Hey, this is Presh Talwalkar.
Can you find the total area of the red spots?
The diagram involves a rectangle with dimensions 10 and 20
The rectangle has one of its diagonals connected.
Inside the rectangle are two circles of equal size wich are tangent to each other and tangent to the rectangle.
If you can solve this problem,
then try this much harder variation.
This was supposedly asked to sixth graders in China.
Can you now find the total area of the red spots?
Give these problems a try and when you're ready keep watching the video for the solution.
We will get started on the first problem.
We will solve for the area of the red regions by subtraction.
We will take the area of this triangle and then we'll subtract out
different portions of the circles.
So we can set up a formula that the area of the red regions is equal to the area
of this right triangle
minus the area of each of these pieces from the different circles.
Now it looks like these two pieces actually join to form one circle.
You can notice this by symmetry and I will illustrate it by putting these two pieces together.
We now solve for the area of each piece.
The area of a right triangle is its base times its height divided by two.
The area of a circle is PI times r squared.
The right triangle has a base of 20 and a height of 10.
The circle has a radius which is equal to half the height of the rectangle
and it's radius is equal to five, so the circle has an area of pi times 5 squared.
This means the area of the red regions is equal to 100
minus 25 pi, or approximately 21.46
Now let's tackle the second part of the problem.
We know the area of the red regions as illustrated is a hundred minus 25 pi.
The second part says what would happen if you remove this region from the lower
left hand corner,
so we can solve for the red regions by solving for the area of this particular piece.
To do that we'll consider a simplified diagram we'll just
consider one of the circles and it will be inside of a square,
with side length 10.
We can again solve this by subtraction.
We consider this right triangle
We're going to remove this portion from the lower right hand corner
and then we're going to remove this portion of the circle
and what remains is the area of the lower left hand corner.
So we'll set up our formula that the area we want is equal
to the area of the rec- of the triangle
minus the area of this piece in the lower right hand corner
minus this piece of the circle.
The area of the right triangle is its base times its height divided by 2
This right triangle has a height of 5
so its area will be 10 times 5 divided by 2
What about the piece in the lower right hand corner?
We can solve this by subtraction. We'll consider one quarter of the square.
This square has a side length of five.
We then subtract out one quarter of the circle.
So we can solve for this piece by taking one quarter of the square and subtracting out one quarter of the circle.
This square has a side length of 5 so it's area is 25
The circle has a radius of 5,
so one quarter of the circle, is PI times 5 squared divided by 4
So we solve for this portion is 25 times 1 minus PI over 4
So we just have one more piece to solve for.
This is the most complicated part.
We'll again consider it by subtraction.
We'll consider this sector of the circle
We'll then remove out this isosceles triangle and that'll leave us the area of the piece we want.
So we can again set up a formula that the region we want is equal to the area
of a sector of the circle
minus the area of this isosceles triangle.
In order to calculate these areas we're going to have to use formulas for these particular shapes.
If we have a circle with radius of R and a sector
having an angle alpha and this is in radians,
then the area of the sector is equal to R squared times alpha divided by 2
For triangle with side lengths A, B and an angle in between them of beta
It's area is equal to A times B times the SIN of beta divided by 2
So,
we now need to consider what are the angles involved and what are the lenghts involved in these formulas.
Well the circle has a radius which is one-half of the square which is this a length of 10
So the radius is 5 and the radius is 5 over here too.
This is also the other side of the triangle so we have an isosceles triangle.
If we consider the right triangle from the lower left hand corner and say this has an angle of theta,
the radius of the circle will be parallel to the side of the rectangle.
So its corresponding angle over here will also be theta.
The other side of this triangle we have an isosceles triangle so it will
also have an angle theta.
This means the central angle of this sector (will be 2 PI-) will be PI minus 2 theta.
Because the angles in a triangle add up to PI radians.
So we can now substitute in these dimensions into our formulas.
We can simplify the formula for SIN of PI minus 2 theta
because SIN of PI minus 2 theta is equal to sine of two theta.
SIN of PI minus any angle is equal to SIN of that angle.
So now we need to consider what is this angle theta.
Well in our original triangle it has a base- it has a height of 5 and a base of 10
So we can use trigonometry to say,
that the angle theta is equal to the inverse tangent of five over 10
which is the inverse tangent of one half.
So we can substitute that in
and we'll simplify your formula for the area of the sector.
Now for the area of the isosceles triangle
We can use the double angle formula for SIN of two theta
and simplify that a bit
and now we need to solve for the sine of theta and the cosine of theta.
Well in this right triangle
we have one side length 10 and a height of 5
So we know that its diagonal can be
solved by the Pythagorean theorem, so we have a (hun-) 10 squared which is a 100
plus 5 squared which is 25
So that means
the square root of these is (eq-) the sum of these is equal to the length of the hypotenuse
so with the hypotenuse is equal to the square root of a 125
So we can solve for the SIN of theta is 5 divided by the square root of a 125
and the COS of theta is equal to 10 divided by the square root of a 125
So substituting that in and simplifying,
we end up with the area of the isosceles is triangle is equal to 10
And now we have a formula for the region of the circle that we want.
It's 25 PI over 2 minus 25
the inverse tangent of one half minus 10
So we have all the different areas of the pieces we want and we can simplify
this formula to get the area of the lower left hand corner is equal to 10 minus
20 5 PI over 4 plus 25 times the inverse tangent of one half.
So we will put this all together.
This is approximately 1.956
So now we're going to put this all together and we can calculate the area of the red regions.
So as illustrated the red regions are a 100 minus 25 PI
In our second problem we want to remove this one piece in the lower left-hand corner
which we just solved for.
So we can substitute that in our formula and then simplify to get the area of
these red regions is a 90 minus 18.75 PI minus 25 times the inverse
tangent of one-half and this is approximately 19.504
Did you figure this out?
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