Can you calculate the Green circle Radius? | (Rectangle) |

[@LawtonDigital]

Thank you! These puzzles are so much more fun than Sudoku.

[@PreMath]

Glad to hear that! You are very welcome! Thanks for the feedback ❤️🙏

[@AzouzNacir]

Let a be the distance between the points of contact of the circle with radius 6 and the circle with radius r with the length of the rectangle, then we get the equations (6+r)²=a²+(6-r)², (3+r)²=(6√6-a)²+(9-r)². Solving these two equations, we geta=4√6 ,r=4.

[@PreMath]

Excellent! Thanks for sharing ❤

[@zupitoxyt]

What an incredible job 👏. Nice question sir 😊

[@PreMath]

Glad to hear that! Thanks for the feedback ❤

[@johnbrennan3372]

Very nice question and solution.

[@PreMath]

Glad to hear that! Thanks for the feedback ❤

[@jamestalbott4499]

Nice! Using our constant friend, the Pythagorean theorem, redefining two variables into a common variable and solving. Great problem for a Saturday.

[@PreMath]

Excellent! Glad to hear that! Thanks for the feedback ❤

[@SnapDragon-y9y]

Sir you videos gives me goosebumps

[@cynthiastandley5742]

I liked this one. Great fun.

[@alanthayer8797]

NICE multiple break downs ! Like these xtra challenging types sir ! Thanku

[@PreMath]

You are very welcome! Thanks for the feedback ❤

[@cyruschang1904]

6√6 = √[(6 + r)^2 - (6 - r)^2] + √[(3 + r)^2 - (9 - r)^2] = √(24r)+ √(24r - 72) 3 = √(r)+ √(r - 3) 9 - 6√(r) + r = r - 3 r = 4

[@PreMath]

Excellent! Thanks for sharing ❤

[@cyruschang1904]

@@PreMath Thank YOU 😊

[@SkinnerRobot]

Wow! Very nice.

[@cyruschang1904]

@ Thank you 🙂

[@alexundre8745]

Bom dia Mestre Obrigado pela aula Desejo-lhe um sábado abençoado

[@PreMath]

Glad to hear that! You are very welcome! Thanks dear❤ Stay blessed ❤🙏

[@wackojacko3962]

@ 10:53 the intense ping pong of math begins. First we square...then take square root to find x ... then square root to find y ... then square to find r ... I'm having a coffee and donut this morning and can't the difference between 'em. I better go back to bed. 😊

[@PreMath]

😀Thanks for the feedback ❤

[@AmirgabYT2185]

r=4 units 🔥

[@PreMath]

Excellent! Thanks for sharing ❤

[@vaggelissmyrniotis2194]

Got it right!!

[@claytonrumley]

Very elegant.

[@PreMath]

Glad you think so! Thanks for the feedback ❤️

[@kimchee94112]

It's hard to see how the manipulation ends up correct. You could start in a different paths and ended up nowhere multiple times like many problems. It's not the math manipulation/mechanics. It's the thought process and visualization that are more important driving the math manipulation. If you don't have a clear thought process you could try this and try that until you ended up with the correct answer. Excellent video once again.

[@PreMath]

Well said!👍 Thanks for the feedback ❤

[@andrehut5983]

Easy. You know the area of the large rectangle, the area of the two circles. Subtract and you get the area of the green circle. From the area, you calculate r.

[@marcgriselhubert3915]

We use an orthonormal center K and first axis (KB), we have Q(0; 3) O(6.sqrt(6); 6) P(a; 12-r) with r the radius of the green circle. *We have VectorQP(a, 9- r) and QP^2 = (a^2) + (81-18.r+(r^2)), but also QP^2 = (r+3)^2 = (r^2)+ 6.r +9, so (a^2) +81 -18.r +(r^2) = (r^2) + 6.r +9; We simplify and get: (a^2) -24.r + 72 = 0 (Eq1) *We also have VectorOP(a -6.sqrt(6); 6 -r) and OP^2 = (a^2) -12.sqrt(6).a + 216 +36 -12.r +(r^2). = (a^2) -12.sqrt(6).a - 12.r +(r^2) +252, but also OP^2 = (r +6)^2 = (r^2) + 12.r +36, so (a^2) -12.sqrt(6).a -12.r +(r^2) + 252 = (r^2) +12.r + 36. We simplify and get: (a^2) -12.sqrt(6).a -24.r + 216 = 0 (Eq2) *In (Eq2) we replace (a^2) -24.r +72 by 0 (which is given y (Eq1)) and get: -12.sqrt(6).a + 144 = 0 and then a = 144/(12.sqrt(6)) = 12/sqrt(6) We replace a by this value in (Eq1) and get: 24.r = ((12/sqrt(6))^2 - 72 = (144/6) - 72 = 96, so finally r = 96/24 = 4.

[@PreMath]

Thanks for the feedback ❤

[@uwelinzbauer3973]

Hallo! I was afraid that we would have to deal with biquadratic or quartic equations and expressions longer than a sheet of paper, but all the difficult things canceled out. Thanks for sharing this nice geometry question! Best wishes ❤️

[@PreMath]

Excellent! You are very welcome! Thanks for the feedback ❤

[@angeluomo]

I was completely with you until 11:13, but then you went down a rabbit hole of complicated algebra and substitution. How about this? y^2=24r and x^2=24r-72. How elegant! So x^2=y^2-72, and y=6*sqrt6-x. Squaring y gives you y^2=216-12*sqrt6*x+x^2. Plug that into the equation and simplify: x=12/sqrt6. Stick this value for x into your equation 2, and you get r=4. Much simpler. The value of y is conveniently 24/sqrt6.

[@PreMath]

Thanks for the feedback ❤️

[@texitaliano64]

Considerando i due trapezi JQPG e GPOE aventi basi rispettivamente JG=a GE=b. La distanza JE=a+b=9+6√(6)-3-6=6√6 La distanza QP=3+r La distanza PO=6+r Utilizzando pitagora abbiamo che: √((6+r)^2-(6-r)^2)=b √((3+r)^2-(9-r)^2)=a a+b=√((6+r)^2-(6-r)^2)+√((3+r)^2-(9-r)^2)=6√6 √((36+12r+r^2)-(36-12r+r^2))+√((9+6r+r^2)-(81-18r+r^2))=6√6 √(24r)+√(24r-72)=6√6 √(6r)+√(6r-18)=3√6 √(r)+√(r-3)=3 √(r)=-√(r-3)+3 √(r)=3-√(r-3) r=9+6√(r-3)+r-3 r=6-6√(r-3)+r 0=6-6√(r-3) 6√(r-3)+r=6 √(r-3)=1 r-3=1 r=4

[@johnbrennan3372]

Another way to do it is to let x=(6root6)-y etc.

[@PreMath]

Thanks for the feedback ❤

[@technologyk9229]

Area of rectangle = l×b So , 9+6√6×12 =185.36 Now , area of rectangle - area of big circle - area of small circle= 44.06 Now , remaining area - unknown radius circle area 44.06-πr2 44.06=πr2 44.06×7/22=r² 14.019=r² √14.019= 4=r approx 🙂

[@rey-dq3nx]

x=√((3+r)²-(9-r)²) x=√(9+6r+r²-81+18r-r²) x=√(24r-72) y=√(6+r)²-(6-r)² y=√36+12r+r²-36+12r-r² y=√24r 9+6√6=3+6+√(24r-72)+√24r 6√6=√(24r-72)+√24r 6√6=2√6√(r-3)+2√6√r 3=√(r-3)+√r (3-√r)² = (√(r-3)² 9-6√r+r=r-3 12=6√r r=4

[@PreMath]

Excellent! Thanks for sharing ❤️

[@unknownidentity2846]

Let's find the radius: . .. ... .... ..... First of all we calculate the second side length of the rectangle: AD = BC = 2*R(big white circle) = 2*6 = 12 Now let's add points R and S such that OPR and PQS are right triangles. In this case P, R and S are located on the same line. By applying the Pythagorean theorem we obtain: OP² = PR² + OR² PQ² = PS² + QS² The green circle has exactly one point of intersection with the big white circle and it has exactly one point of intersection with the small white circle. So we obtain: OP = R(green circle) + R(big white circle) = R + 6 PQ = R(green circle) + R(small white circle) = R + 3 From the given diagram we can conclude: PR = AD − R − 6 = 12 − R − 6 = 6 − R PS = AD − R − 3 = 12 − R − 3 = 9 − R OR + QS = AB − 3 − 6 = 9 + 6√6 − 3 − 6 = 6√6 ⇒ QS = (6√6 − OR) So we finally obtain: OP² = PR² + OR² PQ² = PS² + QS² (R + 6)² = (6 − R)² + OR² (R + 3)² = (9 − R)² + (6√6 − OR)² R² + 12R + 36 = 36 − 12R + R² + OR² R² + 6R + 9 = 81 − 18R + R² + 216 − (12√6)*OR + OR² 24R = OR² 24R = 288 − (12√6)*OR + OR² OR² = 288 − (12√6)*OR + OR² (12√6)*OR = 288 ⇒ OR = 288/(12√6) = 24/√6 = 4*6/√6 = 4√6 ⇒ OR² = (4√6)² = 96 24R = OR² = 96 ⇒ R = 96/24 = 4 Best regards from Germany

[@PreMath]

Excellent! Thanks for sharing ❤

[@CLMN284]

First

[@PreMath]

Excellent! Thanks ❤

[@LuisdeBritoCamacho]

MY RESOLUTION PROPOSAL : 01) AD = BC = (2 * 6) = 12 lin un 02) Draw 2 Horizontal Lines and 2 Vertical Lines, passing by Point Q and Point O. 03) The Horizontal Distance between these two Vertical Lines is equal to (6sqrt(6)) lin un; (3 + 6) - (9 + 6sqrt(6)) = 9 - 9 + 6sqrt(6) = 6sqrt(6) 04) Now I have 2 Different Right Triangles. 05) Let the Radius of the Green Circle equal to X lin un. 06) Divide 6sqrt(6) in two Different Parts : 1) Y and 2) (6sqrt(6) - Y) 07) Using the Pythagorean Theorem I have : 08) (X + 3)^2 = (9 - X)^2 + Y^2. Note that (9 - X) = (12 - 3 - X) 09) (6 + X)^2 = (6 - X) + (6sqrt(6) - Y)^2 10) Now I have a System of 2 Nonlinear Equations with 2 Unknows 11) Solutions : 12) X = 4 lin un and Y = 2sqrt(6) lin un Therefore, MY BEST ANSWER IS : The Radius of the Green Circle is equal to 4 Linear Units.

[@PreMath]

Excellent! Thanks for sharing ❤️