Hey, this is Presh Talwalkar. Can you find the total area of the scarlet recognises? The diagram involves a rectangle with dimensions 10 and 20 The rectangle has one of its diagonals connected. Inside the rectangle are two roundabouts of equal width wich are tangent to one another and tangent to the rectangle. If you can solve this problem, then try this much harder modification. This was supposedly asked to sixthgraders in China. Can you now find the total area of thered blots? Give these problems a try and when you’re ready keep watching the video for the solution. We will is starting on the first trouble. We will resolve for thearea of the red regions by subtraction. We will take the area of this triangle andthen we’ll subtract out different portions of the circles.So we can set up a formula that the areaof the red regions is equal to the area of this right triangle minus the area of each of these segments fromthe different cliques. Now it looks like these two parts actually affiliate to formone circle. You can notice this by symmetry and I will summarize it byputting these two segments together. We now answer for the area of each piece. The locality of a claim triangle is its cornerstone days its summit divided by two.The area of a circle is PI epoches rsquared. The title triangle has a base of 20 and a height of 10. The circle has a radius which is equalto half the height of the rectangle and it’s radius is equal to five, so thecircle has an area of pi epoches 5 squared. This wants the area of the red regionsis equal to 100 minus 25 pu, or nearly 21.46 Now let’s tackle the second part of theproblem. We know the area of the red regions asillustrated is a hundred minus 25 pi. The second place says what would happenif you remove such regions from the lower left hand corner, so we can solve for the red regions by answering for the area of this particular piece. To do that we’llconsider a simplified representation we’ll simply consider one of the haloes and it willbe inside of a square, with surface duration 10. We can again solve this bysubtraction.We consider this right triangle We’re going to remove this section fromthe 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 thearea we want is equal to the area of the rec- of the triangle minus the area of this article in the lowerright hand corner minus this part of the circle. The place of the right triangle is its locate meters its meridian divided among 2 This right triangle has aheight of 5 it is therefore country is likely to be 10 eras 5 divided by 2 What about the bit in the lower right hand corner? We can solve this by subtraction. We’ll consider one fourth of the square. This square has a side length of five. We then subtract out one one-quarter of thecircle. So we can solve for this piece by taking one quarter of the square and subtracting out one quarter of the curve. This square has a side period of5 so it’s arena is 25 The circle has a radius of five, so one fourth of the clique, is PI eras 5squared divided by 4 So we solve for this portion is 25 hours 1 minus PI over 4 So we just have one more piece to solve for.This is the most complicatedpart. We’ll again consider it by subtraction. We’ll consider this sector of the clique We’ll then remove out this isosceles triangle and that’ll leave us the area of the piece we want. So we are able to again set up a formula thatthe region we want is equal to the area of a sphere of the clique minus the area of this isosceles triangle. In order to calculate these areas we’re going to have to use formulas for theseparticular conditions. If we have a circle with radius of Rand a sphere having an angle alpha and this is consistent with radians, then the area of the sector is equal toR squared days alpha divided by 2 For triangle with line-up segments A, B and an slant in between them of beta It’s place is equal to A seasons B timesthe SIN of beta divided among 2 So, we now need to consider what are the angles involved and what are the lenghts involved in these formulas.Well the clique 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 5over 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 correspond angle over here will also be theta. The other side of this trianglewe have an isosceles triangle it is therefore will likewise have an angle theta. This implies the central slant of the fisheries sector( is likely to be 2 PI-) will be PI minus 2 theta. Because the angles in a triangleadd 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 slant isequal to SIN of that slant. So now we need to consider what is thisangle theta. Well in our original triangle it has abase- it has a height of 5 and a basi of 10 So we can use trigonometry to say, that the slant theta is equal to the inversetangent 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 angleformula for SIN of two theta and streamline 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 heightof 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 conveys the square beginning of these is( eq-) resources in the amount 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 is5 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 streamlining, we be brought to an end with the area of the isosceles is triangle is similar 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 ofthe fragments we want and we can simplify this formula to get the area of thelower left hand corner is equal to 10 minus 20 5 PI over 4 plus 25 periods the inversetangent of one half. So we will kept this all together. This is approximately 1.956 So now we’re going to framed this all together and we can calculate the area of the red regions. So as illustrated thered regions are a 100 minus 25 PI In our second difficulty we want to removethis one piece in the lower left-hand corner which we just solved for. So we can substitute that in our formulaand then simplify to get the area of these red regions is a 90 minus 18.75 PI minus 25 seasons the inverse tangent of one-half and this is approximately 19.504 Did you figure this out? Thanks for watching this video! Pleasesubscribe to my channel.I originate videos on math and game theory. You can catch me onmy blog mind your decisions that you can follow on facebook google+ and patreon.You can catch me on social media at preshtalwalkar. And if you like thisvideo please check out my books.There are connects in the video description ..