=0 solution. Generally, dividing by x is not a good habit to get into
@@erenerbay yes, you don't need it for this question. but in some different questions, you might need it. you said that you could divide both sides by x, but it wouldn't work for those questions which asked for the extra x=0 solution. basically, it's just not a good habit to get into
wait wait wait, instead of doing all that, can't you do it like this? all rectangles have equal areas, so if i combine green pink and yellow, it'll be 3x the size of orange, right? we have oranges height and since orange and the combined shape have same width, the height is 3x the one of orange so 3 x 2 is 6. i'd add 6 and 2 i get 8 and 8 square is 64, so 64 is my answer
Easiest way: the top orange rectangle has an area 1/4 of the rectangle formed by removing the blue rectangle from the square. This means the orange rectangle's height of 2 is a quarter that of the preciously described rectangle (and that of the square, since they have the same height).
Yes, but you lose 1 potential result. If he didn't do what he did, he wouldn't get x=0 as an option, and 8*0 = 0^2 => 0 = 0. For this mathematical problem your approach is fine since you've seen the second answer is x=0
Too complicated way to solve this. You can just notice that in order to fill the area taken by yellow, green and pink, you'd need 3 orange rectangles stacked vertically. Which with the other orange would give height of 8. And because the whole thing is a square, the whole area is then 8*8
@@anoopa3334 When I say "vertically", I mean "on top of each other". Like a hamburger. And when they they are stacked like that, it becomes obvious that you just add their heights of 2 for total of 8.
Andy math actually solved this problem like months ago I think, he had a different approach but the shape of the square is still the same ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-bX-f0vZFYjo.html
But Andy's version had a given length of 3. Mindyourdecisions did it 3 years ago with a given length of 4. ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-lrND8AeL23s.html
Sort-of wrong. A simple solution is 4 rectangles are 2x2.5, and one rectangle is 1x5. Total area of square is 25. Each rectangle has 5 area, and it's not difficult to figure out how they all fit together in the square. The question gives no formal information on how the rectangles are arranged into the square, so the arrangement shown in the picture can safely be ignored - it is only there as a generic example for the problem. Having said that, I actually think there are an infinite number of solutions to the problem - X could take any value. This is just a simple solution I came up in my head without doing any paper math. E.g. one could double the sides of all the rectangles in my solution, make X=10, and come up with a similar solution where all areas are 4 times bigger.
...I thought of another way that's even simpler: Transform the problem into a different problem with the same answer. Concentrate on the three blocks at lower left. They form a rectangle with three times the area of one of these equal blocks. The answer would clearly be the same if we divided up that region into thirds in a different way. So divide it into three horizontal strips of equal height, stacked on top of each other. Then, since they have the same width and area, they're obviously congruent to the top rectangle, so they all have height 2 and the side of the square is 8, area 64.
