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

Learn how to find the area of the green rectangle if ABCD is a square and the radii of the two circles are √2 and 2√2 (Square root of 2 and 2 times the square root of two). Quick and easy explanation by PreMath.com

[@tomcruise6738]

Amazing question! ❤️ Greetings from India! 🇮🇳😍

[@PreMath]

Glad you liked it! Thanks dear for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃 Greetings from the USA!

[@sportsworld9820]

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

[@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.

[@veenadabral8237]

Yes

[@ikenga70s]

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

[@rangaswamyks8287]

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

[@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.

[@FlavianTicaDeme]

But it is one of the best and fastest solutions to the problem ! I used exactly the same method and it took about 1 minute, although I was scrupulous and proved that the centers of the two circles and the 2 opposite corners are collinear, so the diagonal is the sum of the two diameters. What followed was trivial ! Congratulations from Romania !

[@javanuwamungu5824]

@@svilponis You are right! He needs some adjustments (with a sqrt(2) factor) on each end of the diagonal. But I think his line of thought is correct but did not express his idea as clearly and precisely as it needs to be! Cheers

[@javanuwamungu5824]

Me too! :)

[@annatygrys9043]

Me too! Nevertheless it is iteresting to learn another way to solve this problem. Thanks!

[@bradleymartinez4876]

I enjoy watching these practical problems

[@maxmantycora5132]

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

[@govindashit6524]

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

[@mohanramachandran4550]

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

[@murdock5537]

Indeed, a much easier way :-)

[@ياخياشتركفيقناتي]

Simply the best from Algeria 🇩🇿

[@ياخياشتركفيقناتي]

Simply the best from Algeria

[@dwarampudilaxmanareddy5670]

one of the toughest problems. Loved your explanation Sir!!

[@PreMath]

Thanks Reddy for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃

[@Waldlaeufer70]

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

[@luigipirandello5919]

Amazing. Easy to understand.

[@charlesbromberick4247]

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.

[@kedarshukla1051]

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.

[@akamai897]

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

[@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

[@GetMeThere1]

I think you have to specify as a given that NT and DB are coincident. They're not coincident just because they APPEAR coincident. Is that correct? EDIT: Then again, I guess they DO have to be coincident, for reasons of symmetry, because ABCD is a square.

[@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 😊

[@vidyadharjoshi5714]

Once you get the value of DB why do other steps ?

[@JLvatron]

Great!

[@johnbrennan3372]

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

[@rivivuel8188]

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

[@charlesmitchell5841]

Good one! Keep em coming.

[@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

[@javanuwamungu5824]

Not at all! I would say it's even easier that way! Once you have the diagonal of a square, you can directly deduce the length of each side of the square by dividing the diagonal by sqrt(2) (by virtue of the Pythagorean theorem) Once you have that, the end result follows (almost) immediately!

[@bernardopontes4472]

Premath!!!! How can I send you a question???? It’s geometry,four circle intersections inside a square, the problem gives you the area of the intersection ( area = 1) and it wants the side of the square

[@philipkudrna5643]

Again very nice and well explained!

[@shreyanshpatel9740]

i did it !! in just 5 min

[@spiderjump]

Why not just find the length of the diagonal of the large square and then calculate the length of the large square . Then the length of the rectangle will length of square minus diameter of smaller circle. And breadth of rectangle is length of square minus diameter of larger circle .

[@PreMath]

Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃

[@Viesto1980]

I did it diferent way. I,ve calculated diagonal of the big square, and then i.ve calculated side of the square.

[@mohanramachandran4550]

Yes it is simple

[@abhishekraj6536]

How Point DNTB are collinear?

[@leomessi2112]

Best explanation sir

[@wardieleppan8443]

Why does DB necessarily pass through N and T?

[@juhivarma7160]

But how do we know that points D, N, T, B are collinear?

[@g.p.2203]

Moreover we must prove that the point of the circles contact also lies on the diagonal BD. I noticed that the author does not pay attention to such questions. It is unacceptable for mathematicians.

[@g.p.2203]

Let's prove it all. The diagonal BD is also the bisector. One of the definitions of a bisector is as follows: a bisector is a set of points inside an angle equidistant from the sides of this angle. The distance from N to AD and CD is the same and equal to the radius, so N is the point on the bisector BD. Similarly with point T. The proof that the point of the circles contact lies on the NT can be found on math-only-math.com/two-circles-touch-each-other.html This point lies not only on NT, but also on the diagonal BD, since for two points N and T there exists no more than one line that contains them both (axiom).

[@josejohnbernardaguilar2928]

@@g.p.2203 needed this, i was skeptic

[@242math]

great job bro, this was challenging to me

[@johnbrennan3372]

I was not e and if points d,n,t,b were collinear. Nice interesting solution

[@mr.knight8967]

Where did you find this type questions, Please help me,. please Please

[@tahasami3409]

Thank for premath very nice

[@JSSTyger]

I tried doing it all in my head...and got 7.

[@adamkhan7234]

How did you kniw that NT and DB were coincicident lines?

[@murdock5537]

because both circles are in a corner of the square. Then do the 45-degree-math...

[@edyap2034]

Please talk about calculus....

[@MsMostanas]

Thank you very match Sir.

[@dummylenovo7772]

How do we know that diagonal DB is coincident with NT? Isn't that an assumption? No such thing is stated in the question.

[@murdock5537]

Because both circles are in a corner of the square. Then do the math with the 45-degree...:-)

[@India-jq7pi]

Thank you sir

[@PreMath]

You are very welcome Gowri! Thank you so much for your continued love and support. Take care dear and stay blessed😃 You are awesome. Keep smiling😊

[@goelchats]

The other perception is: when shorter side of rectangle is tangent to smaller circle then the shorter side will be equal to radius of circle... Thanks and Regards...

[@nirmalajagdish8901]

Vvnice true thanks

[@parthosaha4170]

Sir I have mailed you a geometry problem.Would you please make a video on that?🙂🙂🙂

[@PreMath]

Thanks dear. I'll look into it. Take care😀

[@mtr-math]

Having subtitles makes it difficult to see the text

[@flazen6604]

Acording to this solution showed, the side parallel to AE has two points of tangency..its impossible.

[@bienvenidos9360]

The side of the rectangle meets at the smaller circle's diameter. The height of the rectangle is longer than the smaller circle's radius. The widest part of a circle is at its diameter, so the rectangle doesn't extend into the circle. It only meets at one point.

[@jaaaayt.20]

i like the video

[@theophonchana5025]

3 × square root of 2

[@colendecipulo1303]

dear mr. Premath. I really idolize you to your solutions, but I think this problem is mathematically incorrect. There's no two circle that inscribed in a square with two tangental line(of square sides) atouch each of them.

[@theophonchana5025]

Square root of 2

[@flazen6604]

I think this solution is wrong. The line EP must pass over the center of the smaller circle...Using the fact that the radio is perpendicular to the tangent at the point of tangency. The tangent is perpendicular to EP. And the tangent is pependicular to the radio at the point of tangency...so EP must pass over the center.

[@bienvenidos9360]

P is not located on the line tangent to the circle. It is below it. Since the hypoteneuse of the triangle = 3sqrt2, he made a 45-45-90 triangle, so the sides = 3, so point P is below the point of tangency since it is part of the triangle which extends past the radius (3 is greater than 2sqrt2). It wouldn't make sense to draw a right triangle that doesn't meet at the center of the circle.

[@ياخياشتركفيقناتي]

Simply the best from Algeria

[@theophonchana5025]

2× square root of 2