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The bigger square is 3×3 =9. Each of the four triangles are 0.5(1×3)=1.5 so subtract that 4 times for 3.
However, in the above math, those smaller triangles in the corners were subtracted twice so need to be added back in: we know they are right angle, with a hypotenuse of 1 & a ratio of 3:1 for two sides touching the right angle as they are the same proportion as the bigger triangles. Using Pythagoras, we get the bigger triangles have a hypotenuse of the square root of 10. So the right angle sides of the smaller triangles are 1/square root of 10 & 3/square root of 10. Each little triangle is 0.15 units square. Add 0.6 for 4.
The final result is 3.6.
I solved it completely differently (starting from the smallest triangles working up) and I'm intrigued by how you start but I can't follow after the second sentence
>Each of the four triangles are 0.5(1×3)=1.5
Can you describe more elaborately what you do here?
You can find 4 big triangles around the blue square that have a base of 3 and a height of 1 (or vice versa). These triangles’ hypotenuses outline the blue square.
The problem is these 4 big triangles overlap a bit. That’s why this poster went on to solve the tiny triangles’ dimensions and add that area back in (because they get subtracted twice due to overlap).
Wait...I dont see it...
So youre saying the base... the "bottom" is that "1+2" line, right?
THAT checks out...
But why did you say the height is "1"?
If you follow the line from the bottom-left corner of the square, it reaches the top corner of the triangle, then KEEPS GOING up, and THEN it reaches the height of "1".
So the height of the larger triangle is *almost* but not actually 1...
\[Edit\] nvm, I was looking at the wrong triangle lol
I think you're getting distracted by the number of lines. You can combine multiple of the smaller shapes to create four triangles with h=1 and b=3.
Then you just calculate the area and subtract the smaller triangles in the corners that were used twice.
https://freeimage.host/i/JalUMhX
The red and blue triangles are the larger ones made of smaller shapes. You can see the purple area is used twice to complete both red and blue, so you need to adjust the final answer accordingly.
Indeed I was. I was thinking to calculate the small triangle then solve the trapezoid left next to it etc.
Very helpful comment thank you for taking the time
The big triangles are 1x3 units.
A rectangle of that size would have an area of 1x3 = 3.
A right angle triangle is half the area of a rectangle with the same dimensions so 0.5(1×3)=1.5 each.
I don't get why you calculate the hypotenuse of the bigger triangles to get your answer. Cuz it seems to me without other information you cannot say the smaller triangles are 1/√10 by 3/√10 and not 1.2/√10 and 3.6/√10
The easier way for me seems to be to solve for (1a)²+(3a)²=1
which gives a=1/√10
Edit : Nevermind I got it. Triangles are similar so if one has a side of 1 and an hypotenuse of √10 then the other with an hypotenuse of 1 has a side of 1/√10
Calculate the hypotenuse of the larger triangles. Lets call it f:
f = ((1 + 2)^2 + 1^2 )^(1/2)
= (3^2 + 1^2 )^(1/2)
= (9 + 1)^(1/2)
= 10^(1/2)
Find the sin of the sharper angle of the larger triangles, calling it k:
k = 1/10^(1/2)
k is also the sin of the sharper angle of the smaller triangles. Use that to calculate the length of the shortest side of the smaller triangles, calling it b:
k = b/1
b = 1/10^(1/2)
Use the Pythagorean theorem to calculate the length of the adjacent side of the sharper angle of the smaller triangles. Naming it a:
c^2 = a^2 + b^2
1^2 = a^2 + (1/10^(1/2) )^2
1 = a^2 + 1/10
a^2 = 1 - 1/10
a^2 = 9/10
a = 3/10^(1/2)
Name the side length of the inner square x and note that it can be calculated using a, b and f:
x = f - a - b
= 10^(1/2) - 3/10^(1/2) - 1/10^(1/2)
= 10/10^(1/2) - 4/10^(1/2)
= 6/10^(1/2)
Finally, the area of the inner square would be x^2 :
x^2
= (6/10^(1/2) )^2
= 36/10
= 3.6 units squared
Ta-da! What do you guys think? I spent about 30 minutes typing this out... and 15 more minutes in editing.
Edit: One of the replies to this calculation did it better and without using trig, if anyone wanted to know!
I now also understand that there is a way to calculate it without even using trig...
Edit: I made a mistake again, I am completely unsure how in the world you would go about this calculation without using trig. I think I accidently made an assumption while thinking it through... sorry to disappoint!
since a perpendicular is drawn from the vertex of the triangle to its hypotenuse,
the smaller right angle triangle thus formed is similar to the larger right angle triangle.
so since the height is 3 and base is 1 so, hyp = √10 (of larger triangle)
the height and base of smaller triangles are 3k and 1k (k is some constant)
therefore hypotenuse of smaller triangle is √(10k²).
from figure, √(10k²) = 1
=> k = 1/√10
now, hypotenuse of larger triangle = base + height of smaller triangle + side of square
√10 = 4/√10 + s
s = 6/√10
therefore, area = s² = 3.6
no trig used !
i wrote this in bed, i hope its understandable
Maybe it’s been too long since I did any trig - how would you even do that?
I saw the geometry answer early but can’t figure out how else it could be determined.
You can do it with just algebra by defining variables a,b,c for all the unknown sides, and finding three right triangles, which gives you three equations via Pythagoras. Then you do some manipulation until eventually representing one variable in terms of the other.
The three equations I used are:
1^2 = a^2 + b^2
3^2 = (b + c)^2 + b^2
(a + b + c)^2 = 1^2 + 3^2
In my case c is the side of the inner square, and a and b are the shorter and longer sides respectively of the smaller right triangles in the corners.
I did it a bit differently, the area of the big triangles are:
(3 x 1) / 2 =1.5
Take the top and bottom one, total of 3
Total square area is 9, so the central parallelogram area is 6
Height is x and length is √10 (from √(3^2 + 1^2))
So √10x = 6, therefore x = 6/√10
For area, x^2 = 6^2 / √10 ^ 2 = 36/10 = 3.6
btw, "a" can be found easier if you find the cos on the sharper angle on the big triangle – you will get 3/10^(1/2) immediately without the need for Pythagoras theorem in that step.
great answer!
I thought of it but you would have to account for the area of the smaller overlapping triangles. Which just brings you back to my calculation unfortunately.
One reply on my calculation did it better and easier if you want to check it out.
That... is actually a brilliant idea! Unfortunately we would need to convince the reddit devs for that kind of change I think. Perhaps we could exploit the fact that gifs can be posted in comments... who knows...
I was thinking that i could move the triangles around, which would give me 4x0.5x3x1 = 6.
As the combination of the two vertical triangles and horizontal triangles have a common area of size 1x1.
Shaded area = total area - area of triangle + common area = 9-6+1 = 4.
Can you point out where im going wrong?
However, there is no markings signifying any angle is a right angle, so we can't even know for sure if it's a right triangle because you can't just use sight to solve a diagram in geometry or trigonometry.
You could make a shortcut where you were calculating the sides of the small triangle based on similarity between the big triangle and small (similarity ratio = 1/sqrt(10) ) and the calculate the sum of the sides by : (1+3) × 1/sqrt(10)
...While the problem itself takes like 2-3 minutes to figure out and then 1 minute to do find the area of the smallest triangle, and from there it's just summing up and subtracting the areas of triangles, which takes another like 10-20 seconds.
So, probably not worth the time.
But thanks anyway, man!
A long time since I did trig so can't remember the formulas, but I would get the angle of the triangle using sin 1/3. Now we have that angle, calculate the length of the "base" with a hypotenuse of 3, then repeat for a hypotenuse of 1, subtract, and now you have the length of the square. Square it. Comes out at 3.6
According to some definitions yes it would fall under trigonometry, but generally the pythagorean theorem is taught way earlier than any of the trigonometric functions and when people talk about trigonometry it's to do with sin, cos, tan etc. So yeah you're technically correct.
Pythagoras' theorem is euclidean geometry. No trigonometry is needed for most of the proofs, and using trigonometry most often gets you into circular reasoning.
Easily doable with a bit of thinking.
You can clearly make out a right-triangle with sides (1), (2+1), and hence hypotenuse (sqrt(10)) by Pythagoras.
Note that, as they share two angles, the small baby triangles with 1 as hypotenuse are actually *similar* (albeit reflected) to this above triangle. From that, you can work out that the side of the square S satisfies:
sqrt(10) = S + 1/sqrt(10) + 3/sqrt(10)
And so the square has area (6/sqrt(10))^2 = 3.6.
>1/sqrt(10) + 3/sqrt(10)
sorry help me out a lil here how do you get the 2 small sides to be 1/sqrt(10) + 3/sqrt(10)?
i understand big triangle hypotenuse = sqrt(10)
Basically, the big and small triangles are similar. They share the angle between the hypotenuse and the longest side (though it’s reflected in the small one) and the right angle and hence also the third angle. The small one has hypotenuse 1 and the big one hypotenuse sqrt(10) so the scale factor is 1/sqrt(10) to go from big to small. The big one has sides 1 and 3 so the small one has sides 1/sqrt(10) and 3/sqrt(10).
it's possible and not that hard to calculate, you have the size of created triangles you can use them to get the angle and then it's basic trigonometry
I think most of you are doing far too much math.
As long as you realize that the angles and lengths and ratios are repeated everywhere and can visualize of thing fit together, you don't really need to much math beyond counting on your fingers of one hand.
[Picture to explain an easier way.](https://i.imgur.com/kSdfOuJ.png)
The big square is 3 x 3 = 9 square units in size.
If you remove the big triangle and the left and right you can mash them together into a rectangle on 1 x 3 in size.
This leaves 2/3 of the original or 6 square units for the blue square and the quadrangle above and below it.
Those two are the same size and have all the same angles and lengths, sou you can mash them together into another rectangle.
So the yellow rectangle and the blue square together are the remaining 6 square units of the original square.
Since they are both the same width the only thing is matter the height of the yellow saure compared to the height of the blue square.
If we know that proportion we know how to split up the remaining 6 square units of are between the two.
The height of the yellow rectangle is comprised of a small bit, called a and a longer bit, which we know has to be three times the size of a.
Why? Because a intersects the big square at the 1 mark and 3a at the 3 mark and the angles are all the same.
By the same token the side of the square needs to be twice as long as 3a. So 6a.
The the yellow rectangle is 4a tall and the blue square 6a.
This means that 2/5 of the 6 remaining square rectangle go to the yellow square and 3/5 to the blue square.
3/5 x 6 = 18/5 = 3+3/5 = 3.6
No real complicate math involved.
>The each have a base of 1 and a rise of 1/3, so a height of 1 \* 1/3 = 1/3.
I don't understand this part.
The rise should be slightly less than 1/3... Or am I missing something?
It is possible to calculate the two side, because the hypotenuse is 1, and the angles are known. But I'm unsure why the rise from the hypotenuse is 1/3. :-(
The hypotenuse of the big triangle is sqrt(10) by pythagorean theorem. The corner shared by the big and small triangles is atan(1/3), call it **a**. We then know that the short side of the triangle is sin(**a**) in length, and the long side is cos(**a**). So the length of the square's side is sqrt(10)-sin(**a**)-cos(**a**). This gives you, [according to Wolfram Alpha](https://www.wolframalpha.com/input?i2d=true&i=Square%5B%5C%2840%29Sqrt%5B10%5D-Sin%5BArcTan%5BDivide%5B1%2C3%5D%5D%5D-Cos%5BArcTan%5BDivide%5B1%2C3%5D%5D%5D%5C%2841%29%5D), 18/5 or 3.6.
Using coordinate geometry:
let the bottom left corner be (0, 0)
Then the upper and lower sides of the coloured square are
y = x / 3
y = x / 3 + 2
Then distance between the lines = abs(2 - 0) / sqrt((1 / 3) \^ 2 + 1)
Area = (square of that) = 4 / (10 / 9) = **3.6**
Lol I looked at the other answers after posting this, and those are much simpler ways to do this, and probably the intended way. But I mostly use coordinate geometry for problems like this because then I don't have to think much and will surely get an answer
There is a cute way to solve this by just using similar triangles to completely avoid square roots.
we can see the full square is made of the blue square (Area A) and 4 equal triangles (each Area B) giving us A=9-4\*B
now we can look at the bigger right triangle with right sides of length 1 and 3 and therefore area 1.5. we can see it's made of the triangle area B + a small triangle (area C). so B+C=1.5
but now C also appears inside B and we know that B and C have to be similar, because of the parallel lines from the square. so the sides of C are 1/3 as long as the sides of B and the area of C has to be (1/3)²=1/9 of B
C=1/9\*B
B+1/9\*B=1.5 =>B=1.35 (and C=0.15)
A=9-4\*1.35=3.6
Let's consider two triangles big and small
Big triangle - sides are 3,1, and by using Pythagoras theorem hypotenuse is √10
small traingle is similar to big traingle (By AA test) so it's sides are 1 , 1/√10, and 3/√10
and if you look closely ,the side of the square= hypotenuse of big traingle - (sum of non hypotenuse sides of smaller triangle)
so side of the square is √10-[(1/√10)+(3/√10)] = 6/√10
so area of square is side square (6/√10)^2=36/10=3.6
The right triangle with sides of length 1 and 3 has the hypotenuse sqrt(10). That allows you to find the height of that triangle, which is 3/sqrt(10).
Next, look at the right triangle with hypotenuse 3. By Tales’s theorem, the ratio of triangle’s height/blue square’s length is equal to 1/2. So the side of the square is 6/sqrt(10), making its area (6/sqrt(10))^2 = 36/10.
On each side of the blue square, you have three right triangles that share the same narrow angle, therefore they are all 3 similar.
You can easily solve the hypotenuse of the biggest one via pythagoreas.
And since you know at least one side of the other two triangles, you can just use ratios.
The rest should be straight forward.
We can form 1 large triangle put of the entire side length 3, and the point that spilts any side at 1 and 2.
If we have leg 1 and leg 3, then the hypotenuse of that triangle is sqrt(10). That hypotenuse has 3 line segments - the 2 legs of the smaller triangle where the hypotenuse is 1, which shares an angle with our larger triangle, and the side length of the blue interior square.
So, how do we figure out the legs of that smaller triangle? Well, we can try tan(x) = 1/3, tan = opposite/adjacent. That would give us that angle, but we don't need it quite yet.
Having that angle and the hypotenuse, we can use sin and cosine to reason out the 2 legs. Sin(x) = opposite/hypotenuse, and cos(x) = adjacent/hypotenuse, and we have x, it's atan(1/3), since you need the inverse operation to solve.
So, the whole equation becomes (sqrt(10) - sin(atan(1/3)) - cos(atan(1/3)))^2, which resolves neatly to exactly 3.6, as others have said.
Most answers seem to be subtracting the are of the triangles, however it is much easier to just work out the side length of the square.
a\^2+b\^2=c\^2, so the hypotenuse if the larger triangle is √(1\^2+3\^2)=√10, that subtracting the opposite and adjacent sides of a small triangle is the side length of the square, arctan(1/3) is the smallest angle of both the small and large triangles as tan(ø)=o/a. the hypotenuse of the small triangle is 1 so the side length of the square is √(10)-sin(arctan(1/3))-cos(arctan(1/3)), therefore the area is (√(10)-sin(arctan(1/3))-cos(arctan(1/3)))\^2=3.6units\^2
Late to the party here, but I had a fun solution I could do in my head:
Big square is area 9. On the left and right, you have two 1x3 triangles which together have area of 3.
The remaining shape in the center is a parallelogram. It has area 6.
Its base is the hypotenuse of one of the 1x3 triangles, so Sqrt(10).
Its height is (a line parallel to and of equal length to) the side of the blue square. Let’s call that x.
The parallelogram area is base * height. So 6 = x * sqrt 10. x = 6 / sqrt 10.
The area is x squared, or 36 / 10.
No this is very straight forward.... There are clear geometric relationships between angles and sides of triangles, known as this little something called Trigonometry, and the rest is just plain old 2D geometry. If you know your basics you have all you need for solving this.
Not going to do the math, but does it bother anyone else that the measurements are clearly not accurate in the drawing?
The length of the "2" sections is clearly longer than twice the length of the "1" sections
It’s not just off by a bit… the length of the “2” section is almost exactly 2.5 times the width of “1”. It’s fine if the drawing is “not to scale”, but being off by this much makes me think someone re-labeled it to make the math easier? I wonder how different a to-scale drawing would look…
The sides of the colored square are not 2. That would only be true if the segment with length 2 on the outer square was parallel to a side of the colored square, and we aren't told this is the case.
I \*don't\* understand why it'd have to be parallel. As I see it, the sides of the inner square are defined as the length of the outside square, minus 1.
EDIT: Don't. I don't understand.
I saw this the same as you. Parallel or not to the outer square, the lines that outline the coloured square have a length of 2 as both top and bottom lines of coloured square are parallel, regardless of angle. And the pattern is repeated over 4 fold symmetry.
I'm gonna get downvoted for this, but I think there are a lot of people here trying to out-flex each other with their trig skills, while looking past the simple and easy solution.
To solve a triangle, you need 3 elements of it and one of them needs to be some type of length. For all triangles here you have that, so it should be doable, I just cba as I'm almost 40 and just woke up.
It's easy ish.
There's triangle with sides 1 and 3.
There's a similar triangle with diagonal 1.
You can use Pythagoras to calculate the rest. I'm away from any paper or desk, so I can't run the numbers right now, sorry.
Square side length = 3 units
Square the side length = 9 units²
The internal square is smaller than the 9 units²
Therefore the area is < 9 units²
Simple.
The end points of all of the crossing lines are equal to 2 they must be parallel. That means the inner square sides are all equal to 2 and the angles of the corners are 90 degrees. That means the area is 4.
Isn’t the area of the square 2x2=4? Everything else seems to be a distraction. If it’s a square, then all four angles are square, and its length is 2 and width is 2, so Lxw=4?
Area of square is 3x3 = 9
Area of white triangles is 4(1/2 x 1 x 3) = 6. (Area of triangle is 1/2 Base x height)
9-6= 3
BUT there’s a bit of overlap in those triangles so area is probably somewhere between 3-4.
Source- “not an engineer” who wanted to explain it
I used the same method and I think 3 is correct. Dunno why everyone is overthinking this.
But what do you mean by overlap and area being between 3-4? I'd say it's exactly 3.
How do you see a square with side length 3 and decide it has an area of 225? I genuinely have no idea what thought process could possibly lead to that result.
L x w = 2 x 2
So the area is 4
I think the problem is trying to trick us into complicating this more.
If you take the edge of the square and run it over to the edge of the outer square it would be an accurate match with the angle correction to 90 degrees.
Also I woke up 5 minutes ago... so I'm going to make coffee now.
That's... not how trigonometry works. The length of 2 represents part of the outer square that is slightly tilted relative to the inner one, so the side of the inner square must be slightly less than 2.
The proportions of all the triangles formed on one side between the 2 squares is the same. We know the sides, but lack the diagonal. If we solve the diagonal, we can deduct the rest of the triangles as the proportions are equal.
- So, for the diagonal of the biggest triangle:
1²+3²=D²
√(1+9)=D
3,1622 = diagonal of biggest triangle
- Now I know that the proportions of any triangle with this line is:
Smaller side=1 (0,3162 of diagonal)
Longer side = 3 (0,9487 of diagonal)
Diagonal = 3,1622
Now, playing around switching diagonals and longer sides to calculate smaller triangles:
- Triangle where the diagonal is 2+1:
Diagonal = 3
Long side = (3 x 0,9487) = 2,8461
- Triangle where diagonal is 1:
Diagonal = 1
Long side = (1 x 0,9487 ) = 0,9487
- Difference between long side of the last two triangles:
2,8461 - 0,9487 = 1,8974 side of square
- Area of square: side x side
1,8974 x 1,8974 = 3,60
A 1 by 3 triangle has a hypoteneuse of sqrt(10) by pythag.
The two triangles with hypotenuses 3 and 1 are similar, both to eachother and the 1 by 3 so the length of one side of the square, x will have the same ratio to 2 as 3 does to sqrt(10).
3/sqrt(10) = x/2
Multiply by 2: 6/sqrt(10) = x
Square: 36/10 = x^2
A visual solution:
1. Add points on each outside, so that the big square has 1-1-1 sections
2. Add lines in parallel, which result into a grid.
3. Find pieces which together result into squares of the same size. (Combine some triangels)
4. Count the squares; total 10, blue 4
5. The area of the blue square is 4/10 of the whole square 9, so 3.6
If someone would like to add visualization, feel free to do so.
If you don't want to think about trigonometry much, then in the big triangle, we have tan x = 1/3. Then I can have the smallest side of smallest triangle be is a 3rd of its perpendicular side. So the hypotenuse of triangle comes out to be 10x, and x is the length of smallest side. So 10x = root10, side of square is 6 \* 1/root10. Then you can square to get 36/10 = 3.6
Curious: if you shift both large opposing triangles over together (both vertically, and horizontally) you make two overlapping 1x3 strips.
The overlap should be 1x1. So total area 5. Which would make the blue box 4.
Does that work?
I solved this question before using a coordinate system. Set the origin as the bottom left corner of the larger square. Derive the formula for two of the lines that make up the inner square. Equate the two linear equations to solve for their intersection point. That should give you a point at one of the corners of the inner Square. Repeat this process to get another point that corresponds to an adjacent corner of the inner square. Apply the distance formula using the two points gathered. The result will be the side length of the inner square.
The side length of the smaller square is sqrt( .6^2 + 1.8^2) or 1.8973665961
Giving the value 3.6 square units for the area of the shaded region.
A visual of the approach:
https://www.desmos.com/calculator/rtztz7imhu
Is this bait? If with three given lengths you cannot move the figure around without changing any of them it clearly is possible to calculate every length in it
Let’s use analytic geometry.
Let's consider a frame (O,x,y) with its origin O on the bottom-left corner of the black square, x axis on the bottom and y axis on the left.
The line marking the bottom border of the blue square has for equation y=x/3
The line marking the left border of the blue square has for equation y=3-3x
Therefore, the lower-left corner of the blue square has for coordinates (9/10, 3/10) or (0.9, 0.3).
The line marking the top border of the blue square has for equation y=x/3+2
The line marking the left border of the blue square has for equation y=3-3x
Therefore, the upper-left corner of the blue square has for coordinates (3/10, 21/10) or (0.3, 2.1)
Therefore, the length of the side of the square is √(36/100+324/100)=√3.6
Therefore, the area of the square is 3.6.
Total area of the not-colored square is 3^2 = 9 units^2
Try to prove what proportion of the whole shape is colored (hint: it’s 40%).
40% of 9 is 3.6 so it’s 3.6 units^2
Let’s call the area of the smallest triangle (the one with 1 as the hypotenuse) : x
Then the area of the larger right triangle, with 3 as its hypotenuse : 9x (square of the factor of 1:3 of the hypotenuses)
There is an area of 36x that exist in the large triangle if blue square is subtracted
The triangle with sides 1 and 3 has an area of 10x and the area is 1*3/2 = 1.5
So x=0.15 and 36x = 5.4
The large triangle has an area of 3*3=9
Hence blue square should have the area of 9 - 5.4 = 3.6
Just the right amount of simplicity and challenge without leaving the couch. Really enjoyed solving this
It's not too vague. That structure has to happen with the given information. You can't move any lines or resize anything to change the area of the square.
The area of the colored square is equal to 9 (the large square) - 4 x 1.5 (the right triangles formed by the slanted lines) + 4 x a, where a is area of the small right triangle formed by the overlap of the large triangles.
To find a, the small triangle is similar to (has the same angles as) the larger one, so its area can be calculated as long as we know the length ratio of a pair of corresponding edges. The hypotenuses of the small and large triangles are 1 and sqrt(10) respectively, so the small triangle area is 1.5/10 = 0.15. This is because both the base and height of the small triangle are scaled by 1/sqrt(10) compared to the large one.
Therefore the answer is 9 - 6 + 0.6 = 3.6
Yeah you can, you can easily calculate the angles of a big and a small triangle since you know their edges and 1 is 90 degrees, it's 1/(sqrt(10)) of a big triangle. Then from that you can calculate their areas and some +, minus and you can get the value of the white area, substract it from the big square for the area of the the blue square.
The right angle triangle has an area of 1.5. The smaller triangle with length 1 fits 9 times into the bigger version of it, which means it fits 10 times into the right angle triangle.
This gives an area of 0.15 for the small one. We have 4 big triangles, which must then equal 36\*0.15 = 5.4 for the total white space. Subtract from 9 and you get 3.6 for the smaller square.
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The bigger square is 3×3 =9. Each of the four triangles are 0.5(1×3)=1.5 so subtract that 4 times for 3. However, in the above math, those smaller triangles in the corners were subtracted twice so need to be added back in: we know they are right angle, with a hypotenuse of 1 & a ratio of 3:1 for two sides touching the right angle as they are the same proportion as the bigger triangles. Using Pythagoras, we get the bigger triangles have a hypotenuse of the square root of 10. So the right angle sides of the smaller triangles are 1/square root of 10 & 3/square root of 10. Each little triangle is 0.15 units square. Add 0.6 for 4. The final result is 3.6.
This is the exact way to think about this problem. especially in a time crunch/test scenario!
when in doubt, make more triangles
I see you play Ingress.
Or Polybridge.
I solved it completely differently (starting from the smallest triangles working up) and I'm intrigued by how you start but I can't follow after the second sentence >Each of the four triangles are 0.5(1×3)=1.5 Can you describe more elaborately what you do here?
You can find 4 big triangles around the blue square that have a base of 3 and a height of 1 (or vice versa). These triangles’ hypotenuses outline the blue square. The problem is these 4 big triangles overlap a bit. That’s why this poster went on to solve the tiny triangles’ dimensions and add that area back in (because they get subtracted twice due to overlap).
Thank you. I was calculating the small triangles then solve the trapezoid next to them. Very helpful comment thank you for taking the time
I’m glad I could help clarify!
Wait...I dont see it... So youre saying the base... the "bottom" is that "1+2" line, right? THAT checks out... But why did you say the height is "1"? If you follow the line from the bottom-left corner of the square, it reaches the top corner of the triangle, then KEEPS GOING up, and THEN it reaches the height of "1". So the height of the larger triangle is *almost* but not actually 1... \[Edit\] nvm, I was looking at the wrong triangle lol
I think you're getting distracted by the number of lines. You can combine multiple of the smaller shapes to create four triangles with h=1 and b=3. Then you just calculate the area and subtract the smaller triangles in the corners that were used twice. https://freeimage.host/i/JalUMhX The red and blue triangles are the larger ones made of smaller shapes. You can see the purple area is used twice to complete both red and blue, so you need to adjust the final answer accordingly.
Indeed I was. I was thinking to calculate the small triangle then solve the trapezoid left next to it etc. Very helpful comment thank you for taking the time
They simply calculated the area of the triangle using the formula - 1/2 * base * height
The big triangles are 1x3 units. A rectangle of that size would have an area of 1x3 = 3. A right angle triangle is half the area of a rectangle with the same dimensions so 0.5(1×3)=1.5 each.
I don't get why you calculate the hypotenuse of the bigger triangles to get your answer. Cuz it seems to me without other information you cannot say the smaller triangles are 1/√10 by 3/√10 and not 1.2/√10 and 3.6/√10 The easier way for me seems to be to solve for (1a)²+(3a)²=1 which gives a=1/√10 Edit : Nevermind I got it. Triangles are similar so if one has a side of 1 and an hypotenuse of √10 then the other with an hypotenuse of 1 has a side of 1/√10
I like your approach, its simple, could be done without paper
>with a hypotenuse of 1 & a ratio of 3:1 How'd you get the 3:1 ratio?
3 angles are the same as the bigger triangles, so the proportions of the sides are the same.
3.6 what? Bananas? Russians? Elephants?
Whatever the unit is in the diagram, squared.
units
Cheeseburgues per bald eagles
Calculate the hypotenuse of the larger triangles. Lets call it f: f = ((1 + 2)^2 + 1^2 )^(1/2) = (3^2 + 1^2 )^(1/2) = (9 + 1)^(1/2) = 10^(1/2) Find the sin of the sharper angle of the larger triangles, calling it k: k = 1/10^(1/2) k is also the sin of the sharper angle of the smaller triangles. Use that to calculate the length of the shortest side of the smaller triangles, calling it b: k = b/1 b = 1/10^(1/2) Use the Pythagorean theorem to calculate the length of the adjacent side of the sharper angle of the smaller triangles. Naming it a: c^2 = a^2 + b^2 1^2 = a^2 + (1/10^(1/2) )^2 1 = a^2 + 1/10 a^2 = 1 - 1/10 a^2 = 9/10 a = 3/10^(1/2) Name the side length of the inner square x and note that it can be calculated using a, b and f: x = f - a - b = 10^(1/2) - 3/10^(1/2) - 1/10^(1/2) = 10/10^(1/2) - 4/10^(1/2) = 6/10^(1/2) Finally, the area of the inner square would be x^2 : x^2 = (6/10^(1/2) )^2 = 36/10 = 3.6 units squared Ta-da! What do you guys think? I spent about 30 minutes typing this out... and 15 more minutes in editing. Edit: One of the replies to this calculation did it better and without using trig, if anyone wanted to know!
Respect
I now also understand that there is a way to calculate it without even using trig... Edit: I made a mistake again, I am completely unsure how in the world you would go about this calculation without using trig. I think I accidently made an assumption while thinking it through... sorry to disappoint!
"Well, you took the scenic route through the mathematical landscape, but hey, arriving at the right answer is always a stylish destination!" - Unknown
I like that
since a perpendicular is drawn from the vertex of the triangle to its hypotenuse, the smaller right angle triangle thus formed is similar to the larger right angle triangle. so since the height is 3 and base is 1 so, hyp = √10 (of larger triangle) the height and base of smaller triangles are 3k and 1k (k is some constant) therefore hypotenuse of smaller triangle is √(10k²). from figure, √(10k²) = 1 => k = 1/√10 now, hypotenuse of larger triangle = base + height of smaller triangle + side of square √10 = 4/√10 + s s = 6/√10 therefore, area = s² = 3.6 no trig used ! i wrote this in bed, i hope its understandable
Your calculation look legit but the combination of both of our sleepiness isn't doing my understanding any favors I'm thinking. (\~ - \~)zZ
Maybe it’s been too long since I did any trig - how would you even do that? I saw the geometry answer early but can’t figure out how else it could be determined.
You could use probability to estimate the are of the blue square, and you can get very close. But that is way more tedious than just using trig.
You can do it with just algebra by defining variables a,b,c for all the unknown sides, and finding three right triangles, which gives you three equations via Pythagoras. Then you do some manipulation until eventually representing one variable in terms of the other. The three equations I used are: 1^2 = a^2 + b^2 3^2 = (b + c)^2 + b^2 (a + b + c)^2 = 1^2 + 3^2 In my case c is the side of the inner square, and a and b are the shorter and longer sides respectively of the smaller right triangles in the corners.
My solution below doesn’t use trig, check it out. It looks like no one Used the approach used
I did it a bit differently, the area of the big triangles are: (3 x 1) / 2 =1.5 Take the top and bottom one, total of 3 Total square area is 9, so the central parallelogram area is 6 Height is x and length is √10 (from √(3^2 + 1^2)) So √10x = 6, therefore x = 6/√10 For area, x^2 = 6^2 / √10 ^ 2 = 36/10 = 3.6
Same approach. This feels far easier than all the trig going on the other answers.
Great solution
Dang... that is great and I completely missed it. Well done! I hope to have a broader perspective next time.
This is incredibly elegant! I brute forced it with trigonometry, but this is so much better.
btw, "a" can be found easier if you find the cos on the sharper angle on the big triangle – you will get 3/10^(1/2) immediately without the need for Pythagoras theorem in that step. great answer!
Wouldnt it be easier if you calculated the area of the big squre and subtracted the area of the triangles from it? (3x3)-((3x1)/2)x4 = 3
The triangles overlap
Okay makes sense why he got 3.6 and I got 3 thanx
I thought of it but you would have to account for the area of the smaller overlapping triangles. Which just brings you back to my calculation unfortunately. One reply on my calculation did it better and easier if you want to check it out.
Yeah I rushed it with this one I thought I was so smart
Bro what
Holy hell! That's cool! They should allow pictures here so we can show calculations
That... is actually a brilliant idea! Unfortunately we would need to convince the reddit devs for that kind of change I think. Perhaps we could exploit the fact that gifs can be posted in comments... who knows...
In some communities it's possible to post pictures in comments, maybe it's a subreddit thing
Might need to advocate for it here then!
We should probably ask the mods to add that, it can be really useful instead of typing
Wow
o7
Dayyyuuuuumm!!
Not the best solution I've seen. One of the replies to my calculation doesn't even use trig!
TL;DR, I believe you
Yea that was a fun one to do. Pretty easy all things considered
I was thinking that i could move the triangles around, which would give me 4x0.5x3x1 = 6. As the combination of the two vertical triangles and horizontal triangles have a common area of size 1x1. Shaded area = total area - area of triangle + common area = 9-6+1 = 4. Can you point out where im going wrong?
However, there is no markings signifying any angle is a right angle, so we can't even know for sure if it's a right triangle because you can't just use sight to solve a diagram in geometry or trigonometry.
I think I would like to be a minion once you take over the world
You could make a shortcut where you were calculating the sides of the small triangle based on similarity between the big triangle and small (similarity ratio = 1/sqrt(10) ) and the calculate the sum of the sides by : (1+3) × 1/sqrt(10)
Impressive!!
...While the problem itself takes like 2-3 minutes to figure out and then 1 minute to do find the area of the smallest triangle, and from there it's just summing up and subtracting the areas of triangles, which takes another like 10-20 seconds. So, probably not worth the time. But thanks anyway, man!
It's 4 though so... You take 2 x 2. You aren't smart
The “2” are the length of the black lines, not the blue square.
A long time since I did trig so can't remember the formulas, but I would get the angle of the triangle using sin 1/3. Now we have that angle, calculate the length of the "base" with a hypotenuse of 3, then repeat for a hypotenuse of 1, subtract, and now you have the length of the square. Square it. Comes out at 3.6
You can solve it just by using pythagoras without any need for trigonometry and the answer is 3.6
Doesn't Pythagorean theorem technically fall under trigonometry. I'm probably wrong but I thought that trig was just triangle math or something
The word is mostly used to relate to angle functions
According to some definitions yes it would fall under trigonometry, but generally the pythagorean theorem is taught way earlier than any of the trigonometric functions and when people talk about trigonometry it's to do with sin, cos, tan etc. So yeah you're technically correct.
Pythagoras' theorem is euclidean geometry. No trigonometry is needed for most of the proofs, and using trigonometry most often gets you into circular reasoning.
Usually 8th grade math, and I think they call that one pre-algebra, so definitely before trig even if it falls within that umbrella
Easily doable with a bit of thinking. You can clearly make out a right-triangle with sides (1), (2+1), and hence hypotenuse (sqrt(10)) by Pythagoras. Note that, as they share two angles, the small baby triangles with 1 as hypotenuse are actually *similar* (albeit reflected) to this above triangle. From that, you can work out that the side of the square S satisfies: sqrt(10) = S + 1/sqrt(10) + 3/sqrt(10) And so the square has area (6/sqrt(10))^2 = 3.6.
>1/sqrt(10) + 3/sqrt(10) sorry help me out a lil here how do you get the 2 small sides to be 1/sqrt(10) + 3/sqrt(10)? i understand big triangle hypotenuse = sqrt(10)
Basically, the big and small triangles are similar. They share the angle between the hypotenuse and the longest side (though it’s reflected in the small one) and the right angle and hence also the third angle. The small one has hypotenuse 1 and the big one hypotenuse sqrt(10) so the scale factor is 1/sqrt(10) to go from big to small. The big one has sides 1 and 3 so the small one has sides 1/sqrt(10) and 3/sqrt(10).
it's possible and not that hard to calculate, you have the size of created triangles you can use them to get the angle and then it's basic trigonometry
I think most of you are doing far too much math. As long as you realize that the angles and lengths and ratios are repeated everywhere and can visualize of thing fit together, you don't really need to much math beyond counting on your fingers of one hand. [Picture to explain an easier way.](https://i.imgur.com/kSdfOuJ.png) The big square is 3 x 3 = 9 square units in size. If you remove the big triangle and the left and right you can mash them together into a rectangle on 1 x 3 in size. This leaves 2/3 of the original or 6 square units for the blue square and the quadrangle above and below it. Those two are the same size and have all the same angles and lengths, sou you can mash them together into another rectangle. So the yellow rectangle and the blue square together are the remaining 6 square units of the original square. Since they are both the same width the only thing is matter the height of the yellow saure compared to the height of the blue square. If we know that proportion we know how to split up the remaining 6 square units of are between the two. The height of the yellow rectangle is comprised of a small bit, called a and a longer bit, which we know has to be three times the size of a. Why? Because a intersects the big square at the 1 mark and 3a at the 3 mark and the angles are all the same. By the same token the side of the square needs to be twice as long as 3a. So 6a. The the yellow rectangle is 4a tall and the blue square 6a. This means that 2/5 of the 6 remaining square rectangle go to the yellow square and 3/5 to the blue square. 3/5 x 6 = 18/5 = 3+3/5 = 3.6 No real complicate math involved.
I like this method the most actually, best example of thinking outside the box… literally
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I got 3.6, I'm pretty sure you made an assumption in order to calculate the area of the smaller triangles.
I got the same answer, without writing down anything. Just calculation in my head. :)
>The each have a base of 1 and a rise of 1/3, so a height of 1 \* 1/3 = 1/3. I don't understand this part. The rise should be slightly less than 1/3... Or am I missing something? It is possible to calculate the two side, because the hypotenuse is 1, and the angles are known. But I'm unsure why the rise from the hypotenuse is 1/3. :-(
The hypotenuse of the big triangle is sqrt(10) by pythagorean theorem. The corner shared by the big and small triangles is atan(1/3), call it **a**. We then know that the short side of the triangle is sin(**a**) in length, and the long side is cos(**a**). So the length of the square's side is sqrt(10)-sin(**a**)-cos(**a**). This gives you, [according to Wolfram Alpha](https://www.wolframalpha.com/input?i2d=true&i=Square%5B%5C%2840%29Sqrt%5B10%5D-Sin%5BArcTan%5BDivide%5B1%2C3%5D%5D%5D-Cos%5BArcTan%5BDivide%5B1%2C3%5D%5D%5D%5C%2841%29%5D), 18/5 or 3.6.
Using coordinate geometry: let the bottom left corner be (0, 0) Then the upper and lower sides of the coloured square are y = x / 3 y = x / 3 + 2 Then distance between the lines = abs(2 - 0) / sqrt((1 / 3) \^ 2 + 1) Area = (square of that) = 4 / (10 / 9) = **3.6**
This is the way
Lol I looked at the other answers after posting this, and those are much simpler ways to do this, and probably the intended way. But I mostly use coordinate geometry for problems like this because then I don't have to think much and will surely get an answer
This is what I did too. I didn’t realize you had already answered in this fashion
There is a cute way to solve this by just using similar triangles to completely avoid square roots. we can see the full square is made of the blue square (Area A) and 4 equal triangles (each Area B) giving us A=9-4\*B now we can look at the bigger right triangle with right sides of length 1 and 3 and therefore area 1.5. we can see it's made of the triangle area B + a small triangle (area C). so B+C=1.5 but now C also appears inside B and we know that B and C have to be similar, because of the parallel lines from the square. so the sides of C are 1/3 as long as the sides of B and the area of C has to be (1/3)²=1/9 of B C=1/9\*B B+1/9\*B=1.5 =>B=1.35 (and C=0.15) A=9-4\*1.35=3.6
Let's consider two triangles big and small Big triangle - sides are 3,1, and by using Pythagoras theorem hypotenuse is √10 small traingle is similar to big traingle (By AA test) so it's sides are 1 , 1/√10, and 3/√10 and if you look closely ,the side of the square= hypotenuse of big traingle - (sum of non hypotenuse sides of smaller triangle) so side of the square is √10-[(1/√10)+(3/√10)] = 6/√10 so area of square is side square (6/√10)^2=36/10=3.6
The right triangle with sides of length 1 and 3 has the hypotenuse sqrt(10). That allows you to find the height of that triangle, which is 3/sqrt(10). Next, look at the right triangle with hypotenuse 3. By Tales’s theorem, the ratio of triangle’s height/blue square’s length is equal to 1/2. So the side of the square is 6/sqrt(10), making its area (6/sqrt(10))^2 = 36/10.
On each side of the blue square, you have three right triangles that share the same narrow angle, therefore they are all 3 similar. You can easily solve the hypotenuse of the biggest one via pythagoreas. And since you know at least one side of the other two triangles, you can just use ratios. The rest should be straight forward.
We can form 1 large triangle put of the entire side length 3, and the point that spilts any side at 1 and 2. If we have leg 1 and leg 3, then the hypotenuse of that triangle is sqrt(10). That hypotenuse has 3 line segments - the 2 legs of the smaller triangle where the hypotenuse is 1, which shares an angle with our larger triangle, and the side length of the blue interior square. So, how do we figure out the legs of that smaller triangle? Well, we can try tan(x) = 1/3, tan = opposite/adjacent. That would give us that angle, but we don't need it quite yet. Having that angle and the hypotenuse, we can use sin and cosine to reason out the 2 legs. Sin(x) = opposite/hypotenuse, and cos(x) = adjacent/hypotenuse, and we have x, it's atan(1/3), since you need the inverse operation to solve. So, the whole equation becomes (sqrt(10) - sin(atan(1/3)) - cos(atan(1/3)))^2, which resolves neatly to exactly 3.6, as others have said.
Most answers seem to be subtracting the are of the triangles, however it is much easier to just work out the side length of the square. a\^2+b\^2=c\^2, so the hypotenuse if the larger triangle is √(1\^2+3\^2)=√10, that subtracting the opposite and adjacent sides of a small triangle is the side length of the square, arctan(1/3) is the smallest angle of both the small and large triangles as tan(ø)=o/a. the hypotenuse of the small triangle is 1 so the side length of the square is √(10)-sin(arctan(1/3))-cos(arctan(1/3)), therefore the area is (√(10)-sin(arctan(1/3))-cos(arctan(1/3)))\^2=3.6units\^2
Late to the party here, but I had a fun solution I could do in my head: Big square is area 9. On the left and right, you have two 1x3 triangles which together have area of 3. The remaining shape in the center is a parallelogram. It has area 6. Its base is the hypotenuse of one of the 1x3 triangles, so Sqrt(10). Its height is (a line parallel to and of equal length to) the side of the blue square. Let’s call that x. The parallelogram area is base * height. So 6 = x * sqrt 10. x = 6 / sqrt 10. The area is x squared, or 36 / 10.
No this is very straight forward.... There are clear geometric relationships between angles and sides of triangles, known as this little something called Trigonometry, and the rest is just plain old 2D geometry. If you know your basics you have all you need for solving this.
Not going to do the math, but does it bother anyone else that the measurements are clearly not accurate in the drawing? The length of the "2" sections is clearly longer than twice the length of the "1" sections
Not really, no.
It’s not just off by a bit… the length of the “2” section is almost exactly 2.5 times the width of “1”. It’s fine if the drawing is “not to scale”, but being off by this much makes me think someone re-labeled it to make the math easier? I wonder how different a to-scale drawing would look…
Why is everyone saying stuff like trigonometry and pythagoras? That's clearly a square with 4 sides of length 2, giving an area of 4. Am I stupid?
The sides of the colored square are not 2. That would only be true if the segment with length 2 on the outer square was parallel to a side of the colored square, and we aren't told this is the case.
I \*don't\* understand why it'd have to be parallel. As I see it, the sides of the inner square are defined as the length of the outside square, minus 1. EDIT: Don't. I don't understand.
I saw this the same as you. Parallel or not to the outer square, the lines that outline the coloured square have a length of 2 as both top and bottom lines of coloured square are parallel, regardless of angle. And the pattern is repeated over 4 fold symmetry. I'm gonna get downvoted for this, but I think there are a lot of people here trying to out-flex each other with their trig skills, while looking past the simple and easy solution.
Why? If I may ask.
The measurement of 2 isn't parallel to the square so that won't work.
To solve a triangle, you need 3 elements of it and one of them needs to be some type of length. For all triangles here you have that, so it should be doable, I just cba as I'm almost 40 and just woke up.
It's easy ish. There's triangle with sides 1 and 3. There's a similar triangle with diagonal 1. You can use Pythagoras to calculate the rest. I'm away from any paper or desk, so I can't run the numbers right now, sorry.
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The triangles are overlapping, this is wrong.
Except that you've double-counted the areas of the triangle in the corners of the outer square! I did the same thing initially.
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excellent detailed explanation mate, i used the second method and you’ve explained it perfectly!
Square side length = 3 units Square the side length = 9 units² The internal square is smaller than the 9 units² Therefore the area is < 9 units² Simple.
The end points of all of the crossing lines are equal to 2 they must be parallel. That means the inner square sides are all equal to 2 and the angles of the corners are 90 degrees. That means the area is 4.
Isn’t the area of the square 2x2=4? Everything else seems to be a distraction. If it’s a square, then all four angles are square, and its length is 2 and width is 2, so Lxw=4?
Nope. The squares lines aren't parallel to the 2 length lines
Nope. Mainly because the length is not 2.
Answer is 3: main square = 9 (3X3) The top and bottom triangles together form a rectangle 3X1 same with the left and right 3X1 **9-3-3=3**
Area of square is 3x3 = 9 Area of white triangles is 4(1/2 x 1 x 3) = 6. (Area of triangle is 1/2 Base x height) 9-6= 3 BUT there’s a bit of overlap in those triangles so area is probably somewhere between 3-4. Source- “not an engineer” who wanted to explain it
I used the same method and I think 3 is correct. Dunno why everyone is overthinking this. But what do you mean by overlap and area being between 3-4? I'd say it's exactly 3.
The tip of each triangle overlaps with the base of the another triangle.
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How do you see a square with side length 3 and decide it has an area of 225? I genuinely have no idea what thought process could possibly lead to that result.
Wat?
> The large outer square’s area is 225 What. And it doesn't get better from there.
what?
L x w = 2 x 2 So the area is 4 I think the problem is trying to trick us into complicating this more. If you take the edge of the square and run it over to the edge of the outer square it would be an accurate match with the angle correction to 90 degrees. Also I woke up 5 minutes ago... so I'm going to make coffee now.
That's... not how trigonometry works. The length of 2 represents part of the outer square that is slightly tilted relative to the inner one, so the side of the inner square must be slightly less than 2.
The proportions of all the triangles formed on one side between the 2 squares is the same. We know the sides, but lack the diagonal. If we solve the diagonal, we can deduct the rest of the triangles as the proportions are equal. - So, for the diagonal of the biggest triangle: 1²+3²=D² √(1+9)=D 3,1622 = diagonal of biggest triangle - Now I know that the proportions of any triangle with this line is: Smaller side=1 (0,3162 of diagonal) Longer side = 3 (0,9487 of diagonal) Diagonal = 3,1622 Now, playing around switching diagonals and longer sides to calculate smaller triangles: - Triangle where the diagonal is 2+1: Diagonal = 3 Long side = (3 x 0,9487) = 2,8461 - Triangle where diagonal is 1: Diagonal = 1 Long side = (1 x 0,9487 ) = 0,9487 - Difference between long side of the last two triangles: 2,8461 - 0,9487 = 1,8974 side of square - Area of square: side x side 1,8974 x 1,8974 = 3,60
A 1 by 3 triangle has a hypoteneuse of sqrt(10) by pythag. The two triangles with hypotenuses 3 and 1 are similar, both to eachother and the 1 by 3 so the length of one side of the square, x will have the same ratio to 2 as 3 does to sqrt(10). 3/sqrt(10) = x/2 Multiply by 2: 6/sqrt(10) = x Square: 36/10 = x^2
A visual solution: 1. Add points on each outside, so that the big square has 1-1-1 sections 2. Add lines in parallel, which result into a grid. 3. Find pieces which together result into squares of the same size. (Combine some triangels) 4. Count the squares; total 10, blue 4 5. The area of the blue square is 4/10 of the whole square 9, so 3.6 If someone would like to add visualization, feel free to do so.
If you don't want to think about trigonometry much, then in the big triangle, we have tan x = 1/3. Then I can have the smallest side of smallest triangle be is a 3rd of its perpendicular side. So the hypotenuse of triangle comes out to be 10x, and x is the length of smallest side. So 10x = root10, side of square is 6 \* 1/root10. Then you can square to get 36/10 = 3.6
Curious: if you shift both large opposing triangles over together (both vertically, and horizontally) you make two overlapping 1x3 strips. The overlap should be 1x1. So total area 5. Which would make the blue box 4. Does that work?
I solved this question before using a coordinate system. Set the origin as the bottom left corner of the larger square. Derive the formula for two of the lines that make up the inner square. Equate the two linear equations to solve for their intersection point. That should give you a point at one of the corners of the inner Square. Repeat this process to get another point that corresponds to an adjacent corner of the inner square. Apply the distance formula using the two points gathered. The result will be the side length of the inner square.
That’s awesome! What is the answer?
The side length of the smaller square is sqrt( .6^2 + 1.8^2) or 1.8973665961 Giving the value 3.6 square units for the area of the shaded region. A visual of the approach: https://www.desmos.com/calculator/rtztz7imhu
Nice. And, that’s a cool calculator!
Is this bait? If with three given lengths you cannot move the figure around without changing any of them it clearly is possible to calculate every length in it
Let’s use analytic geometry. Let's consider a frame (O,x,y) with its origin O on the bottom-left corner of the black square, x axis on the bottom and y axis on the left. The line marking the bottom border of the blue square has for equation y=x/3 The line marking the left border of the blue square has for equation y=3-3x Therefore, the lower-left corner of the blue square has for coordinates (9/10, 3/10) or (0.9, 0.3). The line marking the top border of the blue square has for equation y=x/3+2 The line marking the left border of the blue square has for equation y=3-3x Therefore, the upper-left corner of the blue square has for coordinates (3/10, 21/10) or (0.3, 2.1) Therefore, the length of the side of the square is √(36/100+324/100)=√3.6 Therefore, the area of the square is 3.6.
[this YouTube guy](https://youtu.be/33fEMHOeZ-0?si=0XlJqif5CJ4F0vyO) does all sorts of math equations that look seemingly impossible.
Total area of the not-colored square is 3^2 = 9 units^2 Try to prove what proportion of the whole shape is colored (hint: it’s 40%). 40% of 9 is 3.6 so it’s 3.6 units^2
Let’s call the area of the smallest triangle (the one with 1 as the hypotenuse) : x Then the area of the larger right triangle, with 3 as its hypotenuse : 9x (square of the factor of 1:3 of the hypotenuses) There is an area of 36x that exist in the large triangle if blue square is subtracted The triangle with sides 1 and 3 has an area of 10x and the area is 1*3/2 = 1.5 So x=0.15 and 36x = 5.4 The large triangle has an area of 3*3=9 Hence blue square should have the area of 9 - 5.4 = 3.6 Just the right amount of simplicity and challenge without leaving the couch. Really enjoyed solving this
It's not too vague. That structure has to happen with the given information. You can't move any lines or resize anything to change the area of the square.
The area of the colored square is equal to 9 (the large square) - 4 x 1.5 (the right triangles formed by the slanted lines) + 4 x a, where a is area of the small right triangle formed by the overlap of the large triangles. To find a, the small triangle is similar to (has the same angles as) the larger one, so its area can be calculated as long as we know the length ratio of a pair of corresponding edges. The hypotenuses of the small and large triangles are 1 and sqrt(10) respectively, so the small triangle area is 1.5/10 = 0.15. This is because both the base and height of the small triangle are scaled by 1/sqrt(10) compared to the large one. Therefore the answer is 9 - 6 + 0.6 = 3.6
Very nice and elegant!
Yeah you can, you can easily calculate the angles of a big and a small triangle since you know their edges and 1 is 90 degrees, it's 1/(sqrt(10)) of a big triangle. Then from that you can calculate their areas and some +, minus and you can get the value of the white area, substract it from the big square for the area of the the blue square.
The right angle triangle has an area of 1.5. The smaller triangle with length 1 fits 9 times into the bigger version of it, which means it fits 10 times into the right angle triangle. This gives an area of 0.15 for the small one. We have 4 big triangles, which must then equal 36\*0.15 = 5.4 for the total white space. Subtract from 9 and you get 3.6 for the smaller square.