❤❤❤❤❤❤ Thanks Sir Thanks PreMath

@PreMath

4 ай бұрын

You are very welcome! Thanks ❤️

Solved this with the "special case perpendicular intersecting chords theorem". => 4r² = w² + x² + y² + z² 1. AB = 24. 2. Yellow square has a side length of 12 since sqrt(144) = 12 3. Draw a perpendicular form O to AB intersecting AB in half in the point P, at the same time bisecting the upper side of the yellow sqare. 4. Therfor AP = 12 5. Let Q be the upper left vertex of the yellow squere. 6. Let L be the lower left vertex of the yellow squere 7. AQ = 6 QL = 12 QB = 18 8. Elongate LQ unti it touches the Circle above Q and let us call that pont N 9. Apply intersecting chords theorem: AQ x QB = QL x QN (QN is unknown so far) => 6 * 18 = 12 * x => 108 = 12 *x =>108/12=x =>9=x =QN Therefor 4r² = 6² + 18² + 12² +9² 4r² = 36 + 324 + 144 + 81 r² = 585/4 r² = 146,25 Since Area of a circle equals PI*r², there is no need to solve for r. Instead Area = 146,25 * PI = 459,4579....something, rounded to 459,46

@PreMath

4 ай бұрын

Thanks ❤️

@jimlocke9320

4 ай бұрын

For those who aren't familiar with the "special case perpendicular intersecting chords theorem", let M be the midpoint of chord LN, which has length LQ + QN = 12 + 9 = 21, so LM has length 21/2. Construct right ΔLMO. OM has length 6, so OL, the circle's radius r and the hypotenuse of ΔLMO, from the Pythagorean theorem, has length √((21/2)² +(6)² = √(585/4) and r² = 585/4. Proceed to multiply r² by π to get the circle's area.

The best explanation method.

@PreMath

4 ай бұрын

Glad to hear that! Thanks ❤️

Yay! I solved the problem.

@PreMath

4 ай бұрын

Super❤️

Thank you!

@PreMath

4 ай бұрын

You're welcome!❤️

Thank you

@PreMath

4 ай бұрын

You're welcome🌹

At a quick glance, square sides are sqrt(144) =12. Then the red line, GH is 12 + 6 + 6 = 24 cm. Clockwise from the top left of the square ABCD. Draw a vertical line from A to the top of the circle, E, this bisects the red line, GH at F. The two products of the segments of each bisected line, DE and GH are equal. Then 6 * 18 = 12 * FE. Then FE = 6 * 18 /12 = 9. Draw a line from E to O, the circle center and O to D to form a triangle ODE. Then ( 0.5 * DE )^2 + 6^2 = (Radius^2) = (0.5 * (9 + 12))^2 + 36 =146.25 and the circle area is 146.25 * PI = 459.46 cm^2.

The 24 chord can be split at the top left (or right) vertex as 18 and 6 due to the square's sides being 12. 18*6=108 The vertical side of the square is 12, so 12x=108 gives another chord of 12 and 9 (also 108). Move the vertical chord to pass over the centre for 12*12=(x+12)(x+9) where x is the gap between the ends of the newly-moved chord and top and bottom of the circumference and the 12 is the midpoint of the 24 chord. (x+12)(x+9)=144 x^2 + 21x - 36 = 0 (-21+or-sqrt(441-4*-36))=0 (-21+or-sqrt(585)/2 Drop negative result as it would give a negative line length to add. Also, as there is an equal length above and below the new vertical chord, drop the division by 2. Diameter is 21+(sqrt(585)-21) 21+(sqrt(585)-21)=24.1867732... r=12.094 (rounded) 146.265*pi = 459.46 (2dp) sq units. I guess this way needs a calculator handy to be reasonably practical.

Let sdo some geometry

1/ From the center O drop OH perpendicular to vertical side of the square ( from point C), we have: OH = 6 2/ Extend CH cutting the AB and the circle at I and K respectively, we have CIxIK= IAxIB--> IK=9-CK=21 and HK=10.5 Using Pythagorean theorem Sq R= sq 6+ sq 10.5=146.25 Area of the circle=146.25xpi=459.459 sq cm

Let h be the height of the upper isosceles triangle, r^2=h^2+12^2=(12-h)^2+6^2, 144=144-24h+36, h=3/2, then r^2=9/4+144=585/4, therefore the area is 585/4 pi.😊

@PreMath

4 ай бұрын

Thanks ❤️

By the intersecting chords theorem, (24/2 + 12/2) (24/2 − 12/2) = 12x ⇒ x = 9 By Thales and Pythagoras, the square of the diameter of the circle is (12 + 9)² + 12² = 585 Therefore, the area of the circle is π 585/4 ≈ 459.46 cm²

@PreMath

4 ай бұрын

Thanks ❤️

@BKNeifert

4 ай бұрын

How are you getting the diameter or radius from your work? I don't see it. Can you help me? I see you somehow got the right answer, but I'm wondering if that's not from having already known it. Like, you're right, there should be a way to transpose intersecting chords theorem--I went that avenue--but I didn't know how to get to the radius or diameter, so I gave up doing that. This doesn't look right, bro. Like, something's fishy about this work you put out there. Like that's not even intersecting chords theorem. It should be written, (12*x)=(18*6). Which gives you some sloppy numbers, maybe that's why he didn't use it. But I mean you're right, there should be a way to transpose that information to the radius, which is why I'm scanning the comments for a solution to it. I just don't think yours is it.

@ybodoN

4 ай бұрын

@@BKNeifert On PreMath's diagram, drop a perpendicular from C to G on AB and extend to H on the circle. By the intersecting chord theorem, AG × GB = CG × GH ⇒ 6 × 18 = 12 × GH ⇒ GH = 9 ⇒ CH = 21. Since CDH is a right triangle inscribed in a circle, by the converse of Thales theorem, HD is a diameter of the circle. And by Pythagoras, HD² = CD² + CH² ⇒ HD² = 12² + 21² ⇒ HD² = 585. Finally, the area of a circle is π d²/4 where d is its diameter so the area of the circle is π 585/4 cm².

@BKNeifert

4 ай бұрын

@@ybodoN Well, you got the radius squared. I follow you now. Though, you need to multiply 585/4 by pi. That's where I was a little confused. Though, your second walkthrough was much more thorough and better explained. Thanks!

@BKNeifert

4 ай бұрын

@@ybodoN That's amazing how a right triangle, the hypotenuse will always equal the diameter if inscribed in a circle. That's just more evidence of God's design in our universe. Just, how does that happen? Why does that happen? Those two things are unrelated completely, but it shows a relation between a hypotenuse and a diameter. Which shows how intricately nuanced the symmetry of matter is in the universe. That's just amazing. Completely. Pythagoras would probably agree.

3/2 = 1.5, 1.5^2 = 2.25, 144 + 2.25 = 146,25 and 146,25 * π = 459.458

Area of the circle=π(3√65/2)^2=585π/4cm^2=459.46cm^2.❤❤❤ Best regards from cambodia

@PreMath

4 ай бұрын

Thanks dear ❤️🇺🇸

Thanks PreMat. I solved it by extending one side of the square and use two intersecting cords 12*y = 6 * 18 --> y = 9. Now you can say Dia^2 = 21^2 + 12^2. Diameter ^2 = 585. Thanks again

@User-jr7vf

4 ай бұрын

I used a combination of Pythagoras Theorem and Intersecting chords theorem. As it turns out the solution presented by PreMath is simpler than mine.

144=12^2 12/2=6 24-6=18 12＊x=6＊18 x=9 (9+12)/2=21/2 r^2=(21/2)^2+6^2=441/4+144/4=585/4 area of the circle : 585π/4cm^2

Let's use an adapted orthonormal. E beeing the middle of [A,B] we have E(0;0) A(-12;0) B(12;0) C(-6;-12) D(6;-12) The equation of the circle is x^2 + y^2 + ax + by + c = 0 where a, b and c are unknown. B is on the circle, so 144 - 12a + c = 0; B is on the circle, so 144+ 12a + c = 0; D is on the circle so 36 + 144 -6a -12b + c = 0. (No need to use D) The first two equations give that a = 0 and c = -144, then the third equation gives that 12b = 36 and that b = 3. So the equation of the circle is x^2 + y^2 + 3y -144, or x^2 + (y + (3/2))^2 = 144 + (3/2)^2. So we have O(0; -3/2) and R^2 = 144 + (3/2)^2 = 585 /4 The area of the circle (the disk) is then Pi. (585 / 4).

@PreMath

4 ай бұрын

Thanks ❤️

if it is to approximate the result then I can use π as 3.14 - so the answer would be 146.25 * 3.14 = 459.225

Geometrical and Graphical Resolution: The Center of the Circle is C (x ; y) or better C (12 ; y). Two points belonging to the Circumference: A (0 ; 12) and B (6 ; 0) The distance between AC and BC must be equal. Solving for x and y, knowing that x is the middle point of (18 + 6) / 2 = 24/2 = 12 (in the Graphic) It gives me that y = (x + 9) / 2 y = 12 + 9 / 2 y = 21/2 y = 10,5 So the Center of the Circle is C (12 ; 21/2) And the Radius is sqrt(146,25) The Area of the Circle is: A = 146,25 * Pi = 459,458 cm^2 Answer: The Area of the Circle is 146,25 * Pi cm^2 or ~ 459,4158 cm^2

452,16 the area of the circule because 12×12=144cm^2 and r^2×π is 452,16cm^2

@PreMath

4 ай бұрын

Thanks ❤️

A = 146.25π

r°2=6^2+(12-x)^2..x^2=r^2-12^2....r^2=180-24x+x^2=180-24x+r^2-144...24x=36..x=36/24=1,5...r^2=6^2+10,5^2=36+441/4=(144+441)/4=585/4...

@PreMath

4 ай бұрын

Thanks ❤️

I liked solving this ... but I have to say "459" feels too bigor the diagram. Caused me to go back and re-do my work, just to check. Same answer by different calculation. Then, watched video ... and you got the same answer essentially the same way. Ah well ... those crafty diagrams.

6×18/12=9 (21/2)²+6²)π (441+144)π/4 585π/4

Let's do some geometry: . .. ... .... ..... First of all we calculate the side length s of the square: s = √A(square) = √(144cm²) = 12cm May c be the length of the red chord, may r be the radius of the circle and may d be the distance between its center and the red chord. Due to the Pythagorean theorem we can conclude: r² = (c/2)² + d² r² = (s/2)² + (s − d)² (c/2)² + d² = (s/2)² + (s − d)² c²/4 + d² = s²/4 + s² − 2sd + d² c²/4 = 5s²/4 − 2sd 2sd = (5s² − c²)/4 ⇒ d = (5s² − c²)/(8s) d = [5*(12cm)² − (24cm)²]/(8*12cm) = [5*(12cm)² − 4*(12cm)²]/(8*12cm) = (12cm)²/(8*12cm) = (3/2)cm Now we can calculate the area of the circle: r² = (c/2)² + d² = (24cm/2)² + [(3/2)cm]² = 144cm² + (9/4)cm² = 146.25cm² A(circle) = πr² ≈ 459.46cm² Best regards from Germany

@PreMath

4 ай бұрын

Excellent! Thanks ❤️🇺🇸

@BKNeifert

4 ай бұрын

Wow, you're smart. I could never do that. That's very advanced. I can just read it, and see it works. But I could never do it.

@unknownidentity2846

4 ай бұрын

@@BKNeifert Oh, thanks a lot for your very kind post. Best regards from Germany

@BKNeifert

4 ай бұрын

@@unknownidentity2846 It's a learning curve for me. I like Geometry and kind of understand it, so I like to read equations. I can see the nuances of the shape in the math you posted.

@unknownidentity2846

4 ай бұрын

@@BKNeifert That's nothing to worry about. I follow a few English and German math channels since a couple of month already and I also learned a lot. So I can just recommend to face the challenges and to get some new insights, especially if the task is too difficult to master. Then you will definitely improve your skills.

459 + 637 ! Dial pad spells out important messages. 🙂

@PreMath

4 ай бұрын

Thanks ❤️