@@antoniomaurer3746It's congruent because SAS (Side(x) Angle(90°) Side(x)) is known . Since those 2 known sides are the same size (x) then the angles of the other corners are known to be 45° and 45°.
The great/frustrating thing about geometry is, if you can't think of a clever solution, you can always just turn it into a bunch of vector equations and solve it that way. It won't be as elegant as the easier "intended" solution, but not all real world problems have an easy solutions, so in some cases, you're better off just not trying to look for an elegant way to do it, and just plugging everything into vectors. ¯\_(ツ)_/¯ In this case, you'd do that by solving the position of the bottom most corner (C), and the two corners that intersect the circumference (A and B). C•ĵ=0, |B-C|²=4|A-C|², |A|²=|B|²=1, then when you find all 3, just compute ((A-C)×(B-C))•k, or just 2|A-C|², or just |B-C|²/4. Still, I like your method better. :)
That's not a great/frustrating thing about geometry. Because it is NOT a thing about geometry. We translate things from geometry to algebra, and from algebra to geometry. But simple things can get very complicated.
@@barisdogru6437 Geometry is NOT the study of life. That would be Biology. And even if Geometry was the study of life, this wouldn't imply there are easy and hard ways to the same solution. That's just a conjecture.
Very cool problem! I did it with coordinate geometry. Three of the corners of the rectangle are at (-s,s), (0,s), and (s,0). The circle has equation (x-h)² + (y-k)² = 25 and goes through the three listed points. Therefore you get three equations with three unknowns: (s-h)² + k² = 25 (s+h)² + (s-k)² = 25 h² + (s-k)² = 25 These are easy to solve by elimination. You get s = √10, and so area = 20.
Cool! I've done about 10 of these now and it's a blast having all of this coming back to me. I've only figured out 2 of them, but I'm 67, so I'm feeling a bit cocky. I'm pretty sure I've already figured out a way to save 10 minutes of time mowing my yard more efficiently. I'm finally taking control of my life with Mathematics. I was beginning to lose hope. Thanks for the videos!
I struggled with this a bit until I realized that the 3 points of contact with the circle define it, and thus its radius. That the lower left corner is coincident with the diameter chord is irrelevant and a distraction and doesn't affect the answer. I plotted the 3 points on the x-y plane at (-x, x), (0, x), and (x,0) and solved the simultaneous equations for the radius r. With r=5, the area 2x^2 is then 20.
Lovely solution. I approached it by drawing a coordinate system aligned with the rectangle. I gave the rectangle's vertices the coordinates (0,0), (0,2a), (4a,2a), and (4a,0). We know that the circle passes through (0,2a), (2a,2a), and (4a,0). The first two of these have perpendicular bisector x=a, while the last two of these have perpendicular bisector x-y=2a. These lines meet at (a,-a), which must therefore be the center of the circle. Now pick any of the three points of contact; its distance from that center is a*sqrt(10) by the Pythagorean Theorem, which must match the radius of 5. Therefore 10a^2=25, so 2a^2=5, so 8a^2=20. That's the area.
Found it in an easier way by finding the slant of the rectangle then define 1 smaller edge as x. Then you can pass a line equal to x in the middle of the rectangle cutting it into 2 squares then you will see a right triangle with x, x/2 and 5sqrt(2)/2. Then use Pythagorean theorem: x^2 + (x/2)^2 = (5sqrt(2)/2)^2 => x^2 = 10 Area = x*(x+x) = x*2x = 2x^2 = 2*10 = 20 This is briefly explained so sorry if it’s unclear what I did.
1:36 thank you for not losing me there. I'm not smart, but I love to learn to some degree. It really keeps my attention when you make every explanation visible and not imaginative. Thank you sir, I wish I had you as my math teacher.
There's a simpler way to do this without using all the complicated angle theorems. By symmetry you can extend the top right side of the rectangle down to create another chord of length x. You can then draw lines out from the centre of the circle that intersect each chord at right angles. This constructs a square with side length (3/2)x. Then we can create a right angle triangle with sides (3/2)x, (1/2)x and the radius (5) as the hypotenuse. Solving using Pythagoras' theorem gives x = sqrt(10). Hence the rectangular area is 20.
This is my solution: If we extend the longer sides of the rectangle, the intersection points with the circle create 2 parallel cords with lengths which can be shown to be x and 3x and distance between them x. For a cord we have that (c/2)^2+h^2=r^2, where c is the length of the cord, h is the distance from the center of the circle and r is the radius. So for our 2 cords we get that (x/2)^2+(h+x)^2=r^2 and (3x/2)^2+h^2=r^2, from where we find that h=x/2 and the area of the rectangle A=2x^2=(4/5)r^2 therefore A=20 if r=5.
I was an honors math student (including geometry) back in the 1970's. For the life of me I don't ever recall learning the subtended angle on a circle thing. Ever. Head exploded. How exciting.
There is an easier solution for this, draw the diagonal line in the rectangle, connect r on both end of the diagonal line, draw a perpendicular line on the diagonal line (it must be at the exact half of the diagonal line, since both end = r), then you can calculate 1/2 diagonal line = 5 * cos(45) And it's straight forward from here EDIT: the diagonal line in my reply = the green line in video
If a and 2a are the lenghts of rectangle, tracing the other segment from middle upper side to left vertex of the rectangle we have 45º angles and between them it forms a 90º angle and both lines lenghts are √2a. Extend left segment below and the point where it touches the circle with the other vertex of base of rectangle, ir forms a rectangle triangle with a diameter of lenght 10 as hypotenuse. Draw the diagonal of blue rectangle from two points touching the circle, has lenght of √5a, then unite the below point of diameter at circle with most left point of rectangle touching the circle and it forms another 90º rectangle triangle. The left side of this one is a chord who is sustained by a 45º inscribed angle inside blue rectangle, and is sustained also by inscribed angle formed from the diameter and the other side so this one is also 45º, then the other angle must be 45º too so the left side lenght is also √5a and we have an isosceles rectangle triangle with 45º45º90º and sides √5a and hypotenuse 10. Finally by Pythagoras, (√2)(√5a)=10, a=√10, and Blue rectangle area = a(2a)=2a²=2(10)=20.
This is the only use of all the AP and college level math I do. I have never used the pythagorean theorem, Quadratic equations, Matrixes, in real life. the biggest problem i'm using mathing skills is finding the concentration for a solution usually milk or half and half.
I like to try to solve these videos from just watching the thumbnail, was absolutely stumped on this one . Then I watched it and saw the part about the angles being subtended by the minor arc , I definitely didn’t know this so I could breathe a sigh of relief
I feel like these videos should be available for TH-cam kids but personally also want the comments for these videos. So maybe it could be a great idea for Andy to make a separate channel for TH-cam kids but upload the same videos
Hi Andy My guess is that you never did much technical drawing in high school? There is a geometric way which is much shorter and only involves the Pythag theorum. Consider the three points on the circumference. (The corner of the rectangle located on the diametral chord is just a magician's distraction as the diamertral chord can be drawn anywhere.) For some, it may be less confusing just to delete the diametral chord and draw the full circle and reorientate the rectangle so that the long side is horizontal. From geometry the center of any circle lies on the perpendicular bisector of any chord. Let the short side of the rectangle be X and the long side 2X. If these bisectors are drawn the angles and lengths are simple as they are all 45 degrees. From observation the centre of the circle lies X/2 past the rectangle's long side along to the intersection of the two perpendicular disectors. From here, use Pythag to calculate the radius in terms of X. The problem just falls out after that. Cheers
I knew there would be some solution using subtended angles but I was lazy. So I just dropped a perpendicular from the centre to the chord thus dividing the longer side of the rectangle into x and 3x and the shorter side is 2x. Then found its length as sqrt(25-x^2). Then used the Pythagoras theorem on the smaller triangle formed by joining the centre and the with the other end of the rectangle having side lengths 3x, sqrt(25-x^2)-2x, 5. Which gives x^2 = 2.5 and area = 8x^2 = 20.
My teenage self would've ripped off a piece of the test paper, measured the 10, compared it to any of the portions of the rectangle in the hopes they matched up, and made an educated guess based on the multiple choices provided.
make a full circle, flip the rectangle about the noted diameter, this acts to extend the small side till it hits the other edge of the circle, this gives us a right angle triangle with a corner on the edge so its hypotenuse is the diameter by basic circle theorems, so letting the small side be x we have x^2 + (3x)^2 = d^2 x^2 = d^2/10 so x is root(10) so the area is 20
I got the right answer by finding the chord length of the diagonal of the rectangle, which is equal to 2*radius *(sin c/2), where c equal the subtended angle from the center of the circle, which is 90 degrees. *5* sin. 45= 7.07107. The short side and long side are in a ratio of 1:2. Inverse tan 1/2. = 26.565 degrees, so sin 26.565. * 7.07107= 3.16228 and cos 26.565 * 7.07107 = 6.32456. 3.16228* 6.32456= 20
hiiii! i hav a doubt... see aftr forming the right triganle cant we do Tan 45 and find the breadth of the rectangle. and since the lenght is twice the breadth, now we have the breadth too. now we can use the formula for area of rectangle to get the answer? btw luv the way u explain ❤
I have a question, considering the short side as x and the longer as 2x, wouldnt x=10/3? Isn't the sum of the sides of the rectangle forced to be the same length as the diameter?
Take a thread cut it to the diameter of the circle, cut it into 10 equal pieces, use a piece to measure the length and width of the rectangle, multiply the length and width, you got Area of the rectangle.
These math and science videos have become a refuge for me since I've become repulsed by mainstream media, identity politics and such. I can feel my brain healing as I learn. You are gem in an insane world.
i would advise you to try an actually solve the problem before watching the solution then, my brain is very good at pretending like it learned while it actually didn't learn anything lol, at least that's my experience
@@shem7146 Hey great advice. Today I drove out to the middle of a national forest and walked into the woods. It was outstanding I'm doing it again. Also music is good.
Again, this problem can be solved more easily with coordinate geometry. Choose the coordinate system such that its origin is at marked corner of the rectangle and the x and y axes contain the long and short edges of it respectively. Now we can write 3 equations describing the points that lie on the circle. Those points have the coordinates: (0,x), (x,x) and (2x,0). Lets denote the center of the circle as (u,v), then the system of equations is: u²+(x-v)² = r² (x-u)²+(x-v)² = r² (2x-u)²+v² = r². Solve it for x,u,v (r=10 is known): u = -v = r√10/10 x = r√10/20. From this, the blue area is: A = 2x² = r²/5 = 20.
@@pixtane7427 Central/inscribed angles come way later than 7-8th grade. It's taught in 10th grade (for 15-16yrs old students) at least in my country. But you are right in the sense that coordinate geometry is usually taught even later.
![[TH]FFWS SEA Spring 2024 Point Rush - Day 2](http://i.ytimg.com/vi/uTzfNlXO9bg/mqdefault.jpg)
This guy is so chill while teaching , it's almost he's playing a game .😊
He is. With my mind.
You can look at math as a puzzle game
Yes
Learning should be fun. This guy gets that.
Math problems are basically puzzles.
He is Speedrunning math
I know it’s just geometry, but this guy does a really good job of explaining how he gets from one step to another in a way that anyone can understand.
nope i dont even understand the first part, why is that triangle congruent or w/e u call it
@@antoniomaurer3746It's congruent because SAS (Side(x) Angle(90°) Side(x)) is known . Since those 2 known sides are the same size (x) then the angles of the other corners are known to be 45° and 45°.
@@antoniomaurer3746 mid pfp
don't underestimate my incompetence in mathematics
I've learned more about math from a TH-cam channel than 4 years of college. How exciting.
This is middle school math
The great/frustrating thing about geometry is, if you can't think of a clever solution, you can always just turn it into a bunch of vector equations and solve it that way. It won't be as elegant as the easier "intended" solution, but not all real world problems have an easy solutions, so in some cases, you're better off just not trying to look for an elegant way to do it, and just plugging everything into vectors. ¯\_(ツ)_/¯
In this case, you'd do that by solving the position of the bottom most corner (C), and the two corners that intersect the circumference (A and B). C•ĵ=0, |B-C|²=4|A-C|², |A|²=|B|²=1, then when you find all 3, just compute ((A-C)×(B-C))•k, or just 2|A-C|², or just |B-C|²/4.
Still, I like your method better. :)
That's not a great/frustrating thing about geometry. Because it is NOT a thing about geometry. We translate things from geometry to algebra, and from algebra to geometry. But simple things can get very complicated.
Lol. Maths noob complaining he can't solve a simple problem. Go back to school son
I am turdboi
Well, geometry is, after all, the study of life, so there are easy and hard ways to the same solution.
@@barisdogru6437 Geometry is NOT the study of life. That would be Biology. And even if Geometry was the study of life, this wouldn't imply there are easy and hard ways to the same solution. That's just a conjecture.
I really love his way of teaching, you can see his love for maths through it, he also makes it look easy and lovable for others. Keep up !
Very cool problem! I did it with coordinate geometry. Three of the corners of the rectangle are at (-s,s), (0,s), and (s,0). The circle has equation (x-h)² + (y-k)² = 25 and goes through the three listed points. Therefore you get three equations with three unknowns:
(s-h)² + k² = 25
(s+h)² + (s-k)² = 25
h² + (s-k)² = 25
These are easy to solve by elimination. You get s = √10, and so area = 20.
Yep, you are really awesome. It is all about perspective, just rotate the coordinate system and make the right observations.
Cool! I've done about 10 of these now and it's a blast having all of this coming back to me. I've only figured out 2 of them, but I'm 67, so I'm feeling a bit cocky. I'm pretty sure I've already figured out a way to save 10 minutes of time mowing my yard more efficiently. I'm finally taking control of my life with Mathematics. I was beginning to lose hope. Thanks for the videos!
i like how you use simple algebra and concepts to solve these.
Love your energy!
That 'cool property' about the relationship between inscribed angles and arcs was new to me! Thank you for teaching me something new!
love watching these videos, makes me feel smarter than before.
I love these videos ❤ Ty for making them they’re awesome and intriguing + educational
I really like your videos. Always nice problems and the explanation is to the point.
I totally was not taught inscribed angles in all my math career and this was a great concept to learn. Got another tool in my kit thank you kind sir
These videos are really making me consider revisiting a geometry textbook. So much stuff I never learned or don’t remember.
I normally dont interact with channels but man you need to keep making these.
What software or app do you use for these? like for the graphics and demonstrations? or is it just editing? love your vids ❤
It looks like a powerpoint maybe
Microsoft Paint DLC
@@Oropel💀
I am loving it. Please keep up the good work.
I absolutely love these kinds of videos!
I struggled with this a bit until I realized that the 3 points of contact with the circle define it, and thus its radius. That the lower left corner is coincident with the diameter chord is irrelevant and a distraction and doesn't affect the answer. I plotted the 3 points on the x-y plane at (-x, x), (0, x), and (x,0) and solved the simultaneous equations for the radius r. With r=5, the area 2x^2 is then 20.
Very, very, very cool.
Yeah I didn't get that the 3rd point was, well, on the arc
Lovely solution.
I approached it by drawing a coordinate system aligned with the rectangle. I gave the rectangle's vertices the coordinates (0,0), (0,2a), (4a,2a), and (4a,0).
We know that the circle passes through (0,2a), (2a,2a), and (4a,0). The first two of these have perpendicular bisector x=a, while the last two of these have perpendicular bisector x-y=2a. These lines meet at (a,-a), which must therefore be the center of the circle.
Now pick any of the three points of contact; its distance from that center is a*sqrt(10) by the Pythagorean Theorem, which must match the radius of 5. Therefore 10a^2=25, so 2a^2=5, so 8a^2=20. That's the area.
I'm hooked on these videos.
Your videos make me feel smarter haha. They introduce new was of thinking to me
That was incredible, good work
Good job Andy! That was a tought one.
Quick satisfying and digestible. Love it. Got a new sub!
How exciting indeed! Good work sir.
Quick and interesting! Love geometry!👍❤️
Found it in an easier way by finding the slant of the rectangle then define 1 smaller edge as x. Then you can pass a line equal to x in the middle of the rectangle cutting it into 2 squares then you will see a right triangle with x, x/2 and 5sqrt(2)/2. Then use Pythagorean theorem: x^2 + (x/2)^2 = (5sqrt(2)/2)^2 => x^2 = 10
Area = x*(x+x) = x*2x = 2x^2 = 2*10 = 20
This is briefly explained so sorry if it’s unclear what I did.
Can you elaborate? Where did (5√2)/2 come from?
SO COOL!
I never knew that about inscribed angles!
I love you so much rn!
You make math fun!
1:36 thank you for not losing me there. I'm not smart, but I love to learn to some degree. It really keeps my attention when you make every explanation visible and not imaginative. Thank you sir, I wish I had you as my math teacher.
He’s very intelligent and he’s passionate about math! He would make one hell of a math professor in college.
just discovered your channel. i like this content please keep going
Loving your problems and solutions...
There's a simpler way to do this without using all the complicated angle theorems.
By symmetry you can extend the top right side of the rectangle down to create another chord of length x. You can then draw lines out from the centre of the circle that intersect each chord at right angles. This constructs a square with side length (3/2)x. Then we can create a right angle triangle with sides (3/2)x, (1/2)x and the radius (5) as the hypotenuse. Solving using Pythagoras' theorem gives x = sqrt(10). Hence the rectangular area is 20.
This is my solution: If we extend the longer sides of the rectangle, the intersection points with the circle create 2 parallel cords with lengths which can be shown to be x and 3x and distance between them x. For a cord we have that (c/2)^2+h^2=r^2, where c is the length of the cord, h is the distance from the center of the circle and r is the radius. So for our 2 cords we get that (x/2)^2+(h+x)^2=r^2 and (3x/2)^2+h^2=r^2, from where we find that h=x/2 and the area of the rectangle A=2x^2=(4/5)r^2 therefore A=20 if r=5.
66 years of age. Did not take my second Calc course until I was 53. Your channel makes this stuff easy to understand. Well done!
I was an honors math student (including geometry) back in the 1970's. For the life of me I don't ever recall learning the subtended angle on a circle thing. Ever.
Head exploded. How exciting.
this one was particularly wild. bro is NASTY with geometry 🔥🔥
Thank you very much for the explanation, I didn't know how to do it, but thanks to you I even understand it now 😀
Im not interested in math but he got me curious and speaks in a way like it was very interesting to everybody 😅
Definitely good subscription
That's quite an interesting solution Sir.
Good One.
I am very rusty on my math. I’m glad the algorithm brought me here. I will be practicing daily
Same lol, I'm in tertiary studies but completely forgot what the fuck an inscribed angle was
I am subscripting this channel for my son, I am sure he will watch this when he go to school, he's currently 13 months old :)
I don't know why this channel get less views
That's an amazing video
I hope you will get more viewers in future
Love from 🇮🇳🇮🇳India🎉❤
Great Video. Can you please suggest what software you use to create these videos. Looks kinda fun!!!
It took me 15 years to find my favorite TH-cam channel!
I'm not even a fan of math and yet I love every one of your videos.
There is an easier solution for this, draw the diagonal line in the rectangle, connect r on both end of the diagonal line, draw a perpendicular line on the diagonal line (it must be at the exact half of the diagonal line, since both end = r), then you can calculate 1/2 diagonal line = 5 * cos(45)
And it's straight forward from here
EDIT:
the diagonal line in my reply = the green line in video
Epic
Nice, just need to remember that they perpendicular bisector of a chord passes through the center
Love the videos man.
The way this dude smiles while explaining things… if he was my teacher growing up I probably would have cared about math 😂
Idk why are these videos so addictive
Do you have a definitions video? Getting lost on the terms, but the steps are immaculate!
Most excellent.
To be honest, mathematics has always been my favorite subject.
When Andy says "this is a fun one," you know it's gonna be a fun one!
If a and 2a are the lenghts of rectangle, tracing the other segment from middle upper side to left vertex of the rectangle we have 45º angles and between them it forms a 90º angle and both lines lenghts are √2a. Extend left segment below and the point where it touches the circle with the other vertex of base of rectangle, ir forms a rectangle triangle with a diameter of lenght 10 as hypotenuse. Draw the diagonal of blue rectangle from two points touching the circle, has lenght of √5a, then unite the below point of diameter at circle with most left point of rectangle touching the circle and it forms another 90º rectangle triangle. The left side of this one is a chord who is sustained by a 45º inscribed angle inside blue rectangle, and is sustained also by inscribed angle formed from the diameter and the other side so this one is also 45º, then the other angle must be 45º too so the left side lenght is also √5a and we have an isosceles rectangle triangle with 45º45º90º and sides √5a and hypotenuse 10. Finally by Pythagoras, (√2)(√5a)=10, a=√10, and Blue rectangle area = a(2a)=2a²=2(10)=20.
This kind of video should be showing up in everyone's recommended
Extremely impressive. I wish I knew all of this.
If I can't find a thing, I always look behind the fridge.
This is the only use of all the AP and college level math I do. I have never used the pythagorean theorem, Quadratic equations, Matrixes, in real life. the biggest problem i'm using mathing skills is finding the concentration for a solution usually milk or half and half.
I see the blue rectangle. It's right there, in front of a white background. That's the real area of the blue rectangle
Pleasantly exciting,!!
I like to try to solve these videos from just watching the thumbnail, was absolutely stumped on this one . Then I watched it and saw the part about the angles being subtended by the minor arc , I definitely didn’t know this so I could breathe a sigh of relief
Man ive learned more from your videos in yt than my teachers during classes
Beautiful
I feel like these videos should be available for TH-cam kids but personally also want the comments for these videos. So maybe it could be a great idea for Andy to make a separate channel for TH-cam kids but upload the same videos
I used a bit of Pythagoras and then similar triangles. Same answer. I didn’t complete the circle but perhaps doing that was more elegant.
Hi Andy My guess is that you never did much technical drawing in high school? There is a geometric way which is much shorter and only involves the Pythag theorum. Consider the three points on the circumference. (The corner of the rectangle located on the diametral chord is just a magician's distraction as the diamertral chord can be drawn anywhere.) For some, it may be less confusing just to delete the diametral chord and draw the full circle and reorientate the rectangle so that the long side is horizontal. From geometry the center of any circle lies on the perpendicular bisector of any chord. Let the short side of the rectangle be X and the long side 2X. If these bisectors are drawn the angles and lengths are simple as they are all 45 degrees. From observation the centre of the circle lies X/2 past the rectangle's long side along to the intersection of the two perpendicular disectors. From here, use Pythag to calculate the radius in terms of X. The problem just falls out after that. Cheers
I knew there would be some solution using subtended angles but I was lazy. So I just dropped a perpendicular from the centre to the chord thus dividing the longer side of the rectangle into x and 3x and the shorter side is 2x. Then found its length as sqrt(25-x^2). Then used the Pythagoras theorem on the smaller triangle formed by joining the centre and the with the other end of the rectangle having side lengths 3x, sqrt(25-x^2)-2x, 5. Which gives x^2 = 2.5 and area = 8x^2 = 20.
at the end you could've split the rectangle into two squares and each one is x^2, which would make it so you didn't have to do square roots
I was with you all the way up to 'Hey guys'.
I have absolutely no idea what's happening in these videos but I watch them regardless.
Excellent explanation. I hope you're a math teacher.
I love the “how exciting” part
My teenage self would've ripped off a piece of the test paper, measured the 10, compared it to any of the portions of the rectangle in the hopes they matched up, and made an educated guess based on the multiple choices provided.
make a full circle, flip the rectangle about the noted diameter, this acts to extend the small side till it hits the other edge of the circle, this gives us a right angle triangle with a corner on the edge so its hypotenuse is the diameter by basic circle theorems, so letting the small side be x we have
x^2 + (3x)^2 = d^2
x^2 = d^2/10
so x is root(10) so the area is 20
I wish my maths teachers were like you bro.
I got the right answer by finding the chord length of the diagonal of the rectangle, which is equal to 2*radius *(sin c/2), where c equal the subtended angle from the center of the circle, which is 90 degrees. *5* sin. 45= 7.07107. The short side and long side are in a ratio of 1:2. Inverse tan 1/2. = 26.565 degrees, so sin 26.565. * 7.07107= 3.16228 and cos 26.565 * 7.07107 = 6.32456. 3.16228* 6.32456= 20
I wish my math teacher was like him ❤
hiiii! i hav a doubt... see aftr forming the right triganle cant we do Tan 45 and find the breadth of the rectangle. and since the lenght is twice the breadth, now we have the breadth too. now we can use the formula for area of rectangle to get the answer?
btw luv the way u explain ❤
I have a question, considering the short side as x and the longer as 2x, wouldnt x=10/3? Isn't the sum of the sides of the rectangle forced to be the same length as the diameter?
What is ur process of thought when you are approaching these types of problems?
Step 1: ask "how do I solve this?"
Step 2: solve the exercise.
Hahahahahahaha.
Take a thread cut it to the diameter of the circle,
cut it into 10 equal pieces,
use a piece to measure the length and width of the rectangle,
multiply the length and width,
you got Area of the rectangle.
Once you know the properties, solving math problems is just like solving Sudoku but more creatively.
very interesting how you solve these. no wonder I only scraped by math.
Good memories of my geometry teacher.
There is an easier solution to this as well. The sum of two lines at 90 degrees on the central axis of a circle will always equal the diameter.
I still don't see how I was supposed to "guess" this solution. It's like finding a route in a maze where the walls aren't visible.
I like these (videos) because they approach problem solving systematically.
Every triangle is a love triangle when you love triangles - Pythagoras
i wish i found math this interesting when i was in school
These math and science videos have become a refuge for me since I've become repulsed by mainstream media, identity politics and such. I can feel my brain healing as I learn. You are gem in an insane world.
There's another option: stop consuming media and go outside
@@shem7146 I'll try it.
i would advise you to try an actually solve the problem before watching the solution then, my brain is very good at pretending like it learned while it actually didn't learn anything lol, at least that's my experience
@@shem7146 Hey great advice. Today I drove out to the middle of a national forest and walked into the woods. It was outstanding I'm doing it again. Also music is good.
@@Goose____That's what I do, try to solve it before watching the video. I'm getting better with each video.
BTW - There's something about making squares & rectangles with semicircles, maybe a video on this? . . . "Subscribed"
Again, this problem can be solved more easily with coordinate geometry. Choose the coordinate system such that its origin is at marked corner of the rectangle and the x and y axes contain the long and short edges of it respectively. Now we can write 3 equations describing the points that lie on the circle. Those points have the coordinates: (0,x), (x,x) and (2x,0). Lets denote the center of the circle as (u,v), then the system of equations is:
u²+(x-v)² = r²
(x-u)²+(x-v)² = r²
(2x-u)²+v² = r².
Solve it for x,u,v (r=10 is known):
u = -v = r√10/10
x = r√10/20.
From this, the blue area is:
A = 2x² = r²/5 = 20.
He used concepts from like 7-8 grade. I think it is easier for everybody to understand, even if it is not the most efficient way
@@pixtane7427 Central/inscribed angles come way later than 7-8th grade. It's taught in 10th grade (for 15-16yrs old students) at least in my country. But you are right in the sense that coordinate geometry is usually taught even later.
that was awesome
Youre making me like maths again
so beautiful
I must have missed every inscribed angles lecture till now lol