Hey, this is Presh Talwalkar reminding you to Mind Your Decisions
Start with the square whose side length is 10
Draw an inscribed circle and then draw two quarter circles
Solve for the area shaded in blue using only primary school maths
That means trigonometry is allowed but calculus is not allowed
This was asked in China to eleven-year-olds
Pause the video if you'd like to give it a try and when you're ready keep watching to learn how to solve this problem
We'll get started with some useful geometry formulas
The area of a circular sector is one-half times the radius squared multiplied by the central angle theta
If we have a triangle with two sides a and b and an angle C in between them
Its area is 1/2 ab multiplied by the sine of angle C
now if the triangle has the same dimensions as the circular sector
We then get the formula one half R squared sine of theta
if we then subtract this triangle from the circular sector we get the area of a circular segment and that
Will become one half R squared multiplied by the quantity theta minus sine of theta
Next if we have a triangle with sides a, b and c
We can solve for the angle C opposite side C using Al-Kashi's law of cosines
The angle c will be the following. It'll be the inverse cosine of a squared plus b squared minus c squared all over 2ab
So now let's solve the problem
We'll draw a diagonal of the square
Since the square has a side length of 10 its diagonal will be 10 times the square root of 2
This means half of the diagonal will be 5 times the square root of 2
Next we'll draw radii of the inscribed circle
Since the inscribed circle has a diameter of 10 its radius will be 5
Now we'll draw radii of this quarter circle
It'll have a length of 10 because it's exactly the same length as a side of the square
Finally, we'll draw this chord
Let's now focus on one of the triangles in this diagram
We know all three sides of the triangle. So let's solve for this angle A. We'll use Al-Kashi's law of cosines
well substitute in the values we have for this triangle and then solve for the angle a
We'll put that to the side and now we'll solve for this angle B
Well again use Al-Kashi's law of cosines. Well substitute in the values for this triangle and then we'll solve for the value of angle
Now notice that by symmetry, this is also angle B
So this will be C which is equal to two pi minus 2B
We'll calculate the angle C
Again by symmetry this is angle a so we can solve for the value of 2A
We'll now calculate the area of this circular sector
we'll then subtract the area of this triangle and
We'll get the area of this circular segment
We'll simplify this a little bit and we'll put it to the side
we'll now calculate the area of this circular sector and then subtract the area of this following triangle
and then we'll get the area of this circular segment in red
We get this formula and then we'll simplify it and
and now we want to take the area of this red shape from the area of this purple shape
so that'll be pretty much our answer we'll subtract the red shape from the purple shape
But just for fun, let's simplify this a little bit
We have the sine of C. So I'll substitute in the value for C
We'll use some trigonometric identities and then we'll use the double angle identity
Sine of two times an angle will be two times the cosine of the angle multiplied by the sine of the angle
The cosine of the inverse cosine of a quantity will simplify then we have the sine of the inverse cosine of a quantity
We can draw a right triangle
This is the value of the cosine in this triangle so we can solve for the sine of the angle and we substitute that in
so we simplify this and we figure out what sine of C is equal to
We'll substitute that and then we'll do the same sort of thing for sine of two times a
We'll use some trigonometric identities then we'll simplify and then we'll draw a triangle and we'll again calculate the sine of this triangle
This allows us to simplify even further and we get what's sine of 2A is equal to
Now when we take the red shape from the people shape we end up with the blue shape
But we need the value of two of these blue shapes. So we'll multiply everything by two
Next we can substitute in our values of C and A
and then we'll simplify
and we end up with this exact form of the answer
and this is approximately twenty nine point two seven six and don't forget the square centimeters
and that's our answer
Now at this point we've done a lot of work
So it'll always be good to check your answer and we're going to check it using calculus
So we have our shape and we'll put a coordinate system in place
We'll then put this point as ten comma zero
Well, then calculate the equation of this arc
Then we have the equation of this lower part of the circle and then the equation of this upper part of the circle
we can then calculate the intersection of the lower part of the circle with this quarter circle and
Then we calculate the intersection of this upper circle with the quarter circle
We now have three functions and we can calculate the area as follows
So the first part of the shape will be the integral that's going from x1 to x2 of
F of one minus F of two
The remaining will be the integral that goes from X2 to 10 of F of three minus F of two
Now with these integrals we can numerically calculate and get that the answer of one of these blue segments will be
approximately fourteen point six three eight
We want the area of two of them. So we multiply by two and again we get twenty nine point two seven six
So it's a pretty neat problem and I really like that you can approach it both using trigonometry and using calculus
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