You're my favourite teacher Plss pin sir

@PreMath

Ай бұрын

Pinned❤️

Thank you! Liked the introduction of the square, solving the overlapping circles.

@PreMath

Ай бұрын

Glad it was helpful! You are very welcome! Thanks ❤️

@christopherellis2663

Ай бұрын

Yes, I trick I missed

AB = 8 cm R = AB / √2 = 8/√2 = 4√2 cm r = ¼ R = ¼ 4√2 = √2 cm Purple Area : A = A1 - A2 A = ¼πR² - [ 4 (½πr²) - 2(¼.πr²-½r²) ] A = ¼π(4√2)² - 2π√2² + 2(¼π√2²-½√2²) A = 8π - 4π + 2(π/2-1) A = 13,708 cm²

@PreMath

Ай бұрын

Thanks for sharing ❤️

△AOB is an isosceles right triangle. c = a√2 8 = R√2 R = 8/(√2) = (8√2)/2 = 4√2 The radius of the large quadrant is 4√2 units long. The radii of the large quadrant are split into the diameters of two yellow semicircles each. 4r = 4√2 r = √2 The radii of the yellow semicircles are √2 units long each. But two of the semicircles overlap, so we'll have to add the overlapping area once in the equation: Purple region area = Large Quadrant Area - 4 Yellow Semicircle Areas + Overlapping Yellow Area A = (πR²)/4 = [π(4√2)²]/4 = (32π)/4 = 8π A = (πr²)/2 = [π(√2)²]/2 = (2π)/2 = π The combined area of the yellow semicircles is 4π u². Label the point of intersection on the yellow semicircles other than point O as point C. Label the centers of the yellow semicircles as D (vertical wall) & E (horizontal ground). Draw radii CD & CE. They are tangent to each other and form a right angle. This forms square CDOE. Draw diagonals CO & DE. We get an isosceles right triangle △DOE by the Rhombus Opposite Angles Theorem, and CO = DE by the Rectangle Diagonals Theorem. c = a√2 DE = (√2)(√2) DE = 2 So, CO = DE = 2. Diagonal CO divides the overlapping area in half as the line of symmetry. The split areas forms two minor segments of both circles, which are congruent. Segment Area = Small Quadrant (Sector DCO) Area - △CDO Area A = (πr²)/4 = [π(√2)²]/4 = (2π)/4 = π/2 A = (bh)/2 = [(√2)(√2)]/2 = 2/2 = 1 So, the area of the minor segment is π/2 - 1 u². Rewrite as (π - 2)/2. So, the total area of the overlapping yellow region is π - 2 u². Purple region area = 8π - 4π + (π - 2) = 4π + π - 2 = 5π - 2 The area of the purple shaded region is 5π - 2 square units (exact), or about 13.71 square units (approximation).

@PreMath

Ай бұрын

Excellent! Thanks for sharing ❤️

Another excellent Maths lesson Professor …! Well explained , clear solution . Thanks for sharing…..

@PreMath

Ай бұрын

Glad you liked it! You are very welcome! Thanks ❤️

Sir ur a genius i get shocked when i see your question that you apply basic concept to a higher level 😊😊

@PreMath

Ай бұрын

So nice of you dear 🌹 Thanks ❤️

@ 5:50 , I like that square! 🙂

@sandanadurair5862

Ай бұрын

I too enjoyed the introduction of square OCDE. Wonderful construction.

@PreMath

Ай бұрын

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

I really enjoyed solving this.

@PreMath

Ай бұрын

Excellent!🌹 Glad to hear that! Thanks ❤️

watching your videos make my day happier , thanks sir

@PreMath

Ай бұрын

It's my pleasure🌹 Thanks for your continued love and support!❤️

شكرا لكم على المجهودات يمكن استعمال OA=4racine2 ,r=racine2 S=1/4pi OA^2 -2pi r^2+a a=2[1/4 pi r^2-1/2 r^2] S=5pi-2

@PreMath

Ай бұрын

Excellent! Thanks for sharing ❤️

The purple area is A = 3.41. I used tan theta = adjacent / hypoteneuse. So, tan45 = BO/8; BO = .707*8 = 5.6. Divide by 4 and one gets 1.4 for the radius of the semi - circles. 1/2 AO*BO ns one gets 1/2 * 5.6 * 5.6 = 16. the area of one of the half circles is a = 1/2*pi * (1.4)^2

Got it! Many thanks.

@PreMath

Ай бұрын

Excellent! You are very welcome! Thanks ❤️

AB=8→ OA=R=4r=8/√2→ r=√2. En la figura propuesta hay ocho cuartos de círculo amarillo; dos de ellos se superponen y la superficie resultante es r*r → Área púrpura =(πR²/4) -(6πr²/4) -r² =[π(4r)²/4]-(6πr²/4)-r² =5π-2. Gracias y saludos.

@PreMath

Ай бұрын

Excellent! Thanks for sharing ❤️

Can I use the Calculator? Ok. 1) Let "R" be the Radius of the Big Circle and "r" the Radius of the Small Circles. 2) R^2 + R^2 = 8^2 ; 2R^2 = 64 ; R^2 = 64/2 ; R^2 = 32 ; R = sqrt(32) ; R = 4*sqrt(2) So: 3) if R = 4*sqrt(2) ; then r = (4*sqrt(2)) / 4 ; r = sqrt(2) 4) Area of a Quarter of Big Circle = 4A = Pi * R^2 ; 4A = Pi * 32 ; A = (Pi * 32)/4 ; A = 8 * Pi ; A = 8Pi Square Units 5) Area of the Small Semicircle = 2S = Pi * r^2 ; 2S = Pi * 2 ; S = Pi Square Units 6) A = 8Pi sq un and S = Pi sq un 7) Now we have 3 Small Semicircles and a Small Square (Q) with Faces = sqrt(2) lin un. 8) 3S = 3*Pi sq un and Q = sqrt(2) * sqrt(2) ; Q = 2 sq un 9) Purple Region Area; P = A - (3S + Q) ; P = 8*Pi - (3*Pi + 2) ; P = 8Pi - 3Pi - 2 ; P = (5Pi - 2) ; P ~ 13,708 sq un 10) Answer: The Purple Region Area is equal to : (5Pi - 2) Square Units or approx. 13,708 Square Units.

@PreMath

Ай бұрын

Excellent! Thanks for sharing ❤️

Essa foi bacana, obrigado!

What a solution thanks for solution and question both🎉🎉

@PreMath

Ай бұрын

Glad to hear that! You are very welcome! Thanks ❤️

I used an almost-exactly similar solution, with two critically different steps. I set the little-circle radius (arbitrarily) to [1], and then reasoned that whatever answer happened, it would be scaled by the [8] diagonal at the end. [1.1]  Quarter circle radius = 1 ⊕ 1 ⊕ 1 ⊕ 1 [1.2]  Quarter circle radius = 4 [2.1]  Quarter circle area = ¼ π𝒓² [2.2]  Quarter circle area = ¼ π4² [2.3]  Quarter circle area = 4π [3.1]  Yellow circle area = π1² = π So "in my head" I observed that there were 2 half yellow circles, 2 quarter yellow circles and the 1 × 1 yellow square in the bottom left area. Thus… (where mauve is 'light purple', because not scaled yet) [4.1]  Mauve = (4 - (½ ⊕ ½ ⊕ ¼ ⊕ ¼))π - 1 … unit square at end [4.2]  Mauve = (2.5)π - 1 Now to scale it [5.1]  diagonal of radius-4 = √( 4² ⊕ 4² ) [5.2]  diagonal of radius-4 = √(32) [5.3]  diagonal of radius-4 = 4√2 [6.1]  Purple = ( 8 ÷ 4√2 )² • Mauve [6.2]  Purple = ( 2 / √2 )² • Mauve [6.3]  Purple = ( 4 / 2 ) • Mauve [6.4]  Purple = 2 • Mauve [6.5]  Purple = 2 • (2.5 π - 1) [6.6]  Purple = 5π - 2 [6.7]  Purple = 13.708 u² Which is Exactly what the Good Professor also obtained. ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅

@PreMath

Ай бұрын

Bravo! Thanks for sharing ❤️

@robertlynch7520

Ай бұрын

@@PreMath Thank you for noticing my derivation closely enough to warrant a "Bravo!" super-comment. It confirms that all this writing-of-comments I do is not entirely wasted effort. Your videos are quite good, PreMath. Consider (as a suggestion) using the special triangle "20-21-29" in some future 'problems'. I particularly like this one because it is the only (rare) triangles that has the first two numbers separated by [1], and the last factor is a PRIME (29). The first such is -- of course -- 3, 4, 5. The next is -- 20, 21, 29 After that is -- 4059, 4060, 5741 Then ... -- 23660, 23661 and 33461 There arean INFINITE number of these (of course), but rather remarkably (and I haven't the number-theoretic chops to figure out why), they are separated by a mulplicative distance of around 1,353x ... (4059 is 1353 * 3) for the lower one, and (23660 / 20 = 1183) for the second type. And this goes on, and on forever. The program I wrote (using quad-precision floating point, which is HARD) ran for a couple of days and produced numbers up into the hundreds-of-billions, and all the candidates followed this more-or-less scaling pattern. So ... 20, 21, 29 is in a way, even more special than 3, 4, 5. Use it! Yay team! Go Bears (UC Berkeley reference, my alma mater)!!!

@PreMath

Ай бұрын

@@robertlynch7520 Thanks for the suggestion!

@ybodoN

Ай бұрын

​@@robertlynch7520 I too had noticed that 20, 21, 29 is the Pythagorean triangle the closest to an isosceles right triangle. However, I never thought there could be a whole series of them. Your results seem to suggest that as it grows, the triangle tends toward a half-square. Therefore, I suppose the third number should tend to be the average of the first two times the root of two… 🤔I just sent PreMath a problem based on this particular triangle 😉

Let R be the radius of the quarter circle, and r be the radius of the four semicircles. As OA = OB and ∠O = 90°, ∆AOB is an isosceles right triangle, and ∠BAO = ∠OBA = 45°. Triangle ∆AOB: OA² + OB² = AB² R² + R² = 8² 2R² = 64 R = √(64/2) = √32 = 4√2 R could have also been determined by multiplying AB by the sine or cosine of 45°, 1/√2. By observation, r = R/4 = 4√2/4 = √2. The purple area is equal to the area of the quarter circle minus the areas of the four semicircles, plus the two circular segments formed by the overlap of the two semicircles nearest O. Let C be the point of intersection between the circumferences of the two semicircles nearest O, and let P be the center of the semicircle directly to the right of O, along OB. Draw triangle ∆OPC. As PO = PC = r, ∆OPC is an isosceles right triangle. This means that sector OPC covers 90°. Circular segment OC: A = sector OPC - ∆OPC A = (θ/360)πr² - r²/2 A = (90/360)π(√2)² - (√2)²/2 A = 2π/4 - 2/2 = π/2 - 1 Purple area: A = πR²/4 - 4(πr²/2) + 2(π/2-1) A = π(4√2)²/4 - 4(π(√2)²/2) + π - 2 A = 32π/4 - 8π/2 + π - 2 A = 8π - 4π + π - 2 = 5π - 2 ≈ 13.71

@PreMath

Ай бұрын

Excellent! Thanks for sharing ❤️

Radius R of the quarter circle OAB is 8/√2 = 4√2. Area of the quarter circle OAB is ¼ (4√2)² π = 8π ........... ① Radius r of a yellow semicircle is ¼ (4√2) = √2. Area of a yellow semicircle is (√2)² π = 2π .........................  ② Area of the overlapping yellow region is ½ (2π − 4) = π − 2 .............................................................................  ③ Area of the purple region is ① − ② − ② + ③ = 8π − 4π + π − 2 = 5π − 2 square units. Thank you PreMath! 🙏

@PreMath

Ай бұрын

Excellent!🌹 Thanks for sharing ❤️

Quadrant radii are 8/sqrt(2), so 4*sqrt(2). This give quadrant area as 8 pi. Radii of small semicircles are sqrt(2), so each semicircle has an area of pi. Quadrant area - area of two semicircles = 6pi. How to figure out the area of the two overlapping semicircles:? Make a square with the two centre points, the intersection point, and O. This square has sidelengths of sqrt(2), so an area of 2. We then have two small quadrants to add which have radii of sqrt(2). Area of each small quadrant is (2pi)/4, so (1/2)pi. As there are two of them, the combined small quadrant areas is pi. Therefore, the total area of all the yellow regions is pi + pi + pi + 2, so 3pi + 2. This must be subtracted from 8pi, giving 5pi - 2 for the solution. 5pi approximates to 110/7 and 2 is 14/7, giving 96/7. This is 13 and 5/7. 13.7 is a reasonable approximation without using a calculator and using 22/7 as pi. Now to watch the video. Wow! Looks like I got one right :)

@ybodoN

Ай бұрын

For the area of the overlapping semicircles, which we can also call a lens, I simply subtracted the area of the inscribed square from the area of the circle, divided it by four. Doing so, I get the area of the circular segment which also is half of this lens. 😉

@PreMath

Ай бұрын

Bravo! Thanks for sharing ❤️

Rp=8/√2=4√2...ry=4√2/4=√2,calcolo T ,intersezione dei 2 piccoli semicerchi gialli T=(√2,√2)..l'area intersezione risulta π(1-√2/2)...Ay=4π(√2)^2/2-π(1-√2/2)=3π+(√2/2)π...Ap=π(4√2)^2/4-Ay=8π-(3+√2/2)π=(5-√2/2)π

@PreMath

Ай бұрын

Thanks for sharing ❤️

Let's find the area: . .. ... .... ..... Since OAB is a right triangle, we can apply the Pythagorean theorem to calculate the radius R of the quarter circle: AB² = OA² + OB² = R² + R² = 2*R² ⇒ R = AB/√2 = 8/√2 = 4*2/√2 = 4√2 Therefore the radius r of one yellow semicircle turns out to be r=R/4=√2. The yellow area consists of four yellow semicircles, but two of these do overlap. The area of the two overlapping semicircles can be divided into two quarter circles with the same radius and a square with a side length being equal to that radius: A(yellow) = 2*A(semicircle) + 2*A(quartercircle) + A(square) = 2*πr²/2 + 2*πr²/4 + r² = (3π/2 + 1)*r² = (3π/2 + 1)*(√2)² = (3π/2 + 1)*2 = 3π + 2 Therefore the size of the purple area turns out to be: A(purple) = A(quarter circle) − A(yellow) = πR²/4 − A(yellow) = π(4√2)²/4 − (3π + 2) = 8π − (3π + 2) = 5π − 2 ≈ 13.71 Best regards from Germany

@PreMath

Ай бұрын

Wow! Thanks for sharing ❤️

Quadrant 32 pi/4= 25.13... ((32^½)/4)² × pi= one small circle Approx area remaining 12

@PreMath

Ай бұрын

Thanks for sharing ❤️

S=5π-2≈13,708≈13,71😊

@PreMath

Ай бұрын

Thanks for sharing ❤️

hi, can you evaluate my solution because the shaded surface is somehow =13? 10 vdu5:for a=0 to 15:gcola:print a;:next a:vdu1:@zoom%=@zoom%*1.4 20 print "premath-can you find area of the purple shaded region":nu=99:sw=1/nu 30 l1=8:r1=l1/sqr(2):r2=r1/4:xs=sw:goto 60 40 ys=sqr(r2^2-(xs-r2)^2):dgu1=xs^2/r2^2:dgu2=(ys-r2)^2/r2^2: 50 dg=dgu1+dgu2-1:return 60 gosub 40 70 dg1=dg:xs1=xs:xs=xs+sw:xs2=xs:gosub 40:if dg1*dg>0 then 70 80 xs=(xs1+xs2)/2:gosub 40:if dg1*dg>0 then xs1=xs else xs2=xs 90 if abs(dg)>1E-10 then 80 100 print xs,ys:xg1=0:yg1=0:xg2=1:yg2=1:goto 190 110 a11=xg2-xg1:a12=yg2-yg1:a131=xp*(xg2-xg1):a132=yp*(yg2-yg1) 120 a21=yg1-yg2:a22=xg2-xg1:a231=yg1*(xg2-xg1):a232=xg1*(yg1-yg2) 130 a23=a231+a232:a13=a131+a132 140 ngl1=a12*a21:ngl2=a22*a11 150 ngl=ngl1-ngl2:if ngl=0 then print "keine loesung":end 160 zx1=a23*a12:zx2=a13*a22:zx=zx1-zx2 170 zy1=a13*a21:zy2=a23*a11:zy=zy1-zy2 180 xl=zx/ngl:yl=zy/ngl:return:rem print "x=";xl;"y=";yl 190 masx=1200/r1:masy=900/r1:if masxr1/sqr(2) then else 250 240 xor=sqr(r1^2-yo^2) 250 if yo>2*r2 then 270 260 xol=sqr(r2^2-(yo-r2)^2):goto 280 270 xol=sqr(r2^2-(yo-3*r2)^2) 280 da=(xor-xol+xur-xul)/2*dy:ar=ar+da 290 x=xul:y=yu:gosub 210:xb1=xbu:yb1=ybu:x=xur:gosub 210:xb2=xbu:yb2=ybu 300 y=yo:x=xor:gosub 210:xb3=xbu:yb3=ybu:x=xol:gosub 210:xb4=xbu:yb4=ybu:goto 320 310 line xb1,yb1,xb2,yb2:line xb2,yb2,xb3,yb3:line xb3,yb3,xb4,yb4:line xb4,yb4,xb1,yb1:return 320 gosub 310:goto 340:rem an y=x spiegeln *** 330 gosub 110:x=xp+2*(xl-xp):y=yp+2*(yl-yp):gosub 210:return 340 xp=xul:yp=yu:gosub 330:xb1=xbu:yb1=ybu:xp=xur:gosub 330:xb2=xbu:yb2=ybu 350 yp=yo:xp=xor:gosub 330:xb3=xbu:yb3=ybu:xp=xol:gosub 330:xb4=xbu:yb4=ybu:gosub 310 360 xul=xol:xur=xor:yu=yo:next a:agesu=2*ar:print "die gesuchte flaeche=";agesu:goto 370 365 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return 370 gcol8:x=r1:y=0:gosub 210:xba=xbu:yba=ybu:x=0:y=0:gosub 210:xbn=xbu:ybn=ybu 380 gosub 365:y=r1:gosub 210:xbn=xbu:ybn=ybu:gosub 365:gcol 10:x=r1:y=0:gosub 210 390 xbn=xbu:ybn=ybu:gosub 365 7136359 8 9 10 11 12 13 14 15 remath-can you find area of the purple shaded region 1.414213561.41421356 die gesuchte flaeche=13. 0 1 2 3 4 > 5 6 7 run in bbc basic sdl and hit ctrl tab to copy from the results window

PI/2

Need too heavy workload in computation😢, large radius is clearly 4sqrt(2), so small is sqrt(2), overlapping region's area is 2×(1/4×2×pi-2/2)=pi-2, therefore the answer is 1/4×32×pi-2×2×pi+(pi-2)=5pi-2.😅😊😂

@PreMath

Ай бұрын

Excellent! Thanks for sharing ❤️

Sarı alan 1.57. Mor alanda 1.57 birim kare.

Trying to do this in my head is killer. 6π-2?

idk but i got 12.57 as the answer