@@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.
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!
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.
I used coordinates as well. And I even have solved the task just in mind. Now I'm writing that my solution: I placed coordinates another way. These were (0;0), (-1;0) (1;1). So I selected a 1 to be "x" of the task. Having (a,b) as a circle center, that leads to: a^2 + b^2 =rr (a+1)^2 +b^2 =r^2 (a-1)^2 +(b-1)=r^2 The first and the second give a=-0.5 After using of "a" in the first and third equation, that leads to: 2.25 + (b-1)^2 = 0.25 + bb 2 -2b +1=0, b=1.5 So r=√2.5 But in task's units, the radius is equal to 5. So here's a proportion x/1 = 5/√2.5. x=5/5√0.1=5√0.1/0.5=10√0.1=√10
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.
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.
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 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.
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.
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
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.
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.
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 am eating my dinner while watching this, i am 29 year old father of 3 and this is entertaining for me... recalling back all those memories from school lol
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.
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.
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.
Divide the rectangle in two blue squares with sides s. Each square has a symmetry line that goes through the center of the circle. (1/2*s)^2 + (3/2*s)^2 = 5^2 , so s^2 = 10 and the rectangle's area = 20.
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 CZcams 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 !
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!
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.
I used coordinates as well. And I even have solved the task just in mind. Now I'm writing that my solution:
I placed coordinates another way. These were (0;0), (-1;0) (1;1). So I selected a 1 to be "x" of the task.
Having (a,b) as a circle center, that leads to:
a^2 + b^2 =rr
(a+1)^2 +b^2 =r^2
(a-1)^2 +(b-1)=r^2
The first and the second give a=-0.5
After using of "a" in the first and third equation, that leads to:
2.25 + (b-1)^2 = 0.25 + bb
2 -2b +1=0, b=1.5
So r=√2.5
But in task's units, the radius is equal to 5. So here's a proportion x/1 = 5/√2.5.
x=5/5√0.1=5√0.1/0.5=10√0.1=√10
i like how you use simple algebra and concepts to solve these.
Love your energy!
love watching these videos, makes me feel smarter than before.
This guy really enjoys math. I watch the videos just to witness his joy. Good work man!!
I normally dont interact with channels but man you need to keep making these.
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'm hooked on these videos.
That was incredible, good work
I absolutely love these kinds of videos!
I love you so much rn!
You make math fun!
I am loving it. Please keep up the good work.
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
Good job Andy! That was a tought one.
These videos are really making me consider revisiting a geometry textbook. So much stuff I never learned or don’t remember.
Your videos make me feel smarter haha. They introduce new was of thinking to me
That 'cool property' about the relationship between inscribed angles and arcs was new to me! Thank you for teaching me something new!
How exciting indeed! Good work sir.
Quick and interesting! Love geometry!👍❤️
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.
this one was particularly wild. bro is NASTY with geometry 🔥🔥
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!
just discovered your channel. i like this content please keep going
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.
Loving your problems and solutions...
Idk why are these videos so addictive
SO COOL!
I never knew that about inscribed angles!
Man andy these math questions are the best.. i may nkt be able to solve them all, but they definutely get your brain thinking
Quick satisfying and digestible. Love it. Got a new sub!
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💀
Love the videos man.
I'm not even a fan of math and yet I love every one of your videos.
It took me 15 years to find my favorite CZcams channel!
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
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?
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🎉❤
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.
Thank you very much for the explanation, I didn't know how to do it, but thanks to you I even understand it now 😀
That's quite an interesting solution Sir.
Good One.
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 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
When Andy says "this is a fun one," you know it's gonna be a fun one!
I was with you all the way up to 'Hey guys'.
Most excellent.
Beautiful
The way this dude smiles while explaining things… if he was my teacher growing up I probably would have cared about math 😂
Extremely impressive. I wish I knew all of this.
He’s very intelligent and he’s passionate about math! He would make one hell of a math professor in college.
I have absolutely no idea what's happening in these videos but I watch them regardless.
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 kind of video should be showing up in everyone's recommended
Excellent explanation. I hope you're a math teacher.
I see the blue rectangle. It's right there, in front of a white background. That's the real area of the blue rectangle
Man ive learned more from your videos in yt than my teachers during classes
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
Pleasantly exciting,!!
If I can't find a thing, I always look behind the fridge.
To be honest, mathematics has always been my favorite subject.
that was awesome
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.
I love the “how exciting” part
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
I like these (videos) because they approach problem solving systematically.
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.
I wish my math teacher was like him ❤
very interesting how you solve these. no wonder I only scraped by math.
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 wish my maths teachers were like you bro.
Good memories of my geometry teacher.
Do you have a definitions video? Getting lost on the terms, but the steps are immaculate!
i wish i found math this interesting when i was in school
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.
I must have missed every inscribed angles lecture till now lol
Amazing!
Great Video. Can you please suggest what software you use to create these videos. Looks kinda fun!!!
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 am eating my dinner while watching this, i am 29 year old father of 3 and this is entertaining for me... recalling back all those memories from school lol
This looks like a fun one...*insert video*...how exciting.
HOW EXCITING ‼️‼️‼️
I watched this knowing dang well I don’t like math and I enjoyed watching jt
Youre making me like maths again
"This looks like a fun one."
Very.
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.
"how exciting..." love it haha
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.
I wish you were available when I was at school
so beautiful
That's way more complex that I thought it would be
Every triangle is a love triangle when you love triangles - Pythagoras
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.
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.
Divide the rectangle in two blue squares with sides s.
Each square has a symmetry line that goes through the center of the circle.
(1/2*s)^2 + (3/2*s)^2 = 5^2 , so s^2 = 10 and the rectangle's area = 20.
Once you know the properties, solving math problems is just like solving Sudoku but more creatively.