Thank you sir to this beautiful question. 👍👍♥️♥️

@PreMath

2 жыл бұрын

You are very welcome. Glad to hear that! Thank you for your feedback! Cheers! You are awesome 😀

تمرين جميل جيد . رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا . تحياتنا لكم من غزة فلسطين .

i watch the thumbnail, and say to myself, it will be very complicated to find the radius of the green circle !! Next time, i will listen the audio

You could also just add 1/4 of a circle with the area of triangle AOB, since the red element is equally removed from both the white and the green half circle. So: Pi*(11)^2/4 + (11)^2/2 = 155,49... Easier..

Wow beautiful solution👍👍 Thank you so much for sharing😊😊

I went the long way round but I found it simpler, but got it correct, great exercise thanks very much 👍🏻

The solution is 121/4(2+pi) or 155,53 square units. Take a quarter of the circle with radius r=11 and deduct have the square with sidelength r gives 34,5, then deduct this from the half circle. After watching: I did it exactly like shown in the video!

Thank you for another nice video. Yesterday also I liked your video. Thank you.

@PreMath

2 жыл бұрын

You are very welcome. Thank you for your feedback! Cheers! You are awesome Furzaan 😀

i was puzzled until you drew the vertical OA , then - easy.

Thank you teacher for a nice quation with a great solution 🙏.

i enjoy your videos, thank you sir.. much love..

A clear presentation.

Great Video!

Good 👍👍 You can find it by other short method : area of 1/4 of circle + area of right triangle Green area = 121π/4 +121/2 = 155.49

straight forward question. :) solved it!

(Pi x r ^2 ) / 4 + (11x11)/2 = answer… Find the area of the circle with 11 radius. Divided by 4 gives you the quarter of the circle on the left side. Then find the area of the inside triangle, 11x11 / 2. Add both if those gives you the area of the green semi circle.

@PreMath

2 жыл бұрын

Excellent! Thank you for your feedback! Cheers! You are awesome Corman 😀

@phungpham1725

2 жыл бұрын

D corman: Nice solution! The area of the green one is equal to that of a quater circle and the right isosceles triangle AOB.

This one was easy thanks for the problem

very well done, thanks for sharing

@PreMath

2 жыл бұрын

You are very welcome. Thank you for your feedback! Cheers! You are awesome my friend 😀

Thnx solved it Was a logical one

@PreMath

2 жыл бұрын

You are very welcome. Thank you for your feedback! Cheers! You are awesome 😀

amazing! I love this

@PreMath

2 жыл бұрын

Thank you for your feedback! Cheers! You are awesome 😀

The two circles are identical. Green area is same as lower circle minus the portion on the RHS to the chord. CB=2r=22 AOC =2×ABC=2×45=90, This is equal to Quarter of the circle+a right angle triangle ¼πr²+½.11×11=1089/7. &c

The inscribed triangle is a half circle, as the given angel measures 45°. The side length is 22/sqrt(2) = 15.57, area 242 and half of it 121 square units. The half circle area is pi()121/2, thus the white lens is half of the difference between half circle and triangle = 121(pi()/2 - 1)/2. Finally pi()121/2 - 121(pi()/2 -1)/2 = 121/2 (pi() - pi()/2 - 1) = 121/2 (pi()/2 - 1) = 34.53 square units. And 121*pi()/2 = 190.07. Green area = difference 190.07 - 34.53 = 155.54 square units. PD: Hmm... I think it's correct. Maybe a small difference because I didn't use just 3.14 for pi()

Very nice problem, thanks!😃 But, I have a question: how do you know that the green semicircle is equal to the other semicircle?

@PreMath

2 жыл бұрын

I spelled out at the beginning that both are identical! Thank you for your feedback! Cheers! You are awesome 😀

@Chrisoikmath_

2 жыл бұрын

Sorry!!🤭

@mariofabrizi5050

2 жыл бұрын

@@Chrisoikmath_ I missed that bit too.

The area of the white area minus the area of the intersection is equal to the green area. Therefore you can just add the isoceles triangle to the quarter circle.

I got it to 155.533... , but that may be accounted for by taking an alternative path and the discrepancy with assigning a decimal value to pi. I got to 30.25pi - 60.5 for the sector to be subtracted from 60.5pi.

Shaded area= (Area half of the semicircle)+(Area of right triangle with legs equal to the radius)= (πR²/4)+(R²/2)= R²[(π/4)+(1/2)]= (22 /2)²[(π/4)+(1/2)]= 155.533 Greetings to all

Same difference really, but I observed that the area of the clear semi circle less the segment is equal to the green shaded area. Thus the same answer would be the area of a quarter circle plus the area of triangle AOB.

Nice video. In step 1, you didn't prove that line AO connects at a right angle. The proof is that AO & OB are both radius of the semicircle, so that makes triangle OAB an isoceles. This makes angle A the same as B, both 45. This leaves (proves) angle AOB = 90.

@hienvo693

2 жыл бұрын

I agree with your comment 100%. The teacher didn't prove AO is perpendicular with BO.

@JLvatron

2 жыл бұрын

@@hienvo693 It seemed so obvious to me that it was perpendicular, but I realized it needed proof.

@hienvo693

2 жыл бұрын

That's right!

@devondevon3416

2 жыл бұрын

And this is how I get AOB to equal 90 since AO and OB are of equal length

@pavelllamocca5242

2 жыл бұрын

AO is perpendicular to the radius (it is easy to prove). However, it does not prove that O is the middle of the radius

Answer 155.533 Did it differently. Since OB is 11 and 0A is 11 (radius of the circle will touch any point 11 units away) and BA0=45 degrees, angle A0B is 90 degrees. let's label the left of the semi-circle as D, then segment DA = 11 and angel AD0=45 degree The area of rectangle ABD= (22x11)/2 =121 units. Since the area of the semi-circle is 190.06635 (22/7 x11 x 11 x 1/2), then the area of the two chords outside the triangle within the circle = 69.06635 ( 190.06635 - 121). Since both chords are equal then the area of one is 34.5331775. But 34.5331775 is the unshaded area of the shaded semi-circle. Therefore the shaded area is 155.5331775 ( 190.06635 - 35.5331775). Answer 155.5331775. Your answer is slighter smaller than mine because you round off pi to 3.14 for pi and I used 22/7 which is 3.14285714286 for pi.

#triangle #righttriangle

Nice

@PreMath

2 жыл бұрын

Thank you for your feedback! Cheers! You are awesome Jha 😀

ABC (C oposite of B) is isoscel, so (Area Semicircle-Area ABC)/2 is area of "RED"...done !

Nice its a quarter circle and the triangle so pi. 11^2 / 4 + 11^2 /2 so 121 .(pi/4 +1/2) the insight is it is the segment is remove from both.

Zeal for maths is increasing day by day

@PreMath

2 жыл бұрын

Excellent! Glad to hear that! Thank you for your feedback! Cheers! You are awesome 😀

Area of the red segment = Area of sector - Area of right triangle AOB Area of the red segment = 30.25 pi - 60.5

Right triangle AOB

how did you get the radius of the green shaded semi circle is same as the other semicircle?

You omitted explicitly showing that triangle AOB is right angled; this is easy to see, but cannot be automatically assumed. However, you could have saved a step or two elsewhere. There was no need to find out the area of the red segment and then subtract it from the area of the upper tilted semicircle. You could have used the fact that the green area is equal to the white part of the lower upright semicircle (as both are equal to same semicircle minus same red segment), which in turn is equal to the sum of the quarter circle (or sector) on the left and the isosceles right angled triangle on the right. Another slightly lengthier way was to join the point of intersection A with the other end of the diameter, say C, and then use the fact that angle BAC is a right angle and triangle BAC is a right angled isosceles triangle, whose area can be obtained by using Pythagoras. The area of the red segment then, by symmetry, becomes half of the difference between the area of the semicircle and triangle BAC.

@Anonymous-kw7ls

2 жыл бұрын

No. In triangle AOB, OA = OB = Radius. Then two angles will be equal that is 45. Third angle will automatically be 90.

@Red-gw6kz

2 жыл бұрын

@@Anonymous-kw7ls I know, but it still requires explaining, like you did. The narrator kind of stated this fact offhand, as if it was self-evident. Usually, he elaborates every small step in great detail.

Great

@PreMath

2 жыл бұрын

Excellent! Thank you for your feedback! Cheers! You are awesome 😀

Area of right triangle AOB = 0.5 × 11 × 11 Area of right triangle AOB = 60.5

i tried solving it without knowing the other semi circles radius but anyways. the area of a circle is πr^2. r=22/2=11 ->A=121π. we have semi circles so A=60.5π now we want to substract the thing that i dont know the name of but its a circle part minus a triangle. since the angle is 45 degrees, the angle from the middle from the circle to the intersections of the other semicircle is 90*. so the quartercircle has A=90/360•121π=30.25π. now substact the triangle. since it has a 90* and 45* angle, the other angle is also 45* so their opposite sides are equal and 11 which is half the radius. the area of a triangle is A=(bh)/2. b=11 and h=11. so A=60.5. substact that from the quarter circle so we get 30.25π-60.5. substact that from the semi circle. so green A=60.5π-(30.25π-60.5)=30.25π+60.5 that is aproximately 115.5

@PreMath

2 жыл бұрын

Excellent! Thank you for your feedback! Cheers! You are awesome 😀

The answer is 121/4 * ( pi +2) square units.

@PreMath

2 жыл бұрын

Great

Let's say the Green shaded semi circle is not identical to the 22 diameter semi circle,...how do I determine the radius for it to complete the subtraction?

@RichardVerret

2 жыл бұрын

Agreed. That's a huge step, and I don't see anything saying it's ok to assume the green shaded semicircle has a radius of 11 (or diameter 22), same as the non_shaded one in the problem picture. It was only spoken in the statement once the video begins.

@JLvatron

2 жыл бұрын

@@RichardVerret the teacher announces them at the beginning as identical semi circles. And the diameter is given.

@wackojacko3962

2 жыл бұрын

I know...but I would like too know for instance if the Green shaded had a radius of maybe 10.?

@JLvatron

2 жыл бұрын

@@wackojacko3962 Oh, sorry. In that case, you would need some way of knowing or calculating the green radius. That would easily give you the are of the green semi circle. And if the green diameter was not larger than hypotenuse AB, then the question wouldn't work.

@wackojacko3962

2 жыл бұрын

Thank you JLvatron

Nice🌹🌹🙏🙏

@PreMath

2 жыл бұрын

Thanks for visiting Thank you for your feedback! Cheers! You are awesome Engineer 😀

Area of Half circle = pi*11sq/2 = 190.06 Areaof quarter circle = pi*11sq/4 Area of Triangle = 0.5*11*11 = 60.5 Area of segment = Area of quarter circle - Area of Triangle = 34.53 Area of green Part = Area of half Circle - Area of Segment = 155.53

It is not given that the two semicircle have the same diameter. How did you make that assumption?

Ans : [121 Pie/4 + 121/2] square units

@PreMath

2 жыл бұрын

Excellent! Thank you for your feedback! Cheers! You are awesome 😀

@Teamstudy4595

2 жыл бұрын

Thanks Sir

#radius

How do we know r of the green?

155.4 yes

Area of sector = 121 pi × 0.25 Area of sector = 30.25 pi

Interesting: you get 155.49, I get 155.53. it is just about how the calculator rounds the numbers, but still the offset is too much. Any explanation from anyone? Ah, I got it! Because you take 3.14 as pi, where I use pi itself in the calculator. Wow, that really makes some difference, if you also round up pi… 😳

@devondevon3416

2 жыл бұрын

Yes, I got 155.53 too. This is because he rounded pi to 3.14 and you didn't. So he has the area of the shaded region as 7:50 121/4 ( pi +2) = 121/4 (3.14+2) = 121/4 (5.14) = 621.94/4 =155.485 and he rounded it to 155.49 (his answer) I don't know how you arrived at 155.53 ( but I got 155.53 as well using a different method than in the video to solve it. I construct two similar triangles inside the semi-circle, 45-45- 90 degrees given that AO=AB=11 and then it is a milk run from there ). However, if you used 22/7 inside his '121/4 (pi +2)', you get 155.57. These answers are all correct and depend on how much pi was rounded as pi has 31 trillion digits (as for now).

Qual'e il raggio del semicerchio verde?

Hi. How do you know O is the center of the radius?

@lsmith992

Жыл бұрын

And why is it assumed that connecting it to point A results in a right angle? What theorem is at work here, that I am unaware of? Is there a theorem? Usually any theorem involved will be brought in but no mention of any.

The answer is 155.49

How​ did​ we​ know​ that​ the​ two circle are​ same​ radius?

@PreMath

2 жыл бұрын

At the beginning, it was spelled out that that they both are identical!

@suraponwarrarak2239

2 жыл бұрын

@@PreMath Thanks

Radius = 11

Она в уме решается за 5 секунд, но для молодых, неокрепших умов пойдёт!))

@PreMath

2 жыл бұрын

Супер!

@user-rs6qg2kd8t

2 жыл бұрын

@@PreMath Советская система образования)))

1st

@PreMath

2 жыл бұрын

Excellent! You are awesome Mustafiz 😀

theta = 90°

(The green area) = (The semicircle) - (The quadrant) + (The right-angled isosceles triangle). Thus, the answer is 1/2 x π x 11^2 - 1/4 x π x 11^2 + 1/2 x 11^2 = 121/4 x π + 121/2.

@PreMath

2 жыл бұрын

Excellent! Thank you for your feedback! Cheers! You are awesome 😀

Within 2 - 3 second I mentally concluded how to solve this problem

r = 11

Sir can you explain in Hindi

It is unknowable because you didn’t supply the initial condition of the diameter of the green semicircle.

@PreMath

2 жыл бұрын

At the beginning, it was spelled out that that they both are identical! Take care Evan.

@iloveskytops66

2 жыл бұрын

@@vashon100 Wanted to solve it on my own without hints. But thanks I should have listened to the beginning at least