instead of using the law of cosines, side c that is being solved for can be found more quickly with the pythagorean theorem. its touching the bottom of the small square and the top left of the medium square so its length is equal to the square root of 2^2 + 6^2
@@joevano You can draw a vertical line from the lowest point of the diagonal square to the midpoint of the top edge of the first square. This forms a right triangle with known side lengths of 2 and 6.
I’m proud of myself! I did it a little differently (not using law of cosines) and got the right answer all in my head. I’ve been enjoying your videos lately 😁
Did you drop a vertical line from the base of the 4x4 square to the bottom point of the rhombus and use Pythagorean theorem again to get that side length of the largest square? That's what I did.
@@chinanumberone7723 hey I don't know if you had a bad day or something, but being happy and sharing that happiness isn't selfish. I'm sorry that I triggered you. At the same time, I never said anything about people being allowed to comment so I'm confused, would you mind clarifying?
@@jonathanlegendre279 Yeah! I thought that that method would be easier for me and it really does seem so much simpler. I'm glad someone else saw that method too :)
I just used the Pythagorean theorem a bunch. Everything about sines and cosines and such has escaped my mind since school. The diagonal of the medium square is 4*sqrt(2), half of that is the side length of the small square (whose diagonal is 4). So the short side of the blue triangle is 2*sqrt(2). With parts of these squares, the side of the large square forms the hypotenuse of a right triangle with sides 2 and 6. The large square's side length is 2*sqrt(10). Then you get the missing side length of the blue triangle via Pythagoras again, that's sqrt(32). Half base times height [½ * sqrt(32) * 2*sqrt(2)] and you get an area of 8 cm^2.
Solved with geometry: Use a^2 + b^2 = c^2 to calculate the smallest square side length as sqrt8. Noting that we have 3 squares in the diagram, the diagonal across the medium square can be completed up to its top left corner, and be of total length 2sqrt8. We now have an identical triangle to the blue one, with short side length sqrt8 and perpendicular medium side length 2sqrt8. Area = 1/2 x base x height A = 1/2 x 2sqrt8 x sqrt8 1/2 x 2 cancels, so A = sqrt8 squared A = 8 cm^2
Question: why does he check if the smallest square is actually a square, when all of the sides are equal? Did I miss something? Edit: I just realised why he called it a rhombus. 4 equal sides doesn't necessarily mean square.
A generalized solution: Let a=side of the smallest square; the area of the blue triangle is always a^2. Proof: Let ABC the blue right triangle, A=90°, AC=a. Let D the upper right corner of the medium square. We can easily prove that B, A, D are collinear, thus BCD is an isoceles triangle (BC=CD=CE, E is the upper left corner of the medium square), and AB=AD= 2a.
Faster solution without trigonometry: - diagonal of big square (z) = 2c^2 - connect (a) to the middle of the medium square and have a new side called (a + b), and generate a new rectangular triangle made of (z), (a+b) and (b) - z^2 = 2c^2 - z^2 = (a+b)^2 + b^2 - c^2 = a^2 + b^2 It gives at the end that a^2 = 32
I think I can give a creative solution Say the blue triangle with corners ABC, Boeing AB the side of the small square (you solved as x), and AC the other catheto (lets name it y), and BC the hypotenuse (lets name it z) equal to the side of the large square The area of the blue triangle would be 1/2(xy) Lets name the rightmost common point between the medium and small sqare D, and the common point between the medium and large square E You'll now have 2 congruent right triangles, ABC and DBE (AB=BD=x, BC=BE=z, therefore AC=DE=y) Now, you'll have another right triangle ADE, with AD=AE=4, where we can also conclude that DE=2x So, the final blue triangle area is going to be 2x×x/2=x²=8
i have alternative solution instead using cosine law, i only use pytaghorean. We see there is triangle in 1:10, we can use that to find half length of the sides of biggest square =sqrt( sqrt(8) + sqrt(8)/2) = sqrt(10) So the full length is 2*sqrt(10) = sqrt(40) Then continue to this 2:49
Strictly speaking, you have to explain the transition from the right angle in the blue triangle, to the right angle in the "smallest rombus", because even tho it looks that way, the fact the bigger cathetus aligns with the side of the smaller rombus to form a straight line has to be proven
This also implies that if the square with the original defined side length instead has side length x, the blue triangle derived will have area equal to half the square's area, or (x^2)/2.
@@carlosmp2043 Because if we scale it up or down by any multiple k, the areas will remain in proportion. The original square has area 4*4=16 units squares, and we worked out the blue triangle has area 8 units square. Thus, it will always be that the ration of triangular area to square area will be 8/18=1/2
@@themathhatter5290 it should make sense intuitively I agree since everything depends on the original square, but it's still bold to make the assumption that it will always be half, since it could follow some other ratio (Could be for example 4+(x^2)/4 or some other random stuff that would work for x=4 specifically but not other values of x). I feel like a bit of more formal proof would be needed to make that statement but Im not sure
@@carlosmp2043 If you would like to do the whole proof again but with a side length "Z" instead of four, go ahead! It'll be very rewarding, I assure you.
@@themathhatter5290 I believe ya, but trying it with 2, 10 or even a million variables in not sufficient mathematical proof if we play by the rules lol, more empirical proof is needed. Guess I'm just nitpicking here tho, I do agree that it should make sense.
Side of medium square is b=4. 1. Side on small square is a=b sin(45)=2sqrt(2) 2. Mirror small square along right side of triangle. 3. Prove, that new triangle is equal to original one. It's a right angled triangle with hypothenuse, equal to side of big square. And side equal to small square. 4. Since one side is a and other is 2a, S=2a²/2= a²=8 Completely no reason for equations.
i did it by just figuring out the side of the rhombus, then making a right triangle with the shortest side of the blue triangle and half of the side of the big square to figure out that half, multiplied it by two, and did pythagorean on the blue triangle. in my opinion the way i solved it is a little better since it doesn't make use of the law of cosines and thus angles as a way of calculating sides but yours is still nice
Huh, I had modeled the triple-ticked line segment as a vector, basing its horizontal and vertical components off the dimensions of the squares it touches the corners of. Got the same solution, though.
I think Mr.Math kinda overcomplicated it. smallest square side = 4/sqrt(2) diagonal of medium square= 8/sqrt(2) biggest square side= sqrt((8/sqrt(2))^2+(4/sqrt(2))^2) hyp. of blue triangle= biggest square side smallest side of blue triangle= smallest square side then you can easily solve for third side and area of the blue triangle without law of cosines. Yes, I know, I didn't provide proof that the diagonal of the medium square is equal to double the length of the little square but I can guarantee that it is, work it out on paper for yourself.
(note: I'm labelling the squares based on the number of lines that cross to denote equivalent disntances) buffer buffer buffer? I hardly know 'er based on the length of square 1, we can find the side length of square 2 using pythag's theorum (a²+b² = 16, a = b, 2a² = 16, a² = 8, a = √8). from that, we can find the right triangle that would define the length of square 3, as square 2 is rotated 45° compared to square 1. find the (invisible) legs of the isoceles/right triangle using the same method as above (2c² = 8, c² = 4, c = 2) the right triangle described is a 4+2 by 2 triangle (that is, 6 by 2) and we can PT this. we get d = √40. we now have the hypotenus (d) and one leg (a) of the blue triangle. PT but weird. 40 - 8 = 32, e = √32 (√32 * √8)/2 -> (√256)/2 -> 16/2 thus area = 8
0:20 Nice solution! Just a note: we already know the shape with diagonal length 4 is a square because 3 of its 4 lines have the double hash marks. That means 3/4 of the sides are the same length making it a square automatically. No need for the right angles.
Three sides having the same length does not necessarily mean it's a square. A trapezoid could have three identical side lengths with a longer base, as an example. Four identical side lengths could be a rhombus, so we still need to double check to confirm!
@@rudozawa The diagram is drawn incorrectly. My understanding is the perpendicular line means the side length is the same. The rhombus has 4 sides of equal length,, just the notation for it is crossing with the line of the large square.
Instead of using law of cosines we can jst construct a line joining top of the kite like square to the left corner of the square of side 4 cm And then we can find the side of the biggest square... Here one side is known and we can find the second one by Pythagorean theorem and then we can find its hypotenuse which the side of the biggest square by Pythagorean theorem....🧠 :-)
Bro we can do the math without law of cosines. We can prove that the side of triangle of the" 4cm "square that it is the half of the square's diagonal and join the corner with the side of the triangle with an imaginary line. This forms a right angled triangle and we can find the value of the side of the big square.
i got scared with the thumb, at least on my side the measure is cut out of the image clicked to see how you would solve for the area of something without units in under 4min lol amazing video
Ever thought of doing a math compile of rules to functions and thought process behind it... since you added brackets wit ha divisional sign but it was fraction so you didn't multiply it and some people are ocnfused why everything in teh bracket is not computed... the clarity or lack of it in math grammar is sometimes a thing ,but it would be an interesting rule set... and maybe peopel might learn something... Other methods for quadratic equations, to solve such as factoring, binary method I thought was my favourtie way for a x=0 equation.. lol Ther are so amny like bingo box thing.. and many others. lol nothing really teaches... actual reason behind it but its a fucnitonal work rather than teh logistics of pattern systems in question whihc ould make most humans a geunius that besides the point ,the ways of the math.. to solve complex questions... would be interesting. Helpful sets to know... common pairs and bonds.. this owuld help a lot of peopel maybe upgrading also... its liek shortcuts.
Can’t you just say that the side that is sqrt(32) is double the length of sqrt(8) so 2*sqrt(8) which is sqrt(32)? Should be obvious if you flip the blue triangle along its shortest side. Sou you don’t have to calculate the sqrt(40)?
Law of cosines??? My brother in Christ just draw the III line from the rightmost corner of the blue triangle to the topright corner of the medium square, and extend the II line. You now have the blue triangle, but mirrored along the short side. The base is twice as big as the small square and the height is the same, so the surface of the blue triangle is the same as the small square's.
i hope im correct with 2.83 ima watch the video now edit 4 seconds in and i realized i forgor to square the 4, 5.66 edit 2, im stupid and none of my answers are correct because of a non-right triangle
instead of using the law of cosines, side c that is being solved for can be found more quickly with the pythagorean theorem. its touching the bottom of the small square and the top left of the medium square so its length is equal to the square root of 2^2 + 6^2
explain it better pls
Where is the right triangle you are using the Pythagorean theorem on?
@@joevano You can draw a vertical line from the lowest point of the diagonal square to the midpoint of the top edge of the first square. This forms a right triangle with known side lengths of 2 and 6.
@@joevano divide the small square into pieces.
@@rudozawanot seeing it at all… and you do know there are 2 diagonal squares…
I’m proud of myself! I did it a little differently (not using law of cosines) and got the right answer all in my head. I’ve been enjoying your videos lately 😁
Did you drop a vertical line from the base of the 4x4 square to the bottom point of the rhombus and use Pythagorean theorem again to get that side length of the largest square? That's what I did.
Someone’s happy. Only comment on this channel, all because Lily Williams is happy 🙄
Seems like we have a seIfish person folks
@@chinanumberone7723you're just mad that you can't solve it yourself
@@chinanumberone7723 hey I don't know if you had a bad day or something, but being happy and sharing that happiness isn't selfish. I'm sorry that I triggered you. At the same time, I never said anything about people being allowed to comment so I'm confused, would you mind clarifying?
@@jonathanlegendre279 Yeah! I thought that that method would be easier for me and it really does seem so much simpler. I'm glad someone else saw that method too :)
Andy, congrats on the 100K subscribers…..”How exciting!”
Love the vids, keep up the great work.
I just used the Pythagorean theorem a bunch. Everything about sines and cosines and such has escaped my mind since school.
The diagonal of the medium square is 4*sqrt(2), half of that is the side length of the small square (whose diagonal is 4). So the short side of the blue triangle is 2*sqrt(2).
With parts of these squares, the side of the large square forms the hypotenuse of a right triangle with sides 2 and 6. The large square's side length is 2*sqrt(10).
Then you get the missing side length of the blue triangle via Pythagoras again, that's sqrt(32). Half base times height [½ * sqrt(32) * 2*sqrt(2)] and you get an area of 8 cm^2.
Solved with geometry:
Use a^2 + b^2 = c^2 to calculate the smallest square side length as sqrt8.
Noting that we have 3 squares in the diagram, the diagonal across the medium square can be completed up to its top left corner, and be of total length 2sqrt8.
We now have an identical triangle to the blue one, with short side length sqrt8 and perpendicular medium side length 2sqrt8.
Area = 1/2 x base x height
A = 1/2 x 2sqrt8 x sqrt8
1/2 x 2 cancels, so A = sqrt8 squared
A = 8 cm^2
Which is the hypotenuse in this triangle you’re constructing?
@@bectionary right hand side of the largest square.
This is the first video I’ve ever attempted one of these questions and I happened to almost get it correct
I love your videos. For fun or studies they are the best ni doubt, just like you. Keep being amazing.
Question: why does he check if the smallest square is actually a square, when all of the sides are equal? Did I miss something?
Edit: I just realised why he called it a rhombus. 4 equal sides doesn't necessarily mean square.
A generalized solution: Let a=side of the smallest square; the area of the blue triangle is always a^2.
Proof: Let ABC the blue right triangle, A=90°, AC=a. Let D the upper right corner of the medium square. We can easily prove that B, A, D are collinear, thus BCD is an isoceles triangle (BC=CD=CE, E is the upper left corner of the medium square), and AB=AD= 2a.
How exciting indeed🤗
I like this man
Im not sure what i like more, the solving of the equation or how exciting it is!
Simple, clean math
100k subscribers for my favorite youtuber!!! Congratulations Andy. How exciting.
Congrats on 100k Andy!
Thank you! It makes me happy you were looking at the counts! It happened last night.
Got everything first try except the theta angle in the laws of cosine part
Although i understood it after your explanation
Faster solution without trigonometry:
- diagonal of big square (z) = 2c^2
- connect (a) to the middle of the medium square and have a new side called (a + b), and generate a new rectangular triangle made of (z), (a+b) and (b)
- z^2 = 2c^2
- z^2 = (a+b)^2 + b^2
- c^2 = a^2 + b^2
It gives at the end that a^2 = 32
I think I can give a creative solution
Say the blue triangle with corners ABC, Boeing AB the side of the small square (you solved as x), and AC the other catheto (lets name it y), and BC the hypotenuse (lets name it z) equal to the side of the large square
The area of the blue triangle would be 1/2(xy)
Lets name the rightmost common point between the medium and small sqare D, and the common point between the medium and large square E
You'll now have 2 congruent right triangles, ABC and DBE (AB=BD=x, BC=BE=z, therefore AC=DE=y)
Now, you'll have another right triangle ADE, with AD=AE=4, where we can also conclude that DE=2x
So, the final blue triangle area is going to be 2x×x/2=x²=8
Really like your channel. Thanks for the content
i have alternative solution instead using cosine law, i only use pytaghorean.
We see there is triangle in 1:10, we can use that to find half length of the sides of biggest square
=sqrt( sqrt(8) + sqrt(8)/2) = sqrt(10)
So the full length is 2*sqrt(10) = sqrt(40)
Then continue to this 2:49
Explain clearly
Agreed
That one was hard. Nice.
Strictly speaking, you have to explain the transition from the right angle in the blue triangle, to the right angle in the "smallest rombus", because even tho it looks that way, the fact the bigger cathetus aligns with the side of the smaller rombus to form a straight line has to be proven
This also implies that if the square with the original defined side length instead has side length x, the blue triangle derived will have area equal to half the square's area, or (x^2)/2.
How can you generalise that so confidently lol
@@carlosmp2043 Because if we scale it up or down by any multiple k, the areas will remain in proportion. The original square has area 4*4=16 units squares, and we worked out the blue triangle has area 8 units square. Thus, it will always be that the ration of triangular area to square area will be 8/18=1/2
@@themathhatter5290 it should make sense intuitively I agree since everything depends on the original square, but it's still bold to make the assumption that it will always be half, since it could follow some other ratio (Could be for example 4+(x^2)/4 or some other random stuff that would work for x=4 specifically but not other values of x). I feel like a bit of more formal proof would be needed to make that statement but Im not sure
@@carlosmp2043 If you would like to do the whole proof again but with a side length "Z" instead of four, go ahead! It'll be very rewarding, I assure you.
@@themathhatter5290 I believe ya, but trying it with 2, 10 or even a million variables in not sufficient mathematical proof if we play by the rules lol, more empirical proof is needed. Guess I'm just nitpicking here tho, I do agree that it should make sense.
Side of medium square is b=4.
1. Side on small square is a=b sin(45)=2sqrt(2)
2. Mirror small square along right side of triangle.
3. Prove, that new triangle is equal to original one.
It's a right angled triangle with hypothenuse, equal to side of big square. And side equal to small square.
4. Since one side is a and other is 2a, S=2a²/2= a²=8
Completely no reason for equations.
i did it by just figuring out the side of the rhombus, then making a right triangle with the shortest side of the blue triangle and half of the side of the big square to figure out that half, multiplied it by two, and did pythagorean on the blue triangle. in my opinion the way i solved it is a little better since it doesn't make use of the law of cosines and thus angles as a way of calculating sides but yours is still nice
Huh, I had modeled the triple-ticked line segment as a vector, basing its horizontal and vertical components off the dimensions of the squares it touches the corners of. Got the same solution, though.
How exciting
I think Mr.Math kinda overcomplicated it. smallest square side = 4/sqrt(2) diagonal of medium square= 8/sqrt(2) biggest square side= sqrt((8/sqrt(2))^2+(4/sqrt(2))^2) hyp. of blue triangle= biggest square side smallest side of blue triangle= smallest square side then you can easily solve for third side and area of the blue triangle without law of cosines. Yes, I know, I didn't provide proof that the diagonal of the medium square is equal to double the length of the little square but I can guarantee that it is, work it out on paper for yourself.
(note: I'm labelling the squares based on the number of lines that cross to denote equivalent disntances)
buffer
buffer
buffer? I hardly know 'er
based on the length of square 1, we can find the side length of square 2 using pythag's theorum (a²+b² = 16, a = b, 2a² = 16, a² = 8, a = √8).
from that, we can find the right triangle that would define the length of square 3, as square 2 is rotated 45° compared to square 1. find the (invisible) legs of the isoceles/right triangle using the same method as above (2c² = 8, c² = 4, c = 2)
the right triangle described is a 4+2 by 2 triangle (that is, 6 by 2) and we can PT this. we get d = √40.
we now have the hypotenus (d) and one leg (a) of the blue triangle. PT but weird. 40 - 8 = 32, e = √32
(√32 * √8)/2 -> (√256)/2 -> 16/2
thus
area = 8
I used trig and got slightly less than 8 due to rounding errors, it was easier, but it makes sense to do it your way and get an exact answer.
i would have never thought that one of my favorite yt channels would be a guy doing maths
Love your videos
trig is my favorite subsection of math
Man, I can't unsee it being 3d
congrats on the 100k subscribers. it has been a long journey. keep making great videos.
Thank you very much!
Congratulations on 100k
Thank you!!
Just put another double-crossed line from the center of the medium square to its upper left corner and boom!
Amazing 🎉
Congratulations on 100K 🥳🥳
Thank you!!
Which software did you used to make these exciting videos.😊
Nice one! :)
0:20 Nice solution! Just a note: we already know the shape with diagonal length 4 is a square because 3 of its 4 lines have the double hash marks. That means 3/4 of the sides are the same length making it a square automatically. No need for the right angles.
Three sides having the same length does not necessarily mean it's a square. A trapezoid could have three identical side lengths with a longer base, as an example. Four identical side lengths could be a rhombus, so we still need to double check to confirm!
@@rudozawa The diagram is drawn incorrectly. My understanding is the perpendicular line means the side length is the same. The rhombus has 4 sides of equal length,, just the notation for it is crossing with the line of the large square.
Math is so so so beautiful
Wow Really How Exciting 🎉
thank you
good math
Instead of using law of cosines we can jst construct a line joining top of the kite like square to the left corner of the square of side 4 cm
And then we can find the side of the biggest square...
Here one side is known and we can find the second one by Pythagorean theorem and then we can find its hypotenuse which the side of the biggest square by Pythagorean theorem....🧠 :-)
Bro we can do the math without law of cosines. We can prove that the side of triangle of the" 4cm "square that it is the half of the square's diagonal and join the corner with the side of the triangle with an imaginary line. This forms a right angled triangle and we can find the value of the side of the big square.
i got scared with the thumb, at least on my side the measure is cut out of the image
clicked to see how you would solve for the area of something without units in under 4min lol
amazing video
Ever thought of doing a math compile of rules to functions and thought process behind it... since you added brackets wit ha divisional sign but it was fraction so you didn't multiply it and some people are ocnfused why everything in teh bracket is not computed... the clarity or lack of it in math grammar is sometimes a thing ,but it would be an interesting rule set... and maybe peopel might learn something...
Other methods for quadratic equations, to solve such as factoring, binary method I thought was my favourtie way for a x=0 equation.. lol Ther are so amny like bingo box thing.. and many others. lol nothing really teaches... actual reason behind it but its a fucnitonal work rather than teh logistics of pattern systems in question whihc ould make most humans a geunius that besides the point ,the ways of the math.. to solve complex questions... would be interesting. Helpful sets to know... common pairs and bonds.. this owuld help a lot of peopel maybe upgrading also... its liek shortcuts.
The law of cosines… man that brings me back to highschool…
amazing video as always
ez. solven in my head
Can’t you just say that the side that is sqrt(32) is double the length of sqrt(8) so 2*sqrt(8) which is sqrt(32)? Should be obvious if you flip the blue triangle along its shortest side. Sou you don’t have to calculate the sqrt(40)?
i think there's not enough proof to do that
chill bruv my girl is on this app
Can you make a channel called Andy Meth? Where you solve drug math problems on meth?
Hmmm...I have another solution. I will review it. 😃
Law of cosines??? My brother in Christ just draw the III line from the rightmost corner of the blue triangle to the topright corner of the medium square, and extend the II line.
You now have the blue triangle, but mirrored along the short side. The base is twice as big as the small square and the height is the same, so the surface of the blue triangle is the same as the small square's.
i hope im correct with 2.83 ima watch the video now
edit 4 seconds in and i realized i forgor to square the 4, 5.66
edit 2, im stupid and none of my answers are correct because of a non-right triangle
100k ❤🎉
I can’t believe i did everything right and then forgot to divide the final area by 2
There is absolutely NO reason to use the Law of cosines here
I bet you can't solve me! My parents said I was the biggest problem...
i could actually do this
😢😮😮😮😢
ruclips.net/video/6lCwFyE-Nm0/видео.html This is another way to solve this problem which I really liked. Kudos to this Indian guy!
I got 8cm^2
I used different way to caltulate it:
1) 4^2=16 -> 16:2=8 V8
2) V8^2 + 0.5V8^2= 10
3) 2V10^2 - V8^2= 32
4) 2V8*V8:2= 8cm^2
No calculator, no paper🙂
V
How dare you trick us into using trig. Disgusting.
Im like 69
Congrats on 100k subs :D !
Thank you so much! 😀 I can't believe it.
How exciting