This.
The more I look at this shit, the more I think the pink shape is a square.
Also, homosexual OP didn't point out if the area of the shape includes the outline.
Fricking moron.
I made an error in calculations, which falsely allowed me to 'solve' some of the side lengths, it's impossible to get a result without assuming the that the whole shape is a square.
the only geometry that matters is that the middle squares share a side, everything else is not to scale, the top of the green square should be twice as wide as the yellow if we are assuming that all the squares have the same volume.
The rectangles in the middle share a side length, since they have the same area, they must have the same height, call it x. Now the yellow rectangle has a height 2x, meaning it is half as wide as the green and pink ones. This means the orange rectangle has an area 3y, where y is the width of the green and pink ones. But this means 3y = xy, that is x = 3 and the total area is 64, and each rectangle. has an area 12.8.
I mean, it has to be.
That area has the same length as the orange, so since their area is collectively 3 times as big as A, you must necessarily be able to split them into pieces of the same length, and width of 2.
Let's say you morph B, C, and D into rectangles of the same length as A, but with nonequal widths. How could their areas possibly be equal, then?
It should fit 3 of them because there are 3 rectangles with the same area.
If the blue rectangle has a shorter or longer length, then the orange rectangle would necessarily have to also.
And since we are assuming that they all form a square, the scale cannot be argued to be even a pixel off.
just finished writing out a long winded answer where I used algebra to figure out the height of the yellow block but I saw this post and felt like a moron
Fun problem. Here's what I did.
Let x and y be the height and width of the green rectangle.
The purple rectangle has width y and the same area as the green, so its height is x.
The yellow rectangle has height 2x and the same area as the green, so its width is y/2.
The orange rectangle has width 3y/2 and the same area as the green, so its height is 2x/3.
The blue rectangle has height 8x/3 and the same area as the green, so its width is 3y/8.
Solve 2x/3 = 2 to get x = 3.
Since it's a square, 8x/3 = 15y/8, making y = 64/15.
The sides of the square are 8x/3 = 15y/8 = 8.
The area of the square is 64.
It can definitely be simplified, I just wanted to show what I actually did.
One thing we can do with the extra work is rebut the claim the premises are inconsistent.
The yellow, green and pink rectangles combined have an area 3 times that of the orange rectangle. As they have the same width, they must have 3 times the height of the orange rectangle. The yellow rectangle, therefore, has a height of 6. The square has a side length of 8, and an area of 64.
Yeah it's easy when you see you have a rectangle which shares one side with the red rectangle and also has three times the surface area. This means that the other side of that bigger rectangle must be the other side of the red one multiplied by three which gives the area of the whole square. Fun problem.
This is a troll as there is no solution. I walk you through it and why people got 64 but it doesn't work out.
The yellow, green, an purple rectangles can be combined into a single rectangle with three times the area of the orange triangle. Since they share the same length, but one is three times the area of the other, then the height must be three times as long. Hence the height of the yellow rectangle is 2 * 3 = 6. Hence the dimensions of the "square" are 8 x 8 and you get 64. But the blue rectangle does not compute.
Let's call the length of the orange triangle A, and the area of of each rectangle M.
Hence the blue rectangle is (8 - A) * 8 = M
Orange tells us 2 * A = M, or A = M / 2.
So the blue triangle is (8 - M / 2) * 8 = M
Blue is then 64 - 4m = m or m = 64 / 5.
Wait... it works. m = 64/5 and there 5 rectangles with are m, total area is 64 / 5 * 5 = 64. So I guess it is 64.
I got 64
Note, each square is 1/5y^2, so three of the squares is 3/5y^2
Therefore, (y-2)x=3/5 y^2
Since we have 2x=1/5y^2, we can solve with these two equations
it's 64 because the following equations need to be equal : > 2*X = orange area (X the long side) > Y*X/3 = yellow + green + pink area (Y the height) > (2+Y-X)*(2+Y) = blue area
X cannot be zero since the shape wouldn't be a square otherwise
Height of blue is x
Since all rectangles are equal, area of blue is (1/5) x^2
Then width of blue is (1/5)x
Width of orange is then (4/5)x
Area of orange is then (8/5)x
Set areas of orange, blue equal and solve:
They're all rectangles
This.
The more I look at this shit, the more I think the pink shape is a square.
Also, homosexual OP didn't point out if the area of the shape includes the outline.
Fricking moron.
what the frick are you talking about? how are you going to get the area of the outline anyway?
my solution
Not drawn to scale.
yeah I realized my measurements were off because 1024 was the size of the image and not the size of the selection lol. how about now?
What'd you use to draw? MS Paint?
2x+2y=z
all of them are =z
the yellow h is 2 times green and pink's h
it's a lot of work
64
It's 45, each square has an area of 9.
The trick to solving it is that the two rectangles in the middle share a length, so you can solve the lengths in relation to each other.
I made an error in calculations, which falsely allowed me to 'solve' some of the side lengths, it's impossible to get a result without assuming the that the whole shape is a square.
The answer is 64.
dude the statement of the problem says it's a square.
sa = 2(s-a) = (s-2)b = (s-a-b)(s-2)/2
sa = 2s-2a
a = 2s/(s+2)
2s-4s/(s+2) = (s-2)b
b = 2s^2(s-2)/(s+2)
2s^2(s-2)^2/(s+2) = (s-2s/(s+2)-2s^2(s-2)/(s+2))(s-2)/2
(s(s-2)(6s-13))/(s+2) = 0
s = 0, 2, 13/6
Answer is 13/6
13/6 squared I should say.
You're all moronic homosexuals.
It's not actually a square
"FIND THE AREA OD THE SQUARE"
Read the title
>let me just add 6 for no reason at all
Just read the proof under it, moron. You'll know where the 6 came from.
>tfw the length 2 is less than the length x=1/2
>1[Update] [Auto]
Dimension not given. Might as well be a trick question.
it's solvable, just use your brain more
>Imagine Jamal getting you homosexuals to do his homework.
the way I do engineering, I would print it and measure it
doesn't it have to be to scale anyway because it's a square?
the only geometry that matters is that the middle squares share a side, everything else is not to scale, the top of the green square should be twice as wide as the yellow if we are assuming that all the squares have the same volume.
The rectangles in the middle share a side length, since they have the same area, they must have the same height, call it x. Now the yellow rectangle has a height 2x, meaning it is half as wide as the green and pink ones. This means the orange rectangle has an area 3y, where y is the width of the green and pink ones. But this means 3y = xy, that is x = 3 and the total area is 64, and each rectangle. has an area 12.8.
60
Clever attempt, but really you'd get a failing grade because you're assuming 4x2 = s without proving it
Look out guys, we have a middle school math teacher present. Mind your manners.
That's not a way to talk to your teacher, kids.
You'll be getting an F on this quiz, and I'll have to call your parents.
Jokes on you, i am an orphan. My parents died in a colored block problem when I was (5x)(5x) years old.
lol are you moronic
Yeah, 4 * 2 can equal 9, unless your a racist
I mean, it has to be.
That area has the same length as the orange, so since their area is collectively 3 times as big as A, you must necessarily be able to split them into pieces of the same length, and width of 2.
Let's say you morph B, C, and D into rectangles of the same length as A, but with nonequal widths. How could their areas possibly be equal, then?
Yeah, you're right.
It should fit 3 of them because there are 3 rectangles with the same area.
If the blue rectangle has a shorter or longer length, then the orange rectangle would necessarily have to also.
And since we are assuming that they all form a square, the scale cannot be argued to be even a pixel off.
moron
good shit
neat
clever
Wrong
Extremely elegant
just finished writing out a long winded answer where I used algebra to figure out the height of the yellow block but I saw this post and felt like a moron
64
Flawed premise.
Fun problem. Here's what I did.
Let x and y be the height and width of the green rectangle.
The purple rectangle has width y and the same area as the green, so its height is x.
The yellow rectangle has height 2x and the same area as the green, so its width is y/2.
The orange rectangle has width 3y/2 and the same area as the green, so its height is 2x/3.
The blue rectangle has height 8x/3 and the same area as the green, so its width is 3y/8.
Solve 2x/3 = 2 to get x = 3.
Since it's a square, 8x/3 = 15y/8, making y = 64/15.
The sides of the square are 8x/3 = 15y/8 = 8.
The area of the square is 64.
You don't need to the last steps; the problem is already solved when you get x = 3. You can just square 2 + 2x.
It can definitely be simplified, I just wanted to show what I actually did.
One thing we can do with the extra work is rebut the claim the premises are inconsistent.
Why do you think the green and pink shapes are the same and in the same orientation?
>bamboozled by a colored block problem
The yellow, green and pink rectangles combined have an area 3 times that of the orange rectangle. As they have the same width, they must have 3 times the height of the orange rectangle. The yellow rectangle, therefore, has a height of 6. The square has a side length of 8, and an area of 64.
Yeah it's easy when you see you have a rectangle which shares one side with the red rectangle and also has three times the surface area. This means that the other side of that bigger rectangle must be the other side of the red one multiplied by three which gives the area of the whole square. Fun problem.
There is no red rectangle.
>bamboozled
This is a troll as there is no solution. I walk you through it and why people got 64 but it doesn't work out.
The yellow, green, an purple rectangles can be combined into a single rectangle with three times the area of the orange triangle. Since they share the same length, but one is three times the area of the other, then the height must be three times as long. Hence the height of the yellow rectangle is 2 * 3 = 6. Hence the dimensions of the "square" are 8 x 8 and you get 64. But the blue rectangle does not compute.
Let's call the length of the orange triangle A, and the area of of each rectangle M.
Hence the blue rectangle is (8 - A) * 8 = M
Orange tells us 2 * A = M, or A = M / 2.
So the blue triangle is (8 - M / 2) * 8 = M
Blue is then 64 - 4m = m or m = 64 / 5.
Wait... it works. m = 64/5 and there 5 rectangles with are m, total area is 64 / 5 * 5 = 64. So I guess it is 64.
You were right when you weren't wrong
5(2x)
1/5*a^2 = 2*(a-(a/5))
a^2=10*a-a=9*a
a=9
A=a^2=81
*A=64
5 orange
I got 64
Note, each square is 1/5y^2, so three of the squares is 3/5y^2
Therefore, (y-2)x=3/5 y^2
Since we have 2x=1/5y^2, we can solve with these two equations
the square is yellow+green+pink and the area is 36
it's 64 because the following equations need to be equal :
> 2*X = orange area (X the long side)
> Y*X/3 = yellow + green + pink area (Y the height)
> (2+Y-X)*(2+Y) = blue area
X cannot be zero since the shape wouldn't be a square otherwise
you are all idiots. x,y,z, there's no need for this shit here, i solved this in like 20 sec
why is IQfy like this
Height of blue is x
Since all rectangles are equal, area of blue is (1/5) x^2
Then width of blue is (1/5)x
Width of orange is then (4/5)x
Area of orange is then (8/5)x
Set areas of orange, blue equal and solve:
(8/5) x = (1/5) x^2
8x = x^2
x = 0, 8
Area is 64
Smart.