This is the best plateform for learning mathmatics

@PreMath

Жыл бұрын

Glad you think so! Thanks for your continued love and support! You are the best, Ravi. Keep it up 👍 Love and prayers from the USA! 😀

Australian time 13:23. To draw the figure correctly the line AC must bisect the angle at A giving angle DAC = 45 degrees. This is the assumption made.

Nice explanation. Respect from Bangladesh.

In drawing of the diagonal of the square, you didn’t mention why it will be tangent to both semicircles. This was a question for me when I tried to solve it. Thank you

@meaningfulsongs5231

Жыл бұрын

I had to pause at that for a couple of minutes, too. I was able to deduce it, but for a channel that lays things out in such detail, that step seemed a glaring omission.

@timeonly1401

Жыл бұрын

Notice, In the original, given diagram, that all the circles & half circles are pairwise tangent. Drawing the diagonal AC, it's relatively easy to SEE that the given diagram is symmetrical with respect to that diagonal (mirror-images on either side of AC). So, the point of tangency betw the half-circles MUST be on diagonal AC [otherwise, the diagram wouldn't be symmetrical!]. Now, draw the line segment that connects the centers of the half-circles (call it d). Clearly that is 45 degrees to AD and to AB (their centers are equidistant from point A, so segment d would be the hypotenuse of a right triangle with equal-length legs.. hence its a 45-45 right triangle). We know that the diagonal of the square AC is also 45 degrees to AD & AB (the diagonal of a square cuts it into two 45-45 right triangles). We see that diagonal AC & segment d are 45 degrees to square's side AB, but in OPPOSITE directions. Therefore, AC must be perpendicular to d. [two lines each 45 degrees to a 3rd line are either parallel, or perpendicular or the same line; clearly, they're not the same line, nor parallel, so they must be perpendicular.] Done!

MUY BIEN !!!

Thank you so much sir, for making us able to solve these kind of problems.

Nice construction. I think you should be more explict about the sides of the square being tangent to the circle though, because that's essential to showing that the diagonal of the square intersects the center of the circle.

Superb👍 Thanks for sharing😊😊

I look forward to your videos every day

Call the center of the full-circle point O, and call the center of the half-circle whose center is on the square's left side point E, and center of the other 1/2-circle point F. From the center of the full-circle, drop a perpendicular to line segment AD, where this segment intercepts AD.. call it point G. Connecting E & F, we get a 45-45 right triangle AEF, with hypotenuse EF. So, angle AFE is 45 degrees. Connecting all the centers of the circle & half-circles, we get an equilateral triangle with sides of length 2 (so that EO=2), and interior angles of measure 60 degrees. So, angle FEO is 60 degrees. And the sum of measures of those angles AEF + FEO = 45 + 60 = 105. Thus, angle GEO = 180 - 105 = 75 degrees. Consider right triangle GEO (with the right angle EGO, acute angle GEO=75, and hypotenuse EO=2). From defn of sine: GO = (EF) sin(75) = 2 sin(75). Using the sum-angle formula we get GO = 2 sin(30 + 45) = 2 [sin(30)cos(45) + sin(45) cos(30)] = 2 [ (1/2) (√2/2) + (√2/2)(√3/2) ] = 2 [(√2 + √6)/4] = (√2 + √6)/2 Call the length of the sides of the square s. So, s = GO + 1 = (√2 + √6)/2 + 1 s = (√2 + √6 + 2)/2 The area of square A = s², and using the fact that (a+b+c)² = a²+b²+c²+2(ab+bc+ac) , we get: A = (√2 + √6 + 2)²/4 = (1/4) [ 2 + 6 + 4 + 2(2√3 + 2√6 + 2√2 )] = (1/4) [12 + 4(√3 + √6 + √2 )] = 3 + √3 + √6 + √2 Done!

nice work. I made a right triangle with the hypotenuse connecting the center of one of the 1/2 circles to the center of the full circle. That triangle has a side of (.5x -1) and the other side (x-1) where x and a hypotenuse of 2 is the side of the square. Then solve for x where 2^2 = (x-1)^2 + (.5x -1) then x^2 = 8.67 units squared

Thanks

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

Centre of half circle is centre of square also. If we draw a line between these two centre that length will be 1+1= 2 units. Also it will be hypotinuous to the right angled triangle . Hence if ' a' is half side of square and two side of isosceles right angled triangle 2× a square=4 So a = root 2 There fore the side of square will be 2× root 2 Hence area of square will be 8 sq. Units. The author answer 8.6 is wrong.

@martinbrederoo373

Жыл бұрын

"Centre of half circle is centre of square also" This is not true.

thank you for your channel. ❤️❤️❤️❤️❤️❤️

pretty troubling🤔, considering the triangle connecting three centers of circles, the side of the square is 1+2cos 75+2cos 45, and squaring it to get the area 8.6 approximately. 🙂

Why does AC have to go through F?

Si ABCD es un cuadrado, la diagonal AC es eje de simetría de la figura; en caso contrario los centros de los semicírculos no estarían sobre AB y AD. Por otra parte, como los radios son iguales y las circunferencias tangentes entre sí, los centros corresponden a los vértices de un triángulo equilátero con lados de longitud 1+1=2 unidades → Área ABCD = AC²/2 = [(1)+(√3)+(√2)]²/2 = 8.59575 Ud² Gracias y un saludo cordial.

🙌🏻 I did it in one single step Radius of the full circle is 1cm thus diameter will be 2cm add it with the Radius of the first semi circle of 1cm thus it comes approx 3 cm or a little less than 3cm dir to overlap This is now the side of the square thus area is equal to side square = approx 9cm square

Thanks for the video sir!

@PreMath

Жыл бұрын

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

Nice method Sir What software/ app do you use

Very nice question !

@PreMath

Жыл бұрын

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

Lovely explanation!🥂👍❤

@PreMath

Жыл бұрын

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

@bigm383

Жыл бұрын

@@PreMath 😀😃

My result was wrong first time, correct about 8.59. Thanks do much, I am looking forward a new video!

Thanks for sharing.Can you please upload some primary 6 difficult circles questions?Thanks

@adiammu4254

Жыл бұрын

Hi sir,We are from Singapore. Here the primary levels they solve the circles questions without using Pythagoras theorems.Can you please upload more videos. Thank you 🙏

Also if you draw a line between the origins of the two semicircles and also draw two connecting lines between the origin of the full circle and the origins of the semicircles you get an equilateral triangle which means EG must be square root of 3 or 2sin60degrees.

My solution briefly is : since AE=1 & RE=1, hence we get equilateral triangle AER. According to pythagorean theorem AE^+ER^=AR^ Since AE =1, ER =1 hence AR=(/2) square root of 2 consequently AD=(/2) +(/2) =(2/2) Hence AB=BC=CD=DA=2/2 So for ABCD square area We get (AB)(BC)=(2/2)(2/2)= 8

@michelhartman9102

11 ай бұрын

I also got this solution. I don't understand the result of 8.6, it's more complicated!

∆REF is a 30-60-90 triangle

(1+2cos15)^2

Mmm , I was 4% low, but I made a good effort, my method seemed ok but must be slightly flawed, anyway very enjoyable as usual 😃👍🏻

Area of the square ABCD=6+√2+√3+√6 square units

Your method is as usual better than me.😂

I did not understand very 1st step. Joined pojnts A and C. How it is that it will surely pass through point O and center of full circle?

Area of a square= 1/2 diagional squared

Il lato del quadrato l=2sin45+2sin15+1=2,93185...per cui l'area é 2,93185^2

The drawing suggests that the centers of the semicircles are also the centers of the respective sides. But what is wrong. I built on that.

Lets do the blue Areas 1st; 1 full circle +2 half circles = 2 full circles with radius = 1unit of Area -> A=2circles=2×pi×r^2 = 6.28units squared. White area (part i)= 2 Rectangles (adjacent to angles B and D) + 1 small square (adjacent to angle A = 2(0.5×1)+(0.33x0.33) = 1+0.11 = 1.11 Add that to blue circles: 1.11+6.28 = 7.39 All u have left are insignificant "trangles/crescents that can be extrapolated to form a solitary mass that can be guessed as ~1 unit square. Total Area of ABCD = 7.39 + ~1.0 = 8.39. Actual answer = 3 + Root(2) + Root(3) + Root(6) = 8.60? Damn i was close.

if the square has room for one and a half circles, the total radius would be 3. which means the sides would be 3, which mean the area would be 9, wouldn't it?

@bentels5340

Жыл бұрын

Ummm..... Where did you get 1.5 circles? And how did you get a side length 3? If you draw a line from the center of the top semicircle to the center of the whole circle, that line will not be parallel to the side of the square. The semicircle is not stacked on top of the whole circle, so you cannot just add their radii to come up with the side length.

@businesswalks8301

Жыл бұрын

@@bentels5340 if we draw a line from one radius to the next, is the radius of the semi-circle, added to the radius of the whole circle, = 2? if so, then why would we need them to be parallel to the side of the squares to be equal size?

What about Rx2 = 1x2 = 2 2+1=3 Aire of square = 3x3=9

the highest degree of difficulty

It is 8 square unit in a simple and easy way...just 2 simple steps...from time 3.28 , and right triangle AER ,AR=Square root of 2 SO AD = one side of the square= 2*sqrt{2}, THEREFORE area =(2*sqrt{2})^2=4*2=8 unit square....prove me wrong, please, thanks

@martinbrederoo373

Жыл бұрын

I got the same answer. Eventually, I found that this requires the semicircles to be in the middle of the sides. This is not exactly the case: RD is not the same length as AR.

@michelhartman9102

11 ай бұрын

I also got this solution