Thank you so much Sangam dear for your continued love and support. Take care dear and stay blessed😃 You are awesome. Keep smiling😊 Enjoy every moment of your life 🌻
Agree, the solution given in the video is overcomplicated. It is simply a case of forming the right triangle OCE (on his diagram) with side lengths OC=r, CE=r-2 and OE=r-1 and then using Pythagorean theorem and the quadratic formula to solve for r as you have noted.
I arrived at the same quadratic equation and solution using a different approach. Drawing the radius from O to C forms a right triangle OEC with hypotenuse r and legs r-1 and r-2. Applying the Pythagorean Theorem leads to the equation r^2-6r+5=0, with solution r=5, area=25π
Great job John! Yes, there are many approaches for the solution! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
@PreMath I didn't know the chords theorem, so I solved it the same way as @John Alcaraz. Thanks for producing such great videos that keep my brain working ;-)
I had a go at this before watching the video. I think my method, which also correct, is quicker and simpler. I formed a right angled triangle (OCE on your diagram). OC=r, CE=r-2 and EO=r-1. Solving for r using the Pythagorean Theorem reduces to the same quadratic equation and radius of 5.
You can also use analytic geometry with the equation of the circle as x^2+y^2=r^2 then substitute the point (x = - r +2 ,y = r - 1) into the equation of the circle because this point must lie on the circle . You then end up with the same quadratic equation in r that you came up with .
Holy cow! What a convoluted approach. I-no math genius-solved this in a few minutes using the Pythagorean theorem as others have discussed. It’s fun because at first blush the problem looks impossible.
Starting with the diagram at 2:55, draw radii OC and OD. Actually we only need one of them, so focus on the triangle OED. Line segment OE is r - 1 and line segment ED is r - 2. Invoking Pythagoras, we have: r^2 = (r - 2)^2 + (r - 1)^2; multiplying out, r^2 = r ^2 - 4 r + 4 + r^2 - 2r + 1; subtracting r^2 from both sides and collecting terms, 0 = r^2 - 6r + 5; factoring by inspection, (r - 5)(r - 1) = 0; so r can equal either +1 or +5. We reject 1 because r- 2 is a length, and clearly must be a positive number; so r = 5. Surprise! it's a 3-4-5 triangle. Anyway, the square is 4 r^2 = 4*25 = 100; and the circle is pi r^2, or 25 pi. Thank you, ladies and gentlemen; I'm here all week.
Set the center at (0,0), with radius r. Three points on the circle are (-r,0), (0,r) and (2-r, r-1). Using the formula (x-h)^2+(y-k)^2=r^2 with the third point. Gives you the quadratic (after manipulation) of r^2-6r+5=0. Solves to have r=1 or 5. However, r=1 won't work because in the diagram r needs to be greater than 2. r=5 is the solution that works. I do most of these circle problems trying to label 3 points on the circle and then using the standard form for the circle.
Glad to hear that! So nice of you Bismaya dear! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
Great video, thanks! Here is yet ANOTHER way to solve: we start by recognizing that if we translate the rectangle downward to the lower left corner of the square, it is as if it is resting at the origin in quadrant I. The upper right corner of the rectangle allows us to identify the coordinates of a point that rests on the circle: (2, 1). This point is then a solution for the relation that describes a circle: (x - h)^2 + (y - k)^2 = r^2, where (h,k) are the coordinates of the center of the circle. We can see from the picture that the center of the circle is a distance of r from the y-axis and a distance of r from the x-axis, and thus the coordinates of the center are (r,r). So, we can substitute 2 for x and 1 for y, and r for h and k, and solve. In doing so, we arrive at your quadratic again and finish in a similar manner.
Nice task, great way of solution, many thanks! Another way solving the problem (without Chords theorem, a bit easier): (r - 2)² + (r - 1)² = r² (PYTH, famous 3-4-5-triangle), leads to (r - 3)² = 4 r = 5 A = 25π P.S.: finding (r - 2) and (r - 1), you don't need any chords theorem, simply subtraction...🙂
Haven’t watched, but solved it as follows: 1. Draw a radius (r) from the point where the red rectangle touches the green circle. 2. Drop a vertical line from this point to intersect a horizontal radius also drawn. 3. These three lines intersect to form a right angled triangle whose hypotenuse is r. 4. The vertical side of this Δ is (r-1) and its horizontal is (r-2). 5. Using Pythagoras we can write r² = (r-1)² + (r-2)² -> r² = r² - 2r + 1 + r² - 4r + 4 6. Collecting terms 0 = r² - 6r + 5 7. Factorising and finding roots 0 = (r -5)(r-1) -> r = 5 or r = 1. 8. In this case it is obvious from inspection that r = 5, (r = 1 would give a diagram in which the circle was centred at the mid point of the lower edge of the red rectangle.) 9. Use A = πr² to calculate area. A = 25π = 78.54 (4sf)
As someone who often draws on squared paper, I saw that one immediately. If I want to draw a circle freehand, I use the 3 by 1, 1 by 1, 1 by 3 scheme for the quadrant :)
I suppose that r=1 is a meaningful solution: It is the circle inscribed in the 2x2 square that includes the original 2x1 rectangle, since this smaller circle passes the lower-right corner of that 2x1 rectangle and is tangent to its upper and left sides.
i solved it without using chords theorem simply by drawing a line from center to point C and extending it to the edge of the square creating a theta angle between the FO and CO. Use pythagoras theorem. need to label the lengths of the big triangle horizontal of (r-2) and vertical of (r-1). then you can work that out nicely into (r)squared - 6r +5 =0. one trick i used was to add 4 to each side of the equation so you get (r) squared -6r +9 = 4, which translates into (r-3)squared = 4, in other words r=5. slick.
This is similar to the one with the one small red square at the corner. It was only changed to a rectangle. Nice variation. Professor is really an expert. Math is challenging and enjoyable when there are variations to analyze.
I submit that r=1 is a valid solution too. This is because a rectangle that has: 1) its left side on the circle's left tangent; and 2) its top side on the circle's top tangent; and 3) has side lengths 2 and 1; Will also have its bottom right corner touching the circumference of the circle. The difference is, now the red square will overlap the top half of the green circle. Which goes to show that solutions can have unintended and expected interpretations.
I enjoyed this problem. I am not very good at geometric reasoning and tend to use coordinate geometry if I can. In this case I set the origin at the centre and worked out the coordinates of point C to be ( - r +2, r - 1). So the distance OP^2 = r^2 = (r-2)^2 + (r-1)^2, which gives r^2 - 6r + 5 = (r-1)(r-5)=0. r=1 clearly doesn't work, so r =5.
Form a right angled triangle with OC as the hypotenuse of radius r, height r - 1 and base r - 2. Then use Pythagoras. You then get a quadratic to solve giving r = 5.
This is the kind of problem where having neat integer dimensions makes a figure drawn on a grid very powerful. That circle is defined by the point of coordinates (2,1), and tangeancy to the x and y axis : its center must thus lie on the y=x line, try a few different centers and you will see that (5,5) work, thanks to the 3²+4²=5² relation.
So nice of you! Thank you so much Sangam dear for your continued love and support. Take care dear and stay blessed😃 You are awesome. Keep smiling😊 Enjoy every moment of your life 🌻
A to complicate explanation. Just draw a triangle between the center of the circle (O) to the bottom right edge (C) of the rectangle. This is the radius next you draw a parallel line to the top or bottom through the center of the circle and another parallel line to C. The intersection of both parallels is D. You get a triangle OCD with a right angle in D. |OC| = r, |DC| = r -1, |OD| = r - 2. With the good old Pythagoras you get: r² = (r - 1)² + (r - 2)² ⇒ r² = r² - 2*r +1 + r² - 4*r + 4 ⇒ 0 = r² - 6*r + 5 ⇒ 0 = (r - 1)*(r - 5) case 1: r = 5 A = π * r² ⇒ A = 24 π ~ 78,5 Case r = 1 If you not assume that the rectangle touches the circle on the left side or does not intersect with the circle. Otherwise, r = 1 is a valid solution. In this case, the upper half of the circle lies fully in the rectangle.
Place the square at the beginning of the flat coordinate system. Coordinates of the center of an unknown circle (r, r). Points with coordinates (1, 2) (2, 1) belong to this circle. X = R (!) Then (2-x)^2+(1-x)^2=x^2 Now we simplify this equation to a simple quadratic function. x1 = 1, x2 = 5 The end.
Thank you so much Taha dear for your continued love and support. Take care dear and stay blessed😃 You are awesome. Keep smiling😊 Enjoy every moment of your life 🌻
I did this with the circle formula (x - xm)² + (y - ym)² = r² Due to the picture we know the points C(2, -1) and A(xm, 0) lie on the circle. We can also see that r = xm. Setting all this information to the formula, we get the quadratic equation r² - 6r + 5 = 0, as shown in the video. Solutions are r = 1 and r = 5, and only the second solution makes sense according to the picture. So we have r = 5 and A = r² π = 25π.
Let's try that problem: I am going to use the Pythagorean theorem: r² = (r - 2)² + (r - 1)² r² = (r² - 4r + 4) + (r² - 2r + 1) r² = 2r² - 6r + 5 r² - 6r + 5 = 0 (r - 5)(r - 1) = 0 r1 = 5 r2 = 1 (not possible since r must be >= 2) r = 5 A = r²(pi) = 25*(pi) = 78.54 square units P.S. I like the idea of using the chord theorem! :)
Thanks Steven for the feedback.Thank you so much for taking the time to leave this comment. I've given links to some videos as how to solve Quadratic equation below. Actually, I've explained in three different ways. Please check out PreMath channel for more vids. Take care my friend and stay blessed😃 You are awesome. th-cam.com/video/7E8ceozAF-U/w-d-xo.html th-cam.com/video/Sm0wvQSZf1g/w-d-xo.html th-cam.com/video/9D7a0-o7OGo/w-d-xo.html
we have another way most simple without chords theorem juste with pythagore r²=(r-1)²+(r-2)² (triangle ceo ) r²=r²+1-2r+r²+4-4r=2r²+5-6r r²-6r+5=0 delta=16 2 solutions with only one real because r>2 r=5 and area like you
disPLacemenT of rectangle to mid section of Circle × 3 makes extenTo centre ----- its 6 -- 1 reacH centre --- radius 5 ----- volumE 5 × 3 cubE ----- 5 5 5 ---- 125 cubic uniT EM gonnA to some You tube music
As a kid, I liked to draw schemes and maps and plans. And one trick to get a good quarter of a circle was to use a 5x5 square, and draw the line through (0,0), (3,1), (2,4) and (5,0) as all three points are a distance of 5 away from the point (0,5) (3² + 4² = 5² and yada yada). So I saw your picture and immediately knew: The radius of the circle has to be 5!
So nice of you dear! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
Sorry, but this was a 1 second problem. A circle's radius is the same to every point of the circle. So it is the same to the top and side as the 2,1 point. The line from the 2,1 point to the center of the circle is like the hypothenusa of a straight corner with sides 1 and 2 shorter than the length of the hypothenusa. Which matches the simplest example of the Pythagoras triangle, with sides 5, 4 and 3 - and nothing else. So the hypothenusa, and the radius, must be 5.
(r-2) u r right it's a chord, but (2r-1) it's not a chord. therefore, u .cannot apply Chords Theorem there. u r taking a shortcut over there a d it's wrong
2(r-2) = Chord. but (2r-1) it's not a chord since it crosses the centre of a circle. it's a diameter so u r confused when to apply Chords Theorem apparently
The way of your explaination is just superrbbb 😛🤗🤗
Thank you so much Sangam dear for your continued love and support. Take care dear and stay blessed😃 You are awesome. Keep smiling😊 Enjoy every moment of your life 🌻
@@PreMath thankuu so much for your such beautiful words
Unnecessarily complicated, needing theorems that nobody remembers.
Circumnavigation of the globe when a walk across the street would do.
As some viewers have pointed out, in the case, it is much more simpler and straightforward to start with (r-1)^2 + (r-2)^2=r^2.
Well said!!!
Yep. If you calculate it with the coordinate system, you can leave the whole rubbish away.
Agree, the solution given in the video is overcomplicated. It is simply a case of forming the right triangle OCE (on his diagram) with side lengths OC=r, CE=r-2 and OE=r-1 and then using Pythagorean theorem and the quadratic formula to solve for r as you have noted.
Wow . Its a 345 triangle
Thank you. I was about to point that out as well.
I arrived at the same quadratic equation and solution using a different approach. Drawing the radius from O to C forms a right triangle OEC with hypotenuse r and legs r-1 and r-2. Applying the Pythagorean Theorem leads to the equation r^2-6r+5=0, with solution r=5, area=25π
Yes, It is an alternative solution and it does work,, well done
Great job John! Yes, there are many approaches for the solution!
You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
Youŕ solution is much simpler and elegant.
me2
@PreMath I didn't know the chords theorem, so I solved it the same way as @John Alcaraz. Thanks for producing such great videos that keep my brain working ;-)
Again you are simply the best
I had a go at this before watching the video. I think my method, which also correct, is quicker and simpler.
I formed a right angled triangle (OCE on your diagram). OC=r, CE=r-2 and EO=r-1.
Solving for r using the Pythagorean Theorem reduces to the same quadratic equation and radius of 5.
Highly recommended. Easy to understand. Thank you so much for saving my time.
You can also use analytic geometry with the equation of the circle as x^2+y^2=r^2 then substitute the point (x = - r +2 ,y = r - 1) into the equation of the circle because this point must lie on the circle . You then end up with the same quadratic equation in r that you came up with .
Exactly! That's what I did to solve the problem (using a different basis though)
It is clear ,--one need not respond in words..the children feel happy smiling & appreciate ur expression
Holy cow! What a convoluted approach. I-no math genius-solved this in a few minutes using the Pythagorean theorem
as others have discussed. It’s fun because at first blush the problem looks impossible.
新たな発見につながる良問とは思えないけど、設問がシンプルで楽しくて好き。
(r-2)^2 + (r-1)^2 = r^2
r^2 -4r + 4 + r^2 -2r + 1 = r^2
r^2 -6r +5 =0
(r-1)(r-5)=0
∴ r=1 or r=5
しかも答えが整数になるのも素晴らしい。👍。
有名なピタゴラス数、3:4:5に根付いた問題だから答えが整数になるのか。納得した。
You can also use the pythagorean theorem with triangle OEC: (r-1)^2 + (r-2)^2 = r2
Interesting. All your examples are really challenging.
Starting with the diagram at 2:55, draw radii OC and OD. Actually we only need one of them, so focus on the triangle OED. Line segment OE is r - 1 and line segment ED is r - 2.
Invoking Pythagoras, we have:
r^2 = (r - 2)^2 + (r - 1)^2; multiplying out,
r^2 = r ^2 - 4 r + 4 + r^2 - 2r + 1; subtracting r^2 from both sides and collecting terms,
0 = r^2 - 6r + 5; factoring by inspection,
(r - 5)(r - 1) = 0;
so r can equal either +1 or +5. We reject 1 because r- 2 is a length, and clearly must be a positive number; so r = 5. Surprise! it's a 3-4-5 triangle.
Anyway, the square is 4 r^2 = 4*25 = 100;
and the circle is pi r^2, or 25 pi.
Thank you, ladies and gentlemen; I'm here all week.
You can't exclude a negative length as a solution. If a length is negativ, it points in the other direction.
Set the center at (0,0), with radius r. Three points on the circle are (-r,0), (0,r) and (2-r, r-1). Using the formula (x-h)^2+(y-k)^2=r^2 with the third point. Gives you the quadratic (after manipulation) of r^2-6r+5=0. Solves to have r=1 or 5. However, r=1 won't work because in the diagram r needs to be greater than 2. r=5 is the solution that works. I do most of these circle problems trying to label 3 points on the circle and then using the standard form for the circle.
The coordination of the point is (2-r,r-1), then use (2-r)^2+(r-1)^2=r^2, Solve r
Your explanations are so easy to understand. Very helpful to me in my studies.
Glad to hear that!
So nice of you Bismaya dear! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
This one was great. Once you used the intersecting chords theorem it was straightforward.
Great video, thanks!
Here is yet ANOTHER way to solve: we start by recognizing that if we translate the rectangle downward to the lower left corner of the square, it is as if it is resting at the origin in quadrant I. The upper right corner of the rectangle allows us to identify the coordinates of a point that rests on the circle: (2, 1). This point is then a solution for the relation that describes a circle: (x - h)^2 + (y - k)^2 = r^2, where (h,k) are the coordinates of the center of the circle. We can see from the picture that the center of the circle is a distance of r from the y-axis and a distance of r from the x-axis, and thus the coordinates of the center are (r,r). So, we can substitute 2 for x and 1 for y, and r for h and k, and solve. In doing so, we arrive at your quadratic again and finish in a similar manner.
I did the pythagorean solution, didn't notice the option of intersecting chords. I was relieved I got the correct answer. Thanks
Nice task, great way of solution, many thanks!
Another way solving the problem (without Chords theorem, a bit easier):
(r - 2)² + (r - 1)² = r² (PYTH, famous 3-4-5-triangle), leads to
(r - 3)² = 4
r = 5
A = 25π
P.S.: finding (r - 2) and (r - 1), you don't need any chords theorem, simply subtraction...🙂
Can also be done with pithagorean theorem where (r-1)^2 + (r-2)^2 = r^2
Haven’t watched, but solved it as follows:
1. Draw a radius (r) from the point where the red rectangle touches the green circle.
2. Drop a vertical line from this point to intersect a horizontal radius also drawn.
3. These three lines intersect to form a right angled triangle whose hypotenuse is r.
4. The vertical side of this Δ is (r-1) and its horizontal is (r-2).
5. Using Pythagoras we can write r² = (r-1)² + (r-2)² -> r² = r² - 2r + 1 + r² - 4r + 4
6. Collecting terms 0 = r² - 6r + 5
7. Factorising and finding roots 0 = (r -5)(r-1) -> r = 5 or r = 1.
8. In this case it is obvious from inspection that r = 5, (r = 1 would give a diagram in which the circle was centred at the mid point of the lower edge of the red rectangle.)
9. Use A = πr² to calculate area. A = 25π = 78.54 (4sf)
After 2:34 you can also use Pythagoras wirh the triangle OEC. OC=r, OE=r-1, OC=r-2. This leeds also to (r-1)(r-5)=0
I definitely feel like I lost out on my education. You made it look so easy.
As someone who often draws on squared paper, I saw that one immediately. If I want to draw a circle freehand, I use the 3 by 1, 1 by 1, 1 by 3 scheme for the quadrant :)
I suppose that r=1 is a meaningful solution: It is the circle inscribed in the 2x2 square that includes the original 2x1 rectangle, since this smaller circle passes the lower-right corner of that 2x1 rectangle and is tangent to its upper and left sides.
I agree. The bottom corner or the rectangle touches the circumference.
i solved it without using chords theorem simply by drawing a line from center to point C and extending it to the edge of the square creating a theta angle between the FO and CO. Use pythagoras theorem. need to label the lengths of the big triangle horizontal of (r-2) and vertical of (r-1). then you can work that out nicely into (r)squared - 6r +5 =0. one trick i used was to add 4 to each side of the equation so you get (r) squared -6r +9 = 4, which translates into (r-3)squared = 4, in other words r=5. slick.
This is similar to the one with the one small red square at the corner. It was only changed to a rectangle. Nice variation. Professor is really an expert. Math is challenging and enjoyable when there are variations to analyze.
could the crossing chords rule be used in this problem ?
Good review of the chord theorem, which is also a good review of the use of the equation of a circle.
I submit that r=1 is a valid solution too. This is because a rectangle that has:
1) its left side on the circle's left tangent; and
2) its top side on the circle's top tangent; and
3) has side lengths 2 and 1;
Will also have its bottom right corner touching the circumference of the circle.
The difference is, now the red square will overlap the top half of the green circle.
Which goes to show that solutions can have unintended and expected interpretations.
I enjoyed this problem. I am not very good at geometric reasoning and tend to use coordinate geometry if I can. In this case I set the origin at the centre and worked out the coordinates of point C to be ( - r +2, r - 1). So the distance OP^2 = r^2 = (r-2)^2 + (r-1)^2, which gives r^2 - 6r + 5 = (r-1)(r-5)=0. r=1 clearly doesn't work, so r =5.
Why go all the way using Chords theorem when you can easily use Pythagoras theorem
(r - 1)^2 + (r - 2)^2 = r^2
which yields same answer r = 5
Excellent explanation
Form a right angled triangle with OC as the hypotenuse of radius r, height r - 1 and base r - 2. Then use Pythagoras. You then get a quadratic to solve giving r = 5.
This is the kind of problem where having neat integer dimensions makes a figure drawn on a grid very powerful.
That circle is defined by the point of coordinates (2,1), and tangeancy to the x and y axis : its center must thus lie on the y=x line,
try a few different centers and you will see that (5,5) work, thanks to the 3²+4²=5² relation.
Totally brilliant!!
Such a simple questio, you went round the bush to solve it.
I am again first here to see ur amazing video ... Keep rocking sir ... and Keep blessing us ❤❤ You have just an *Einstein Brain*....
So nice of you!
Thank you so much Sangam dear for your continued love and support. Take care dear and stay blessed😃 You are awesome. Keep smiling😊 Enjoy every moment of your life 🌻
triangle OCE has dimensions r, r-1, r-2, which obviously implies 5,4,3 (the most well known hypotenuse) > circle area = 25 pi
5:33 Triangle CEO
CE² + EO² = CO²
(r-2)² + (r-1)² = r²
r=5 or r=1(reject since r must be > 2)
Thus,
Area=πr²
Area=25π
Your videos are awesome just increase the speed of explaining
A to complicate explanation.
Just draw a triangle between the center of the circle (O) to the bottom right edge (C) of the rectangle. This is the radius next you draw a parallel line to the top or bottom through the center of the circle and another parallel line to C. The intersection of both parallels is D.
You get a triangle OCD with a right angle in D. |OC| = r, |DC| = r -1, |OD| = r - 2.
With the good old Pythagoras you get:
r² = (r - 1)² + (r - 2)²
⇒ r² = r² - 2*r +1 + r² - 4*r + 4
⇒ 0 = r² - 6*r + 5
⇒ 0 = (r - 1)*(r - 5)
case 1:
r = 5
A = π * r²
⇒ A = 24 π ~ 78,5
Case r = 1
If you not assume that the rectangle touches the circle on the left side or does not intersect with the circle. Otherwise, r = 1 is a valid solution. In this case, the upper half of the circle lies fully in the rectangle.
awesome, walked the same way...🙂. One remark: For r = 1, the center of the small circle is c (so only 1/4 of the circle lies in the rectangle).
I advice my all friends to see PreMath.
🇮🇳🇮🇳🇮🇳
Thanks dear Govinda! You are very generous. Love and prayers from USA.
Nice one,thanks
Place the square at the beginning of the flat coordinate system.
Coordinates of the center of an unknown circle (r, r).
Points with coordinates (1, 2) (2, 1) belong to this circle.
X = R (!)
Then
(2-x)^2+(1-x)^2=x^2
Now we simplify this equation to a simple quadratic function.
x1 = 1, x2 = 5
The end.
Thank for premath
Thank you so much Taha dear for your continued love and support. Take care dear and stay blessed😃 You are awesome. Keep smiling😊 Enjoy every moment of your life 🌻
I did this with the circle formula (x - xm)² + (y - ym)² = r²
Due to the picture we know the points C(2, -1) and A(xm, 0) lie on the circle.
We can also see that r = xm. Setting all this information to the formula,
we get the quadratic equation r² - 6r + 5 = 0, as shown in the video.
Solutions are r = 1 and r = 5, and only the second solution makes sense according to the picture.
So we have r = 5 and A = r² π = 25π.
Let's try that problem:
I am going to use the Pythagorean theorem:
r² = (r - 2)² + (r - 1)²
r² = (r² - 4r + 4) + (r² - 2r + 1)
r² = 2r² - 6r + 5
r² - 6r + 5 = 0
(r - 5)(r - 1) = 0
r1 = 5
r2 = 1 (not possible since r must be >= 2)
r = 5
A = r²(pi) = 25*(pi) = 78.54 square units
P.S. I like the idea of using the chord theorem! :)
An easier way using Pythagoras theorem is a simpler solution.
Indeed
Pathagorous theorem can also be applied
(r-1)^2+(r-2)^2=r^2
Using same labelled diagram
But nice effort
*4 steps:*
1. Flip vertically. Left bottom point - origin of coordinates.
2. (2-r)²+(1-r)²=r²
3. r=5;
4. S=25π.
Just build a right triangle OEC. Its sides are R, R-1, R-2. Use Pythagoras, then solve a quadratic equation for R.
The real challenge is calculating the area of the rectangle
😂😂😂
I watched and liked the video
I had trouble following the computing of the quadratic equation. But I followed the logic of everything else.
Thanks Steven for the feedback.Thank you so much for taking the time to leave this comment. I've given links to some videos as how to solve Quadratic equation below. Actually, I've explained in three different ways. Please check out PreMath channel for more vids.
Take care my friend and stay blessed😃 You are awesome.
th-cam.com/video/7E8ceozAF-U/w-d-xo.html
th-cam.com/video/Sm0wvQSZf1g/w-d-xo.html
th-cam.com/video/9D7a0-o7OGo/w-d-xo.html
@@PreMath thanks!
Very nice question and your explanation is simply outstanding.
Thank you sir
(r-2) sqaure = 1 * (2r - 1). Product of 2 intersecting chord.
seems complicated... OCE is a right triangle so r^2=(r-1)^2+(r-2)^2...
It can be easily solved in less than a minute
Bravo Ohiovic! Very smart👍Take care dear and stay blessed😃
The ratio of the area of the circle to the area of the square is π/4 or 78.5%
Magic of maths
we have another way most simple without chords theorem juste with pythagore
r²=(r-1)²+(r-2)² (triangle ceo )
r²=r²+1-2r+r²+4-4r=2r²+5-6r
r²-6r+5=0
delta=16 2 solutions with only one real because r>2
r=5 and area like you
I used Pythagorean theorem on your OCE: r, r-2, r-1.
Thank you sir for your teaching ability
Good explanation
Thank you so much
disPLacemenT of rectangle to mid section of Circle × 3 makes extenTo centre ----- its 6 -- 1 reacH centre --- radius 5 ----- volumE 5 × 3 cubE ----- 5 5 5 ---- 125 cubic uniT
EM gonnA to some You tube music
As a kid, I liked to draw schemes and maps and plans. And one trick to get a good quarter of a circle was to use a 5x5 square, and draw the line through (0,0), (3,1), (2,4) and (5,0) as all three points are a distance of 5 away from the point (0,5) (3² + 4² = 5² and yada yada).
So I saw your picture and immediately knew: The radius of the circle has to be 5!
I usually just think by the video cover but not click in it😂
great problem
36pi or 113.1
Vvnice true thanks
thx for sharing
So nice of you dear! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
r=5.....A=pi25
Sorry, but this was a 1 second problem. A circle's radius is the same to every point of the circle. So it is the same to the top and side as the 2,1 point.
The line from the 2,1 point to the center of the circle is like the hypothenusa of a straight corner with sides 1 and 2 shorter than the length of the hypothenusa.
Which matches the simplest example of the Pythagoras triangle, with sides 5, 4 and 3 - and nothing else.
So the hypothenusa, and the radius, must be 5.
👌👍👏🌷
C’est très énervant !
400：314
All square : green circle!
2r = 1 + [r-2]^2
S=25π≈78,5
Yes
(r-2) u r right it's a chord, but (2r-1) it's not a chord. therefore, u .cannot apply Chords Theorem there. u r taking a shortcut over there a d it's wrong
2(r-2) = Chord. but (2r-1) it's not a chord since it crosses the centre of a circle. it's a diameter so u r confused when to apply Chords Theorem apparently
Cord formula looks so wrong. How can a diameter be equal to its cord? The math is right but just looking at it in figure feels odd.
The answer is 2
Wow
Lol im a circle
Are you an Indian???
Ur voice is not academic
Again you are simply the best