@@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, you mean why is it isosceles? Congruence implies 2 or more triangles equal to each other. It's isoceles because opposite sides of rectangle are congruent. Since it was given that the leg of the triangle is congruent to length of tge remaining part of the rectangle, which is congruent to the width. The given is indicated with the tic marks. Width in diagram is twice length.
@@TibRibno, it's high school geometry with a creative twist. I'm a high school math teacher who has taught geometry. These problems require a bit of creativity (adding auxiliary lines and a semicircle) and then applying the theorems learned in h.s geometry (not middle school). I gave these kind of problems to my Honors Geometry classes as extra credit. Of course, I would teach similar problems in class. The exact creative problems I taught were on the exams, but not as extra credit. I didn't bother teaching problems Luke this in my general (non honors) because I knew I would end up losing their attention. They had enough of a challenge just learning new geometry concepts not learned on middle school.
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 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 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
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 ❤
Is there a relationship between the rectangle and the circle if opposite angels are at the same half of the circumference of a circle while a 3rd angel at diameter
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 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 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 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?
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
@antoniomaurer3746, you mean why is it isosceles? Congruence implies 2 or more triangles equal to each other. It's isoceles because opposite sides of rectangle are congruent. Since it was given that the leg of the triangle is congruent to length of tge remaining part of the rectangle, which is congruent to the width. The given is indicated with the tic marks. Width in diagram is twice length.
I've learned more about math from a UA-cam channel than 4 years of college. How exciting.
This is middle school math
@@TibRibno, it's high school geometry with a creative twist. I'm a high school math teacher who has taught geometry. These problems require a bit of creativity (adding auxiliary lines and a semicircle) and then applying the theorems learned in h.s geometry (not middle school). I gave these kind of problems to my Honors Geometry classes as extra credit. Of course, I would teach similar problems in class. The exact creative problems I taught were on the exams, but not as extra credit. I didn't bother teaching problems Luke this in my general (non honors) because I knew I would end up losing their attention. They had enough of a challenge just learning new geometry concepts not learned on middle school.
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!
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 like how you use simple algebra and concepts to solve these.
Love your energy!
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
love watching these videos, makes me feel smarter than before.
I really like your videos. Always nice problems and the explanation is to the point.
I love these videos ❤ Ty for making them they’re awesome and intriguing + educational
I am loving it. Please keep up the good work.
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.
These videos are really making me consider revisiting a geometry textbook. So much stuff I never learned or don’t remember.
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!
this one was particularly wild. bro is NASTY with geometry 🔥🔥
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
That 'cool property' about the relationship between inscribed angles and arcs was new to me! Thank you for teaching me something new!
I'm hooked on these videos.
just discovered your channel. i like this content please keep going
Good job Andy! That was a tought one.
That was incredible, good work
How exciting indeed! Good work sir.
Your videos make me feel smarter haha. They introduce new was of thinking to me
It took me 15 years to find my favorite UA-cam channel!
I love you so much rn!
You make math fun!
Quick and interesting! Love geometry!👍❤️
I absolutely love these kinds of videos!
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!
When Andy says "this is a fun one," you know it's gonna be a fun one!
SO COOL!
I never knew that about inscribed angles!
I'm not even a fan of math and yet I love every one of your videos.
The way this dude smiles while explaining things… if he was my teacher growing up I probably would have cared about math 😂
Loving your problems and solutions...
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🎉❤
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 :)
Idk why are these videos so addictive
I was with you all the way up to 'Hey guys'.
That's quite an interesting solution Sir.
Good One.
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.
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?
If I can't find a thing, I always look behind the fridge.
Love the videos man.
I have absolutely no idea what's happening in these videos but I watch them regardless.
Thank you very much for the explanation, I didn't know how to do it, but thanks to you I even understand it now 😀
This kind of video should be showing up in everyone's recommended
Do you have a definitions video? Getting lost on the terms, but the steps are immaculate!
Man ive learned more from your videos in yt than my teachers during classes
To be honest, mathematics has always been my favorite subject.
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.
Excellent explanation. I hope you're a math teacher.
He’s very intelligent and he’s passionate about math! He would make one hell of a math professor in college.
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 must have missed every inscribed angles lecture till now lol
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 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
Great Video. Can you please suggest what software you use to create these videos. Looks kinda fun!!!
BTW - There's something about making squares & rectangles with semicircles, maybe a video on this? . . . "Subscribed"
I love the “how exciting” part
Extremely impressive. I wish I knew all of this.
Every triangle is a love triangle when you love triangles - Pythagoras
I like these (videos) because they approach problem solving systematically.
very interesting how you solve these. no wonder I only scraped by math.
This looks like a fun one...*insert video*...how exciting.
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 ❤
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 wish i found math this interesting when i was in school
"how exciting..." love it haha
I wish my maths teachers were like you bro.
I watched this knowing dang well I don’t like math and I enjoyed watching jt
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
Youre making me like maths again
Good memories of my geometry 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
Pleasantly exciting,!!
Is there a relationship between the rectangle and the circle if opposite angels are at the same half of the circumference of a circle while a 3rd angel at diameter
I wish you were available when I was at school
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 the rectangle is standing upright, will there be some values of x which gives twice the height?
"This looks like a fun one."
Very.
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
Question, isn't the rectangle's corner inside the circle an inscribed angle as well?
Damn, he damn did it! That was weirdly soothing!!! Huhh
Most excellent.
I wish my math teacher was like him ❤
Then the math teacher is like: You need to prove the rules/formulas you have used.
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.
Where do you get these questions from???
That's way more complex that I thought it would be
is there an another method to solve this?
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
that was awesome
Once you know the properties, solving math problems is just like solving Sudoku but more creatively.
Beautiful
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?