This video offers a lucid demonstration of how fundamental theorems can elegantly resolve spatial puzzles with minimal friction. It is a refreshing example of pedagogical clarity that prioritizes logical flow over mere calculation.
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A Nice Geometry Problem – Can You Find Red Shaded Area?
Added:Hello everyone, welcome back to another interesting geometric problem.
Here we have given a quarter circle and there is a rectangle inside the quarter circle.
And we have given that this part of the quarter circle is three units and this part is six units.
Here our target is to find out this one area of the quarter circle.
To find out this one area of the quarter circle here we need the area of the quarter circle, the area of rectangle and then their difference.
So first we'll try to find out the area of the rectangle. So for that first of all we will join this center with this one point on the quarter circle and the rectangle. So this figure will become Now as this is the radius of this quarter circle so let us suppose this is small r.
And this is also the radius of this quarter circle and this part is six units. So what will be this one length, this side of the rectangle? This will be r minus six units.
And here this is the radius and this part is three. So what about this one side on the rectangle?
This will be r minus three.
Now as this is a rectangle so therefore here there will be a right angle and there will be also a right angle. So therefore this figure is a right angle triangle. And here in this figure we can apply the Pythagoras theorem where this side is r, this is r minus six and this is simply r minus three.
So by applying the Pythagoras theorem, the square of perpendicular plus square of base is equal to square of hypotenuse.
So therefore from this figure here we can write by Pythagoras theorem this will become this is r minus six whole square plus r minus three whole squared equal to r squared.
So, let's simplify this equation for the a value of r. So, here we will expand these two terms using a minus b whole squared identity.
So, this will become r squared minus two times r times six, which is 12 r.
Plus six squared is 36.
Plus and this will become r squared minus two times three is six. So, this is six r plus three squared is nine.
is equal to r squared.
Now, there's r squared in both sides. We can cancel r squared with r squared. So, this will become r squared minus 12 r minus six r. This is minus 18 r.
Plus 36 plus nine, it is simply 40 five.
is equal to zero.
Now, we have to factorize this one quadratic equation.
Here, we cannot factorize this quadratic equation. Here, this is r squared and we can write this 18 r. This is simply two times r times nine.
So, here we need the value of nine squared.
So, for that here we'll add 36 to both sides to make this number as a nine squared, 81.
So, this will become this is r squared minus 18 r plus 45.
Here, we'll add 36 to both sides.
So, this side will be also 36.
So, further this is r squared minus 18 r.
Here, 45 plus 36, this is 81.
is equal to 36.
And further we can write this left hand side here. This is r squared minus This can be written as 2 * r * 9.
Plus here and here 81 can be written as this is 9 squared is equal to 36.
And here this is algebraic identity. We know that a minus b whole squared is equal to a squared minus 2 * ab plus b squared.
So here this left hand side is in this one form. So we will write this left hand side of this expression so this become r minus 9 whole squared.
So this is r minus 9 whole squared is equal to 36.
And we'll take square root on both sides.
So here square and square root will be cancelled. This will become only r minus 9 is equal to and this will become plus minus 6.
So here we have two values of r minus 9.
r minus 9 is equal to plus 6 and r minus 9 is equal to negative 6.
So this become r is equal to this is 6 plus 9 is simply 15.
And this will become r is equal to this become minus 6 plus 9. So minus 6 plus 9 it is simply 3.
So here we have two values of r.
What about this one value r is equal to 3? If we substitute r is equal to 3 here so this become 3 minus 6 and 3 minus 6 is minus 3. And this side this is the width of the rectangle which cannot be negative.
So therefore here r minus 3 is not possible for the radius. So we'll take out only r is equal to 15.
If we substitute r is equal to 15 here so this side of the rectangle will become this become 15 minus 6, which is 9.
And this side will become 15 minus 3.
So, this will become 12.
We'll try to find out the area of this rectangle having length 12 and width 9.
So, therefore, the area of rectangle will become area of rectangle that will become its length time width.
So, length is 12 and its width is 9.
So, 12 * 9 it is about 108 square units.
So, this is the area of rectangle. Now, we'll try to find out the area of this quarter circle having radius R is equal to 15.
So, therefore, the area of the quarter circle will become area of quarter circle that will become pi R squared divided by 4.
So, let's substitute the value of R.
This become pi times here our R is 15, so this become 15 squared divided by 4.
And this is 15 squared is 225 times pi by 4.
And if we calculate this one value, so this gives him about 176.
1 7 1 square units.
Now, we'll take the difference of these two areas, area of quarter circle and area of rectangle. So, our final area will become Here, our final required area that will become it will become the area of the quarter circle minus area of rectangle.
So, here the area of the quarter circle, it is about 176.7 square units.
And the area of rectangle, it is 108 square units.
And subtracting these two numbers, this gives them the approximate area.
So, that will be about 68.71 square units.
And that's the final required area.
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