Math puzzle – Can YOU find the area?

hello my lovelies it's Suzanna and today I want to show you how to find out how large the red shaded area is we have this quad Circle here they give us the radius of this quad Circle and we want to find this red shaded area so I call it a and we can find it by taking the area of this quad Circle first so of this whole thing I call it a q for Quad Circle but then we have to subtract these white areas of these half circles so I call this area here A1 this area here A2 and we subtract both of them of our area of our quad Circle this is our plan so let's start by finding the area of our quad Circle here on the next page if we want to find the area of the Quad Circle we can first take a look at the formula if you want to find the area of a full circle the formula is pi * the radius squared so if we only have a quarter Circle we just take a quarter of this formula so 1 over4 of < * R 2ar our radius is given by 6 M so we can uh plug the six here into our radius then we have 6 squar so 6 * 6 = 36 and then if we simplify this a little bit we have 1 * 36 which equal 36 over 4 and my Pi 36 over 4 = 9 and this is my result for our area of our quadri Circle so I plug it in here for this area it's 9 pi and then I try to find a one let's go to another page and see if we can find the area of this half circle we take the same uh formula but this time it's a half circle so we only take half of a full circle so 1 / 2 and multiply it by pi * R 2ar what is my radius this time of this half circle from the center of my circle circle to the edge of the circle it's half of this entire side here so it was 6 M so my radius of my half circle is only 3 m so I plug three into my radius here then I have 3^ SAR which equals 9 and then for my area A1 I get 1 * 9 which equal 9 over 2 piun and this is the result for my A1 so I inserted here 9/ 2 pi and the only thing that's missing is my A2 so let's try and find A2 it's a half circle so it's the same again right so the formula is 1/ 2 * < * R 2 what is my R here in this case from the center of my half circle to the edge of it I don't know how large this radius here is I just call it R I only know that the entire side here is of length six and half of it would be of length three but my radius here is just in the middle of nowhere so I don't know how long this line is not yet but we haven't used an important Point yet and that is this touching point of these two circles every time you solve such a problem and you have two things that touch each other this is an important thing to look at so if we draw the line from the center of this half circle to this touching point this is also the radius of my half circle right and if I do the same with this half circle here so from this touching point to the center of my circle this is also my radius so I know that it's of length three and now I've built a line here from these two centers of my half circles and why is it a line why is it straight really what's the explanation for this well we have this touching Point here where our circles touch each other right so it is one point that they have in common if I draw the tangent line at this point here it should look something like this so it is a tangent line to this circle and it is the same tangent line to this circle because they touch each other at one point and the radius if I draw it from the center to this touching point is always perpendicular to the tangent line to my circle at this point and that this is the same for this half circle here so this radius is also always perpendicular to the tangent line to the circle at this point so I have 90° 90° so I have 180° between these two lines which means it is just one straight line okay so I have this one straight line here and with this I've built a triangle here and it is a right triangle because of my quad Circle and my R what I want to find is part of this triangle so let's try to find the lengths of the other sides so this side here is just half of the entire side so it's of length three and what about this length here I only know that the entire side here is of length six so if I just subtract this part here of the entire side then I get what I need so I take the entire side this six and subtract this R and this is what I need for my triangle so let's use the Pythagorean theorem here in our right triangle first of all we have to find the hypotenuse of our right triangle that is the side that lies across my right angle so this here is my hypotenuse and the Pythagorean theorem says then take one of the other side so take your 6 - R and square this side plus take the other side and square it so the three and then you get the hypotenuse squared so this is r + 3 in parentheses and square this thing and this is an equation we can solve it for R and then we found our radius for our half circle here so let's go to another page and solve this equation for R we have parentheses squared here so to simplify this we can just take these parentheses and multiply them by themselves then we have this + 3^2 = 9 and the same on the other side we also have parentheses squared so we take the parentheses and multiply them by themselves okay let's multiply these two which means that we take every element of the first parentheses and multiply it by every element of the second so 6 * 6 = 36 6 * R is -6 r r * 6 = -6 R and R * R is + r² don't forget the + 9 here and the same on the other side we want to multiply so we multiply each element by each element R * r = r 2ar r * 3 is 3 R 3 * R again is 3 R and 3 * 3 = 9 so let's simplify on the left side we start with our R SAR first then we have our R parts so -6 - 6 of them is -12th of my R and then the numbers 36 + 9 = 45 the same on the other side we start with our R sared then we have 3 r + 3 R which equal 6 R and then the + 9 okay this is what we have so far it looks like this is a quadratic equation but if we take a closer look we have r s on the left side and R 2 on the right side so it will cancel out so if we subtract R 2 on both sides we will get rid of it so also take the part with R to the left side so I subtract the 6r uh in this step as well on both sides so that I get R 2 - R 2 cancels out -12 r - 6 of them is -8 R plus my 40 5 and on the other side these two cancel out as well these two cancel out as well and I only have my nine to solve for r i want to get rid of the 45 so I subtracted on both sides of the equation then I have my -8 R here these two cancel out on the other side I have 9 - 45 which equals -36 and to solve for r i divide both sides now by -8 so that this cancels out and only my R is left and here I have -36 over -18 which equals 2 I found my radius my radius is of length two so now I can plug it into my formula here for my radius so I insert the two in here so I have 2^ SAR which equals 4 my area A2 is then 1 * 4 so 4 over 2 and then the pi and 4 over 2 equal 2 so 2 pi is my result for my area A2 I inserted here and now I only have to calculate this I have 9 pi - 9/ 2 piun - 2 pi so 9 / 2 = 4. 5 I think it's easier to calculate 99 - 4. 5 is 4.

5 - 2 is 2. 5 and then I just have to add my Pi this is my area the exact value and if you want to find the decimal number then your area is around 7.