You give us such interesting problems. When I was in high school I lost interest in math because I was interested in why things work and not as much as how they worked. I lost interest in math because it was all repetition without foundation. You combine the why and how in succinct steps creating great interest in the why as well as to the how. Keep up the good work.

@PreMath

4 ай бұрын

Thanks for your continued love and support! ❤️

AD = BC = 6 and OP = 4. So QO = ED = 2 and AE = AO = EO = 4 ⇒ AEO is an equilateral triangle and AEF is a 30° - 60° - 90° right triangle. Therefore, AF = 4√3 and the area of ⊿AEF = ½ 4²√3 = 8√3. The area of the semicircle is ½ π4² = 8π. So the yellow region is 8 (π + √3) u².

Wow nice sharing thanks for sharing brother full watch❤

Nice One.👍😊

@PreMath

4 ай бұрын

Thanks ❤️

First view and a very nice solution with good explanation thanks sir

@PreMath

4 ай бұрын

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

Thank you!

@ 3:10 is the moment of enlightenment! 🙂

@PreMath

4 ай бұрын

Glad to hear that! Thanks ❤️

Yellow region area=8π+8√3square unit=38.99 square units.❤❤❤ Thanks.

@PreMath

4 ай бұрын

Excellent! You are very welcome! Thanks ❤️

Very neat solution

@PreMath

4 ай бұрын

Glad to hear that! Thanks ❤️

1/ OQ//=1/2AE--> AE=4=1/2EF so the triangle AEF is a half of an equilateral one of which the side = EF=8 Area of triangle AEF=64sqrt3/8=8sqrt3 Area of the yellow part=8sqrt3+8 pi

@PreMath

4 ай бұрын

Thanks ❤️

Belíssimo problema de geometria. Muito obrigado por me ensinar a resolvê-lo.

At a quick glance, Split the yellow region into three segments, OAF, OFE and OEA. Bisecting AF at G, OG = 6-4 =2. AG = sqrt(16-4) = sqrt(12)=sqrt(3)*2. Angle FOA =2 * Cos^-1(0.5) = 120. area of OAF, 1st segment = sqrt(3) * 2 * 2 = 6.93 . Then Angle OFE = 360 - 120 -60 =180 . Then 2nd segment =PI *16 * 1/2 =25.13 Then Angle AOE = 60 and 3rd segment = 2 * sqrt(3) * 2 =6.93. Yellow region area = 6.93 * 2 + 25.13 = 38.9 area units.

Nice❤

@PreMath

4 ай бұрын

Thanks 🔥

Yellow Region Area = △AEF Area + Semicircle Area Draw a segment through O and P that intersects segment AB at point G. This segment, GP, includes radius OP, which is perpendicular to tangent CD by the Tangent to a Circle Theorem. So, segment GP should also be perpendicular to segment AB by the Perpendicular Transversal Theorem. GP = BC. So, GP = 6 units. Because OP = 4, GO = 2 units. Draw radius OF, it should also be 4 units long. Based on these segments, AG = 2√3 and FG = 2√3 by definition of a special 30°-60°-90° right triangle. So, AF = 4√3. Diameter FE = 8 units. So, AE = 4 units by definition of a special 30°-60°-90° right triangle. A = 1/2 * b * h = 1/2 * 4 * 4√3 = 2 * 4√3 = 8√3 So, the area of △AEF is 8√3 square units. A = 1/2 * π * r² = 1/2 * π * 4² = 1/2 * 16 * π = 8π So, the area of the semicircle is 8π square units. Yellow Region Area = 8π + 8√3 So, the area of the yellow region is 8π + 8√3 square units (exact), or about 38.99 square units (approximation).

The rectangle diagonal = 4 + 4 + x Draw a vertical line down from the circle center to the point of intersect 6/(8 + x) = 4/(4 + x) 6(4 + x) = 4(8 + x) 2x = 32 - 24 = 8 x = 4 The diagonal line divide the rectangle corner into 30 and 60 degree Connect the circle center to the two intersecting points on the lower left and upper right, thus split the yellow area into 4 quadrants 1) a 60-60-60 degree equal lateral triangle on the left. Area = 4(2✓3)/2 = 4✓3 2) a 120 degree circle segment on the bottom and a 60 degree one on the left, the combined area = half of the circle area = 4(4)(π)/2 = 8π 3) a 30-120-30 degree triangle on top, the area = 2(2✓3) = 4✓3 The yellow area = 8✓3 + 8π = 8 (π + ✓3)

Sir plz solve it. Q. P, Q, and R are arbitrary points on the sides BC, CA, and AB respectively of tri- angle ABC. Prove that the three circumcentres of triangles AQR, BRP, and CPQ form a triangle similar to triangle ABC.

Let the larger section of the circle outside the rectangle be X and the smaller section be Y. The yellow section is the area of rhe circle minus sections X and Y. By observation, OP is 4. If OP is extended to intersect AB at R, then as ∠OPC and ∠FRO are both 90°, PR = BC and OR = 6-4 = 2. As OR = OA/2, ∆ORA is a 30-60-90 triangle, ∠AOR = 60°, ∠RAO = 30°, and AR = 2√3. The area of Section X is the area of the sector corresponding to the minor arc AP (which covers 60+60 = 120°) minus the area of ∆AOP (which is ∆ORA plus ∆FRO). Let OG be parallel to AF and intersect AE at G. By observation, ∆AGO and ∆OGE are congruent with ∆ORA. Therefore AE is 4, ∠EOA is 60°, and ∆EOA is equilateral. The area of Section Y is the area of the sector corresponding to the minor arc AE minus the area of ∆EOA. Section X: A = (120/360)π4² - 2(2√3)2/2 A = 16π/3 - 4√3 Section Y: A = (60/360)π4² - 4(2√3)/2 A = 16π/6 - 4√3 = 8π/3 - 4√3 Yellow area: A = π4² - (16π/3 - 4√3) - (8π/3 - 4√3) A = 16π - 16π/3 + 4√3 - 8π/3 + 4√3 A = 16π - 24π/3 + 8√3 A = 8π + 8√3 ≈ 38.99

Hers my take... Half circle area(AC) + the other two triangles area (AOE and AOF) AC = r²π/2 AEO = r²Sin(60)/2 AOF = r²Sin(120)/2 this give us Area(A) = r²π/2 + r²Sin(60)/2 + r²Sin(120)/2 -> A = 1/2 (r²π + r²Sin(60) + r²Sin(120) ) -> A = r²/2 (π + Sin(60) + Sin(120) ) -> A = r²/2 (π + √3) -> [ _Sin(60) + Sin(120) = √3_ ] A = (πr²)/2 + (r²√3)/2 -> A = (πr² + r²√3) / 2 -> *A = r²(π + √3) / 2* if we put in the value of r=4 we get (16π + (16)√3) /2 = 8π + 8√3 = *8(π+√3)* Notes: _( Sin(60) + Sin(120) ) = √3_

I started by creating a rectangle AEGF, where point G is the intersection of AC on the circle.

At a quick glance : From symmetry EF= 8 and intersects the center of the circle. The yellow shaded area is the area of semicircle EPFO and area of triangle EFA. AF = 2 * sqrt(4^2-2^2)=6.93. AE= sqrt(8^2- 6.93^2)= 4. The area of EAF = 2* 6.93 =13.86. The area of the semi circle EPFO = 0.5 * PI * 4^2 = 8 * PI. Hence the area of the yellow shaded area is 13.86 + 8 * PI = 39 square units.

Very NICE … and almost exactly how I solved it too. I didn't immediately get that the ½-circle was at the lower right, but in the end, similar. Height of rectangle relative to the circle's center? Had to be [6 - 4] or 2 units above the center. Once you 'got that', it is easy to see that the left intersection of the rectangle and the circle must be 4 units apart. Then, as you say, the area calculation becomes   Area = (area of lower arc) + (area of right arc) + (area of left triangle)   Area = ½(¹²⁰⁄₃₆₀)2π𝒓² ⊕ ½(⁶⁰⁄₃₆₀)2π𝒓² ⊕ ½(4 × 4√3)   Area = ¹⁶⁄₃π ⊕ ⁸⁄₃π + 8√3   Area = 8π + 8√3   Area = 8(π + √3)   Area = 39.989 u² Same result with a little trip down a longer path. Thank you again! ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅

AE=2OQ directly due to line joining midpoint of two sides

Well-designed puzzle to let us to decode the secrets, the yellow region is simply composed by a semicircle of radius 4 and a beautiful right- angled triangle 4×4sqrt(3), so the answer is 8pi + 8sqrt(3)=8(pi+sqrt(3)).😊

I would like to give a solution suitable for exam answer: 1. Use Thales theorem to confirm O is the centre of circle. Hence OA, OE, OF and OP are radii = 4. 2. Get OQ = PQ - OP = BC - OP = 6 - 4 = 2 as PQ & BC equal opposite sides of rectangle BCPQ constructed using tangent theorem for OP // BC and OQ as extension of OP. 3. OP // BC and hence // AD as opposite sides of rectangle to give similar triangles AEF and QOF. 4. Get AE = OQ x 2 = 2 x 2 = 4 by similar triangles AEF and QOF with given side ratio EF:OF = 8:4 = 2:1 5. Get AF of triangle AEF by Pythagoras theorem = SQR(EF^2 - AE^2) = SQR(8^2 - 4^2) = SQR(48) = 4 x SQR(3) 6. Get area of triangle AEF = 1/2 x AE x AF = 1/2 x 4 x 4 x SQR(3) = 8 x SQR(3) 7. Total area = area of semicircle + area of triangle AEF = 1/2 x 4^2 x pi + 8 x SQR(3) = 8 x pi + 8 x SQR(3).

@hongningsuen1348

4 ай бұрын

Correction: Use Thales theorem to confirm centre O lies on mid-point of EF.

It's not clear to me which method(s) to try, so maybe take pot luck and see what happens :) EDIT: I couldn't find a way of being sure that EF formed the diameter at first, so I had to find a way around that. IThis led to me taking a more round-about route. Try intersecting chords. Make a vertical diameter from P that intersects with AF. 6*2 = (4-x)(4-x) (radii - x) 12=16-8x+x^2 x^2-8x+4=0 (8+or-sqrt(64-4*4))/2 (8+or-sqrt(48))/2 (8+or-4*sqrt(3))/2 Discard the positive result (for a change) because x refers to the distance between the midpoint of AE and the left edge of the circle. (4-2*sqrt(3)). That's about 0.536, so looks okay - but stick with the surd for now. Call the midpoint of AE, T. Right triangle OAT has sides of 4 (hypotenuse), and 2. sin(-1)(2/4)=30 degrees, which looks convenient.

Sir I have one idea 💡. we will do a live stream on your birthday in that stream we will solve as many problems as your age number will be. 😊😅❤❤

easy

AE^2=FE^2-FA^2

At 5:40, let's focus on ΔFQO instead of ΔAQO,. We can observe that one side, OQ, is half as long as the hypotenuse, so this is a special 30°-60°-90° right triangle and the long side is √3 times as long as the short side, so length QF = 2√3. Alternatively, we can compute this length using the Pythagorean theorem, as done in the video for ΔAQO. We then observe that ΔFQO and ΔEAF are similar, as at 7:30. Continuing, I got the side lengths AF and AE slightly differently, using ratios of corresponding sides. Side AF of ΔEAF is twice as long as corresponding side QF of ΔFQO. So, side AF has length 4√3. By similarity, side AE of ΔEAFis twice as long as corresponding side OQ of ΔFQO, and has length 4. We now have the base and height of ΔEAF and can compute its area as (1/2)(4)(4√3) = 8√3. In the video, the side lengths of ΔAQO are used later to compute side lengths of ΔFQO, which is found to be similar to ΔEAF at 7:30. By starting with ΔFQO, we get its side lengths sooner.

@PreMath

4 ай бұрын

Thanks ❤️

Nothing really different to find the aunknown area. We can note that in triangle QAO the sinus of angle QAO is QO/AO = 2/4 = 1/2, so angle QAO = 30°, so angle BAC = 30° and AC = 12 It is then easy to see that (QP) and the parallel at (QP) containing F are sharing the big rectangle ABCD in 3 equal rectangles. It is exactly the same with the parallel at (AB) containing O and the parallel at (AB) containing E.

@PreMath

4 ай бұрын

Thanks ❤️

S=38,989

A sector 1= 1/6 π r² A sector 1 = 1/6 π 16 A sector 1 = 8/3 π A sector 2 = 1/3 π r² A sector 2 = 1/3 π 16 A sector 2 = 16/3 π A ∆1 = 1/2.4.4. √3/2 A ∆1 = 4√3 A ∆2 = 1/2.4.4. √3/2 A ∆2 = 4√3 A 1 = A sector 1 - A ∆ A 1 = 8/3 π - 4√3 A 2 = A sector 2 - A ∆ A 2 = 16/3 π - 4√3 The Yelliow Region = 16π - (8/3 π - 4√3) - (16/3 π - 4√3) = 16π - 24/3 π + 8√3 = 8π + 8√3 = 8 (π + √3) Square Units = 38.99 Square Units

Ok, now I'm ready to give my answer. Given that AE = 4 ; OP' = 2 AP' = sqrt(12) = 2*sqrt(3) Area of triangle [AEF] = AE * AF / 2 = 4 * 2sqrt(3) / 2 = 8*sqrt(3) ~ 13,86 su Area of Semicircle [EPF] = Pi*r^2 / 2 = 16*Pi / 2 = 8*Pi su ~ 25,13 su Yellow Area = 8*sqrt(3) + 8*Pi = 8*(sqrt(3) + Pi) su ~ 38,99 su Answer: The Yellow Region is equal to approx. 38,99 su

Si calcola l'angolo ACB=π/3,tramite la similitudine dei triangoli..Ayellow è la somma di due triangoli e di due settori circolari..=4cos30sin30+4cos60sin60+1/6π4^2+1/3π4^2=8√3+32/4π...ah ah spero di non sbagliarmi.. ovviamente i due settori circolari è la metà del cerchio(180gradi)

@PreMath

4 ай бұрын

Excellent! Thanks ❤️

Thank you!