Can also be solved using the chord theorem. First chord is horizontal diameter line, with line segments of 2*x and 10. Second chord is right-hand vertical side of square, intersecting first chord at right angle. Both of its line segments = x. Solving for 2x * 10 = x * x yields x = 20. Don't even need quadratics.
we can do it easily with some geometry formula and tricks we learnt in our childhood we already know a formula from geometry to graphically find geometric mean, u know if you have two chords perpendicular to each other then lets take one chord horizontal and the other verticle, if the verticle chord splits the horizontal into two parts x and y, then the height of the upper half of the verticle chord will be equal to *√(xy),* you have learnt it in ur childhood.. since the lower half of the circle is not used irl we draw a semi circle.. anyway so heres how u can use it: lets take the side of the square = s, lets take the radius of the circle = r, First of all by relation we can already see that, *_s + 10 = 2r ---(A)_* again, by the the geometric mean formula we can say, *_(s/2) = √(10*s)_* or, s²/4 = 10*s or, s² = 40*s or, s² - 40*s = 0 or, s(s-40) = 0 or, _s = 0 or 40_ Since 0 is a NONO, so *we take s = 40.* Now if, s = 40, then obviously- *40 + 10 = 2r (from equation A)* or, r = 50/2 = 25 Therefore, *r = 25* • hence area of the square = *_1600_* cm² • & that of the circle = *_π*625_* cm² Oh and by this method u can also find the general solutions for all parameters, u can try it out yourself how it is derived but heres the answer: _given the length of the shorter part of the horizontal chord = "d",_ (1) *_s = 4d_* (2) *_r = (5/2)*d_* (3) *_A₁ = 16*d² [area of the square in terms of d]_* (4) *_A₂ = (25/4)*π*d² [area of the circle in terms of d]_* [putting d = 10 in these general solutions will give u back the results I got] U can also find in terms of s from relation (1). Also heres something extra for fun: w.k.t formula for area of a segment = _(1/2)*(r²)*(Θ - sinΘ),_ where theta is in radian, and is the angle subtended by the chord. Try to find out the general solution for the area of the region not covered by the square inside the circle, in terms of "d" using some pythagorean theorems inverse trig and iQ. good luck. (Edit: thanks for the likes and also for replying man, have a nice day ❤️)
I found my own solution: 1. Joining the two extreme points on a diameter of a circle to a third point anywhere on the circumference will always form a right-angle. So, I use the horizontal diameter and use the intersection of one of the corners of the square with the circumference as my third point. 2. Now the constructed right-angle can be split into two triangles using the vertical square edge; one triangle inside the square and the other smaller one outside. 3. We know the side lengths of the triangle within the square are 2x and x, giving a tangent of 1/2. Therefore, the tangent of the angle in the outer smaller triangle must also be 1/2 to make up the full compliment of 1 for the right-angle. 4. The tangent being 1/2 in the smaller triangle means this triangle must also have an adjacent side length twice that of the opposite side, which in this case we know is 10 cm. 5. Therefore, x = 20 cm. I'm quite proud of finding that. Feeling like an ancient Greek with this kind of method. :D
Your solution is more elegant than mine. I used an isosceles variant. Regardless, similar triangles are the key. Go Ancient Geeks! I'm glad I'm not the only one who thinks with a compass, a straight edge, and the basic principles of geometry.
Great video once again! At first glance, this problem looks near impossible, as you're only given one piece of information, but using the radius and 1/2 the length of the square with pythagoras to get that quadratic is really quite satisfying! Great stuff! :)
Intersecting Chords seems the easiest way. For square side length L, and circle diameter D, D=L+10[cm], intersecting chords gives us: (L/2)^2 =L*10[cm] So L=40[cm], and the area of the square is: L^2=1600[cm^2]
I agree, I just eliminated it because it's not in 'the spirit' of the problem as I saw it. However, you are right that it works algebraically and is not strictly forbidden! :)
Waaaaaaaaaaaaay too difficult solution, you don't need r to solve the problem: You just need to realize that upper right corner of the square forms a right triangle with the middle line (Thales' theorem). Then you use the Pythagoras: 5x² + (x²+10²) = (2x + 10)², which simplifies straight to x = 20 (without even using the quadratic formula).
Aha! Thale's theorem, thanks I didn't know it had a name. But we can find the answer even quicker after using Thale's: the vertical side of the square splits our "Thalian" chords into two right-angle triangles. The one inside the square we know has base 2x and height x, hence a tangent of 1/2. Therefore, the other triangle must also have tangent 1/2 to add up to the full 1 required for the right-angle. So, the smaller triangle is similar to the large, therefore x = 2 × 10 cm = 20 cm.
What’s this problem even for? “I need to get this pipe to make water flow into this box - but I have a 10 cm gap. Is enough water going to be able to flow into this box?” Don’t answer that! I was trying to make a point, and that might’ve been an immediate contradiction.
I also asked questions like this until I had an epiphany. I was learning about waves and was confused to see the trigonometric functions pop up there. "How can something about solid triangles be useful for describing waves?", I thought. That's when it struck me that what makes mathematics so powerful is how the same generalised mathematical relationship can have a wide range of applications. So, wherever we find interesting mathematical relationships we should pocket them because they could prove useful in all kinds of unexpected situations.
Just for fun really :) Agree with the other reply on here; thinking about new problems is never a waste of effort as long as we are learning and devloping our skills!
Can also be solved using the chord theorem.
First chord is horizontal diameter line, with line segments of 2*x and 10.
Second chord is right-hand vertical side of square, intersecting first chord at right angle.
Both of its line segments = x.
Solving for 2x * 10 = x * x yields x = 20.
Don't even need quadratics.
Yup, that's exactly how I did it. It's extremely simple with that method.
we can do it easily with some geometry formula and tricks we learnt in our childhood we already know a formula from geometry to graphically find geometric mean, u know if you have two chords perpendicular to each other then lets take one chord horizontal and the other verticle, if the verticle chord splits the horizontal into two parts x and y, then the height of the upper half of the verticle chord will be equal to *√(xy),* you have learnt it in ur childhood.. since the lower half of the circle is not used irl we draw a semi circle.. anyway so heres how u can use it:
lets take the side of the square = s,
lets take the radius of the circle = r,
First of all by relation we can already see that, *_s + 10 = 2r ---(A)_*
again, by the the geometric mean formula we can say,
*_(s/2) = √(10*s)_*
or, s²/4 = 10*s
or, s² = 40*s
or, s² - 40*s = 0
or, s(s-40) = 0
or, _s = 0 or 40_
Since 0 is a NONO, so *we take s = 40.*
Now if, s = 40, then obviously-
*40 + 10 = 2r (from equation A)*
or, r = 50/2 = 25
Therefore, *r = 25*
• hence area of the square = *_1600_* cm²
• & that of the circle = *_π*625_* cm²
Oh and by this method u can also find the general solutions for all parameters, u can try it out yourself how it is derived but heres the answer:
_given the length of the shorter part of the horizontal chord = "d",_
(1) *_s = 4d_*
(2) *_r = (5/2)*d_*
(3) *_A₁ = 16*d² [area of the square in terms of d]_*
(4) *_A₂ = (25/4)*π*d² [area of the circle in terms of d]_*
[putting d = 10 in these general solutions will give u back the results I got]
U can also find in terms of s from relation (1). Also heres something extra for fun:
w.k.t formula for area of a segment = _(1/2)*(r²)*(Θ - sinΘ),_ where theta is in radian, and is the angle subtended by the chord.
Try to find out the general solution for the area of the region not covered by the square inside the circle, in terms of "d" using some pythagorean theorems inverse trig and iQ. good luck.
(Edit: thanks for the likes and also for replying man, have a nice day ❤️)
Why did you have to add iq? So the people who couldn't solve it should feel bad and stupid?
Nice method- thanks for watching and adding to the discussion :)
I found my own solution:
1. Joining the two extreme points on a diameter of a circle to a third point anywhere on the circumference will always form a right-angle. So, I use the horizontal diameter and use the intersection of one of the corners of the square with the circumference as my third point.
2. Now the constructed right-angle can be split into two triangles using the vertical square edge; one triangle inside the square and the other smaller one outside.
3. We know the side lengths of the triangle within the square are 2x and x, giving a tangent of 1/2. Therefore, the tangent of the angle in the outer smaller triangle must also be 1/2 to make up the full compliment of 1 for the right-angle.
4. The tangent being 1/2 in the smaller triangle means this triangle must also have an adjacent side length twice that of the opposite side, which in this case we know is 10 cm.
5. Therefore, x = 20 cm.
I'm quite proud of finding that. Feeling like an ancient Greek with this kind of method. :D
Your solution is more elegant than mine. I used an isosceles variant. Regardless, similar triangles are the key. Go Ancient Geeks! I'm glad I'm not the only one who thinks with a compass, a straight edge, and the basic principles of geometry.
I had thought of a similar method but I couldn't get it to work nicely. This is awesome!! :)
its the guy from maths challenge
hes a legend
Great video once again! At first glance, this problem looks near impossible, as you're only given one piece of information, but using the radius and 1/2 the length of the square with pythagoras to get that quadratic is really quite satisfying! Great stuff! :)
Thanks- I love this type of problem because you start with almost nothing and have to do it all yourself
Intersecting Chords seems the easiest way. For square side length L, and circle diameter D, D=L+10[cm], intersecting chords gives us:
(L/2)^2 =L*10[cm]
So L=40[cm], and the area of the square is:
L^2=1600[cm^2]
wow that was a very beautiful solution thanks!
x = 0 is a perfectly cromulent solution: circle has a radius of 5 and the square has degenerated to a point.
I agree, I just eliminated it because it's not in 'the spirit' of the problem as I saw it. However, you are right that it works algebraically and is not strictly forbidden! :)
Waaaaaaaaaaaaay too difficult solution, you don't need r to solve the problem: You just need to realize that upper right corner of the square forms a right triangle with the middle line (Thales' theorem). Then you use the Pythagoras: 5x² + (x²+10²) = (2x + 10)², which simplifies straight to x = 20 (without even using the quadratic formula).
Clever, I really like that method
Aha! Thale's theorem, thanks I didn't know it had a name. But we can find the answer even quicker after using Thale's: the vertical side of the square splits our "Thalian" chords into two right-angle triangles. The one inside the square we know has base 2x and height x, hence a tangent of 1/2. Therefore, the other triangle must also have tangent 1/2 to add up to the full 1 required for the right-angle. So, the smaller triangle is similar to the large, therefore x = 2 × 10 cm = 20 cm.
Really nice idea, I didn't know that was called Thales' theorem :)
Cough cough Pinner high cough cough
🤘
What’s this problem even for?
“I need to get this pipe to make water flow into this box - but I have a 10 cm gap. Is enough water going to be able to flow into this box?”
Don’t answer that! I was trying to make a point, and that might’ve been an immediate contradiction.
I also asked questions like this until I had an epiphany. I was learning about waves and was confused to see the trigonometric functions pop up there. "How can something about solid triangles be useful for describing waves?", I thought. That's when it struck me that what makes mathematics so powerful is how the same generalised mathematical relationship can have a wide range of applications. So, wherever we find interesting mathematical relationships we should pocket them because they could prove useful in all kinds of unexpected situations.
Just for fun really :) Agree with the other reply on here; thinking about new problems is never a waste of effort as long as we are learning and devloping our skills!