Cut each green area lengthwise with a chord. You now have four identical areas. Distribute these around the inside of the circle. Interior to them they define a square with diagonal 14. Area of the green regions equals area of the circle minus area of the enclosed square, which turns out to be 49pi-98.

@engralsaffar

Жыл бұрын

That is exactly what I did 👍🏼

@PreMath

Жыл бұрын

Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@mariangorski

Жыл бұрын

Exelant idea and simple calculations. Congratulations

@jimlocke9320

10 ай бұрын

Extremely well done! I also found the four identical areas. The area of each is equal to a quarter circle with radius 7 less that of a 45°-45°-90° special right triangle with side length 7. So, each area is (1/4)π(7²) - (1/2)(7²) = 49π/4 - 49/2 which can be written as (49π - 98)/4. Multiply by 4 and I get the same answer that you and PreMath got, 49π - 98. However, you were clever enough to place the 4 identical areas inside one circle and simplify the calculations! One thing missing is that I didn't see a proof that the combined green and brown areas combined form a semicircle, which is a critical requirement for all 3 solution methods to produce a correct answer.

I am sorry but when you assumed the circles intercest in the middle , when you said its not to the scale. What if the center of the middle circle is below the line you drew.

What an awesome problem and an elegant solution! Blessings to you!

Beautiful sharing ❤😊

@PreMath

Жыл бұрын

Glad you think so! Thanks for your continued love and support! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

It's so cool when dropped perpendiculars and similarities between shapes are revealed too simplify solutions. I absolutely love it.!🙂

@PreMath

Жыл бұрын

Excellent! Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

Here's a different and possibly simpler method: Consider the left-hand green area. Draw a chord cutting it in half. The portion of the left-hand green area below and to the left of the chord is equal to one-fourth the area of the upper circle minus the area of an isosceles right triangle with legs of length seven. That is, 49π/4 - (1/2)(7)(7) = 49π/4 - 49/2 = 49(π/4 - 1/2); But the total green area is four times this, so we have (4)(49)(π/4 - 1/2) = 49(π - 2) = 49π - 98. Cheers. 🤠

@PreMath

Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@Abby-hi4sf

Жыл бұрын

Will you please label the triangle? I appreciate if you corelate it with Premath video @2:27, if it is not much trouble? Thanks in advance

Amazing 👍 Thanks for sharing😊😊

Loved it!

Thanks for video.Good luck sir!!!!!!!!!!!

@PreMath

Жыл бұрын

You are very welcome! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

S=49(π-2)≈56

Very good solution. You could also consider that, if yuo call T the intersection of the three circles (the midpoint of segment AB) the green area is four times that of the circular segment subtended by the chord DT (or CT). But the area of the circular segment is that of the quarter of circle DAT minus that of the triangle DAT. So, called As the area of the segment: As = (1/4)(π·49) - (1/2)·49=(49/4)·(π-2). So the green area is 49(π-2).

The given conditions should have included a statement that the middle circle intersects the tangent point of the two lower circles. Just eyeballing it as such isn't really enough.

@peterwallace8441

Жыл бұрын

Absolutely. The diagram as drawn doesn't give a unique solution.

Thanks for the problem! I solved it similar to the answers in the comments by finding the equation for 1/2 the green area in one circle giving ((π-2)/4 )r^2 or (π-2)r^2 for the total green area. At the end substituting 7 for r.

so the general formula for this case is πr^2-2r^2

@sandanadurair5862

Жыл бұрын

Good observation

@PreMath

Жыл бұрын

Excellent! Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

Mind blowing 😮

@PreMath

Жыл бұрын

Thanks for your feedback! Cheers! 👍 You are awesome, Zubair. Keep smiling👍 Love and prayers from the USA! 😀

Another masterpiece... I did it the other way, but more than one way is good to see🤓👍🏻

@PreMath

Жыл бұрын

Glad to hear that! Thanks for your feedback! Cheers! You are awesome, Ian. Keep smiling👍 Love and prayers from the USA! 😀

Thanks Professor!👍

@PreMath

Жыл бұрын

You are very welcome! Thanks for your continued love and support! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

@bigm383

Жыл бұрын

@@PreMath 🥂😀

Joining points A, B, and O. AB = 14. The height of O above line AB = 7. Therefore, angle ABO = 45 degrees. Therefore, the green areas each subtend an angle of 90 degrees with respect to their centres. Then calculating the area of a 1/4 circle, subtracting the area of a 7 x 7 right angled triangle, then multiplying by 4. [ ( Pi x 49 /4 ) - (1/2 x 7 x 7 ) ] x 4 = 55.94.

@PreMath

Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

Thanks professor!! Mensuration also cover so help for exam .. ❤

@PreMath

Жыл бұрын

You are very welcome! Nice suggestion... Thanks for your feedback! Cheers! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

Thanks premath

@PreMath

Жыл бұрын

You are very welcome! Thanks for your continued love and support! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

V nice presentation sir🎉😊

@PreMath

Жыл бұрын

So nice of you, Ramesh Thanks for your feedback! Cheers! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

May I know ,who inspired you to enlighten in scopes of math

@PreMath

Жыл бұрын

I'm indeed very blessed and grateful for that 🙏

As a formula: the green shaded area is _r² (π - 2)_ where _r_ is the radius of the circle.

Let's call H the intersection point of the three circles, then DH segment, which divides the intersection of the circle with center O and the circle with center A into two equal parts, can be calculated with the Pythagorean theorem DH =sqrt(DO^2+ OH^2)= 7sqrt(2). DH is also the chord of the circle with center A and we can calculate the area of the relative circular segment by subtracting the area of the right triangle BOA (7*7*0.5=49/2) from the area of the circular sector BOA (7^2)/ 4*pi = 49/4*pi and multiply the result obtained by 4 4*(49/4*pi - 49/2) This leads to 49pi - 98

@PreMath

Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

4*(Area of arc - Area of triangle) = 4*49*(pi/4 - 1/2) = 55.94

Quarter. Circle. - Triangle. =1/4 Green Area ( 1/4. * π * 7 * 7 ) - ( 7 * 7 ÷ 2 ) = 1/4 Green Area 38.4845. - 24.5 = 13.9845 13.9845 * 4 = 55.938 Ans = 55.938 square units

@PreMath

Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

1/4 of the green area is 1/4 circle area (49pi/4) minus 7x7 right-angled triangle area (49/2) => green area is (49pi/4-49/2)*4 = 49pi-98

@PreMath

Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

👍👍👍👍👍👍

All solutions become easy mental calculations if you take π value as 22/7.!

(14²/2) - (14² - 7²Pi) =7²(Pi - 2) =55.94 Gracias y saludos.

@PreMath

Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

2:12 How come AB and CD are straight lines and ABCD is a rectangle🤔 We need some proof, Billy, as the Treasure Island meme says 😊

Güzel soru.

@PreMath

Жыл бұрын

Thanks dear You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

49pi-98

Green Area = semicircle Area- Brown , radius (r)=7, Area circle =76.969, 112-76.969=21.03 Brown Area, 49π-98=55.938

@PreMath

Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

There ia a point missing in the demo, you have first to justify/demonstrate that both top 2 intersection points and O are colinear. No given data can tell us that withouth some kind of reasoning. That point is important to justify talking about a semi-circle. As coarllary, A, B and the bottom intersection point are also colinear and therefore both lines are parallel.

A(green) =4((1/4pi49-49/2))=pi49-98

@PreMath

Жыл бұрын

Excellent!! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

56 square units (where pi = 22/7)

49pi-98=55.938

how about this : Area of ​​a circle - the square inside the circle. So we have 49π - (7√2)² ≈ 55.938 ≈ 55.94

4 times area of segment

Draw radii from point A to the two points on Circle A intersected by Circle O. We'll name those points P (intersecting Circles O and A) and W (intersecting all three circles). By observation, ∠PAW is 90°, so ∆PAW is an isosceles right triangle. By observation, as above construction can be done twice in Circle O and once in Circle B, connecting the green areas with the respective circle centers an additional three times, and all three circles-and thus all four sectors and triangles-are congruent, the total green area is equal to the difference between the Sector PAW and the Triangle ∆PAW, multiplied by four. Triangle ∆PAW: A = bh/2 = 7(7)/2 = 49/2 Sector PAW: A = (θ/360)πr² = (90/360)π(7²) = 49π/4 Green area: A = (49π/4 - 49/2)4 = (4)(49/2)(π/2 - 1) A = 98(π/2 - 1) = 49π - 98 ≈ 55.94

Зелёная площадь - это разность полукруга и коричневой. Но коричневая - это прямоугольник минус две четверти (т. е. тот же полукруг). Раскрываем скобки, получаем полукруг минус прямоугольник плюс полукруг. Т. е. полный круг минус прямоугольник со сторонами r и 2r (2r²). S=πr²-2r²=49π-98.

@PreMath

Жыл бұрын

Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

@zawatsky

Жыл бұрын

@@PreMath всегда пожалуйста.

This is wrong I guess my answer is 56 exact I’ve learned alot form you my solution is like this.. we form a square around circle a which will side of diameter of circle =14 area 196 now subtract the area of a circle 154 to get your 4-semi green area =42 then divide by 2 to get one whole green area =21 now take area of semi circle which is 77 subtract green area - 77-21 to get answer exact 56… if I am wrong please correct

@poopytowncat

Жыл бұрын

_"the area of a circle 154"_ No. The area of a circle is π * radius squared or π * 49 which will never be an exact whole number. If you switch your calculator to "scientific" mode you will see a π key. If you get the area of a circle and semi circle right by using π your method gives the right answer.

@rishabharya1072

Жыл бұрын

@@poopytowncat sry I am not a mathematician of sort I just use my head to count these ,it fun daily puzzle to me I learned that π has value of 22/7 in school which is not that accurate now I learned.😊

@poopytowncat

Жыл бұрын

@@rishabharya1072 -- I'm an old retired guy now but I enjoy the math puzzles and learning what I can. I remember learning 22/7 once a long time ago before we had calculators. I see it's really close to π - only about 0.05 percent off. And the 7 in the 22/7 and 7 in the problem do give an exact number. Tricky! IMO use π (or pi) in your math until the very end when you need a decimal number answer.

2x(7^2pi/2-7^2)=98x(pi/2-1)=55.938 approximately. 🙂

@PreMath

Жыл бұрын

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

Simpler way. Get the area of the rectangle (98 sq) and substract 1/4 area from each bottom circle (38.485 x 2 = 76.97) and that leaves 21.03. Calculate 1/2 of the area of the top circle (76.97) and subtract 21.03 from that and you get 55.94 sq units.

@PreMath

Жыл бұрын

Excellent! Thanks for your feedback! Cheers! You are awesome. Keep smiling👍 Love and prayers from the USA! 😀