Find the Area of the Green Rectangle | Step-by-Step Tutorial

[tomcruise]

Amazing question! ❤️
Greetings from India! 🇮🇳😍

[sportsworld]

Highly recommended. Easy to understand. Thank you so much for saving my time.

[rangaswamyks]

Super sir.. I loved the way you solved this problem.. There was a beauty in it.. Only the persons who loves mathematics can only come to recognise that beauty.. Thank you so much for sharing such a cute problem

[ikengas]

Sir this is beautiful to behold! Great Job. You explanations render Mathematics poetic.

[botfeeder]

I did it by finding the diagonal d of the big square, then the side S of the square as d/sqrt(2), and then got the dimensions of the rectangle by subtracting the diameters of the circles from S . But that only differs slightly from the way Professor did it.

[GeorgeFoot]

We ought to calculate the green rectangle's shorter dimension (AE) first, and check that it is greater than the smaller circle's radius before calculating the long dimension of the green rectangle.

[bradleymartinez]

I enjoy watching these practical problems

[maxmantycora]

OTHER WAY:
N and T and the intersection U of the two circles are on (BD) and DB=DN+NU+UT+TB.
DN is a diagonal of a square of side 2sqrt(2). So DN=2sqrt(2)*sqrt(2)=4
NU=2sqrt(2)
UT=sqrt(2)
TB is a diagonal of a square of side sqrt(2). So TB=sqrt(2)*sqrt(2)=2
Then DB=DN+NU+UT+TB = ... = 6+3sqrt(2) = 3sqrt(2)*(1+sqrt(2))
and DB is a diagonal of a square. So AB=3*(1+sqrt(2)) = 3+3*sqrt(2)
Then AF=AB - 2*sqrt(2)= 3+sqrt(2)
And AE=AB - 4*sqrt(2)= 3-sqrt(2)
Finally, Area of Green Rectangle is = AF*AE= (3+sqrt(2))*(3-sqrt(2)) = 3² - (sqrt(2))²= 9-2 =7

[govindashit]

Your explanation is to usefull for all students.
thanks again.
L💗VE from
West-Bengal

[mohanramachandran]

Simple to find
Diagonal. DB = DN + NT + TB
DB = 4 + 3√2 + 2 = 6 + 3√2
Diagonal. ÷ √2 = Side Value
AB = ( 6 + 3√2 ) ÷ √2

[charlesbromberick]

The key in so many problems is to say, "OK, profe, let x equal .... Then often times everything falls in place. As my favorite math professor, Alfius George Davis, once said to Shri Agarwal, "Whad ya try?" Thanks for another fun problem.

[Waldlaeufer]

I tried it like this:

x = side of the square
a and b = sides of the green rectangle

A = a * b
A = (x - 2 * sqrt(2) * (x - 4 * sqrt(2))

Here I got stuck. Then I added your value for x into the equation:

A = (3 * sqrt(2) + 3 - 2 * sqrt(2) * (3 * sqrt(2) + 3 - 4 * sqrt(2))
A = (3 + sqrt(2)) * (3 - sqrt(2))
A = 3² - (sqrt(2)²
A = 9 - 2 = 7

[theoyanto]

Nearly had it, I was 1 away 😃 ... keep 'em coming 👍🏻
Oh wait a minute...Just found my error, used the wrong triangle, ... so that's good 😊

[johnbrennan]

Was not sure if points d, n, t, b were collinear. That was key to solving it.

[kedarshukla]

ET must be a tangent of the circle of radius 2.2^1/2.Hence the line joining the centre and the tangent point must be equal to the radius as above.

[akamai]

Thank you for doing these. They are helping me to understand better procedures.

[math]

great job bro, this was challenging to me

[ΚΑΝΑΒΟΣΜΑΝΩΛΗΣ]

Dear sir, I had in mind to start directly with the length of diagonal of the square which is =6+3*sqrt(2) but it seems more complicated...Have a nice day Emmanuil Kanavos

[rivivuel]

I understand. To solve geometrical problems always use right-angled triangles. 😏👍🤓

[micke_mango]

In my opinion it would have been more elegant without trigonometry, avoiding the angles completely.
It is not necessary to know that the triangle is isosceles, just write expressions for both sides independently (in terms of the side length of the square) and you'll notice that they are equal.
You wouldn't even have to know the side length of the square (or the triangle), just write an expression of the rectangle area in terms of the side length of the square and the radii and solve the second degree equation using the pythagorean theorem