The 2nd. method is a very mess !! I prefer to use the 1st.one. Thanks

If you end up using trig sub, why not just do a polar integral from the start?

THX for the nice problem and solutions. The purpose of a nagging musical background is beyond me though!

Well I get a strange answer of 3π

10:04

Second solution is sick 😂

1) semicircle 2) first circle piece 12pi-9*sq(3) 3) secont circle piece 6pi-9*sq(3) 4) fcp-scp=6pi 5) 6pi/2=3pi From Greece -- no good English

I just set up a triangle and a rectangle of the red area, using standard trig and geometry. Got those areas. Then added the amount from the segment area of sector AOB. ( the space between the chord and the arc AB of sector AOB . As I say, I added up everything and the total came to 9.39. Pretty damn close to suggested solution. Short and sweet.

You lost me @ 9:48. I threw up @ 11:30.

We can use an Integral to solve this. 01) sin(30º) = 1/2 02) sin(60º) = sqrt(3)/2 03) Interval of the Integral [3sqrt(3) ; 3] 04) Equation of the Circle : X^2 + Y^2 = 36 05) Y^2 = 36 - X^2 ; Y = +/- sqrt(36 - X^2) 06) INT sqrt(36 -X^2) dX between 3sqrt(3) and 3. 07) Area = 9,425

A = A₁- A₂ A = ½[½R²(α-sinα) - ½R²(β-sinβ)] A = ¼R²(120°-sin120°)-(60°-sin60°) A = ¼6²(⅔π-√3/2-⅓π+√3/2) A = 3π cm² ( Solved √ )

Position of points A and B: h₁= x₂ = R.cos30° = 3√3 cm h₂= x₁ = R.cos60° = 3 cm x = x₂ - x₁= 3√3-3 = 2,19615cm Area of right trapezoid: A₁ = ½(h₁+h₂).x = ½(3√3+3)*2,19615 A₁ = 9 cm² Area of circular segment: A₂ = ½R²(α-sinα) = ½6²(30°-sin30°) A₂ = 0,4248 cm² = 3π-9 Red shaded area: A= A₁+A₂= 9,4248cm² = 3π (Solved √)

O es el centro del círculo; ABCD es el área sombreada; OB y AD se cortan en E; G es el extremo del radio horizontal; OB y AG se cortan en H → Los triángulos EHA rojo y el EDO blanco son congruentes y el semi segmento circular HAB rojo lo es con el CBG blanco; Ambas superficies rojas adosadas al trapecio rojo DEBC completan el sector circular GOB→ Área sombreada =π*6²*30º/360º =3π. Buen rompecabezas. Gracias y un saludo.

i did it by using trignometry by extending it to semi circle and then dividing the resulted area by 2.

At 1:10, designate the point where the arc intersects the line through CD as point E. Construct OA and OB. We note that sector OAE is 1/6 of the circle's area and OBE 1/12. We can find the red area by computing the area of sector OAE and deducting the areas of ΔOAC and the black area bounded by BD, DE and arc BE. That black area is the area of sector OBE less the area of ΔOBD. So the red area is area (sector OAE - ΔOAC - (sector OBE - ΔOBD)) = sector OAE - sector OBE - ΔOAC + ΔOBD. As found in the video, ΔOAC and ΔOBD are congruent, so they cancel out. Sector OAE - sector OBE is sector OAB, or 1/12 the area of the circle, = (1/12)πr² = (1/12)π(6²) = 3π, as The Phantom of the Math also found.

The area equality or substitution in the first method is very elegant! That's how such a problem is meant to be solved. The second method is awful! So it's a good contrast. The lazy one does well to think creatively before begining work on the first idea that pops up. First chess master Steinitz said that the difficult thing with chess is, when one has come up with a good move, to keep looking for a better one.

I calculated the (area of) circle half-segment beyond AC, then subtracted the circle half-segment beyond BD. Half-segment beyond AC is the circle sector below OA minus right triangle OAC. Same for half-segment BD, sector below OB, triangle OBD. Sector below OA = 2/3 of the quarter circle's area, and sector below OB is 1/3. Relative sides of 90-60-30 triangles known by heart.

Great sir, really great!!!

Great problem 😃

Area of triangle: A = ½b.h + ½b.h = ½(13.r)+½(14.r) A = 13,5 . r Area of triangle, Heron's formula: A²= s(s-a)(s-b)(s-c) A²=21(21-13)(21-14)(21-15) A = 84 cm² = 13,5.r --> r=56/9cm Área of semicircle: A = ½πr² = 60,815 cm² Red shaded area: A = A₁- A₂ = 23,185 cm² (Solved √)

S= √25 = 5 cm ; α = atan(1/2) Intersecting chords theorem : S/cosα .s/cosα = (½S)² s = 0,8 (¼S²)/S = S/5 = 1 cm A = s² = 1 cm² ( Solved √ )

7.556 Answer Let the radius of the quarter circle = R then the radius of the half circle = R/2 The area of the quarter circle = R^2/4 pi The area of the half circle = (R/2)*2 /2 pi or R^2/8 pi The combined area of the two = R^2/4 pi + R^2/8 pi = 3R^2/8 pi The area of the circle = 8 R Hence, the AREA of the shaded region = 8R - 3R^2/8pi Draw a line from the center of the half-circle through where the half-circle and quarter-circle touches to the lower bottom of the rectangle to form a triangle. This is the hypotenuse of length R + R/2 = 3R/2 (since it is the combined length of the two radii) The length of the smallest base is R/2, and the longest is 8 Employ Pythagorean (3R/2)^2 = (R/2)^2 + 8^2 9R^2/4 = R^2/4 + 64 9R^2/4 - R^2/4 = 64 8R^2/4 = 64 2R^2 =64 R^2 = 32 R = sqrt 32 Hence, the radius of the circle is the square root of 32 Since the AREA of the shaded region in terms of R is 8R - 3R^2/8 pi (see above, Hence, 8R - 3 (32/8) pi) 8R - 12 pi 8 (sqrt 32) - 12* 3.1416 45.2548 -37.699 =7.555833

thanks Easy. The sum of areas of white triangles is :24+16+4=44, the area of the general square is 8^2=64. The area of red triangle = 64-44=20

How do you know that the circle is exactly a 1/2 circle???

Yes but there can he another method.find the hypotaneous . Area = 1/2 bh for larger triangle. Draw a tangent that divides the triangle into two halves . This tangent is equal to the diameter of two circles. Find the oenth of tangent . With we fan know the radius. Combined area = πr^r + πr^2 = 2πr^2.

Using Multiple-angle formulae is an another way 😀

2

Subtitles ruining all

All good - but, unlike the opening description, the requested area is NOT the intersection between the triangle and the semicircle, that would be just the semicircle .... but we know what you mean!

i got the solution much faster... by estimating 😉 first i determined the radius of the circle to be (√50/2), then the side of the small square would be slightly less then (√50/2 - 5/2), which if you put in a calculator comes out to be (1.035...). this was the extent of my math ability so i stopped there and said "its probably 1" which in turn would make the area 1^2=1

Just find the equivalent right triangle with the same area. b = 6+2 = 8cm h = 2+2+¼4 = 5cm A = ½b.h = ½*8*5 = 20 cm² ( Solved √ )

Good video sir Thanks a lot.

ABC=(14r/2)+(13r/2)=27r/2 Perímetro de ABC =13+14+15=42→ Semiperímetro =42/2=21 → Según fórmula de Herón: ABC =√[21*(21-13)*(21-14)*(21-15)=84. 27r/2=84→ r=56/9→ Área sombreada =84-(πr²/2)=84-(1568π/81) =23,18497... Gracias y un saludo.

Nice. My first thought was to mirror the triangle along the 15 line, but it does not seem to help much.

There is one Degree of Freedom. It's easy to see that in the two extreme cases, the solution is 2pi. 1) maximizing the left part, yields a quarter of a circle with radius sqrt(2). 2) maximizing the right part yields half a circle with radius 1.

I found the area of ​​the triangle the same way, but I had trouble finding the radius of the circle, but now I know how to find it, math has become fun !

Yes, as others have mentioned: EBQ=atan(9/12) PBQ is half that, and is atan(1/3), so QB =3r =12-3r r=2 12=r(3+1/Tan(atan(3/4)/2)) r=12/(3+1/Tan(atan(3/4)/2))‎ = 2 iOS 18 maths notes help!

The use of limit results very elegant!

Elegant.

❤

Thanks. Connect B to O and have triangle BCO in which it is similar to triangle ABC.( two angled and side BC), and mark CD as X. So : 2R-X/2=R-X/R===> R =1 and area of rectangke 2×1=2

I cheated. I used trig. One only needs to notice that only the right triangle needs to be solved is that which is formed by right triangle PQB. The acute angle of PQB is simply the 1/2 angle of the CAB acute angle. so ... r/(12-3r) = tan(1/2 angle of CAB) In the example, r = 2. The rest is just using the pi-r-squared formula. I wonder if there is a way to do this puzzle from pure geometry without Pythagoras? Maybe, using similar to the trig solution I show above. After all, angles and half-angles and their relationships are well defined with respect to the corresponding ratios of the sides of the right triangle no?

I have the impression that there is something wrong: at 3:57 √(506*a) must be an integer, ok, so you wrote a=506*(x^2). For me it would be a*506=x^2. Since 506 is the product of three prime numbers the smallest square under the root is 506^2. That gives a=506. The same for y, at 5:29, y=506. The others 2 solutions (0,2024) and (2024,0) obviously don't need any calculus.

Как то это сложно все. 12 - 4r = 2r, r = 2. Все решение. Для тех, кто не понял (хотя трудно таких представить). Внутренняя общая касательная (она же радикальная ось) отсечет от треугольника подобный ему треугольник, для которого одна из окружностей, - правая, - вписанная. Так как и тот и другой треугольники подобны треугольнику со сторонами 3,4,5, у которого радиус вписанной окружности 1, то стороны отсеченного радикальной осью треугольника можно записать, как 3r, 4r, 5r, а разница между большими катетами равна 2r. И никаких длинных уравнений. :)

drop a perpendicular to BC from P, PD. PDB and POB triangles are congrunet. If b^ = 2x tan 2x = 3/4. PBD = tan x. OB is 12 - 3r Using tan formula, 3/4 = 2 tan x / 1 - tan^2 x -> 3tan^2 x + 8 tan x - 3 = 0, tan x = 1/3 or -3 since x > 0, tan x = 1/3 tan x = Po / OB = r/(12- 3r) = 1/3, r = 2 Area = 8pi

The long side of the equation has equation y = 9-3x/4. the circumference of the right hand circle has (y-r)^2 + (x-3r)^2 = r^2. It is straightforward to determine r from these equaqtions and the area of the circles follows easily.

Assume triangle vertices are A, B, C, labeled counterclockwise from right angle. Left circle center is O, right circle center is P, all as in video. (Paused video here) Let the point of tangency between circle O and AC be S, and the points of tangency between circles O and P and AB be M and N respectively. Let the point of tangency between circle P and BC be V, and the point of tangency between the two circles will be T. Let the radius of the two congruent circles be r. As AC = 9 = 3(3) and AB = 12 = 4(3), it's clear that ∆ABC is a 3:4:5 Pythagorean triple right triangle and BC = 5(3) = 15. As NB and BV are tangent to circle P and intersect at B, NB = BV = x. As SA and AM are tangent to circle O and intersect at A, SA = AM. As ∠OSA = ∠AMO = 90° as S and M are points of tangency and OS and OM are radii, and as ∠SAM = 90°, ∠MOS = 90° as well and SAMO is a square. As the point of tangency between two circles and their centers is always collinear, OP contains point T, so OP = OT + PT = 2r. As PN and OM are both perpendicular to AB and thus parallel to each other, OMNP is a rectangle with a width of 2r (OP, MN) and height of r (OM, PN). As AB = 12 and AN = r+2r = 3r, then as x = NB, x = 12-3r. VC thus equals 15-(12-3r) = 3+3r. Draw PV, and extend NP to L on BC. As ∠BVP = ∠LNB = 90° (radius at tangent), and ∠VLP = ∠BLN = ∠BCA (corresponding angles, as NL is parallel to AC), then ∆PVL and ∆LNB are similar to ∆ABC. On ∆PVL, as PV = r corresponds to the long leg of the 3:4:5 triangle, then VL = 3r/4 and LP = 5r/4. BL = BV + VL BL = x + 3r/4 = 12 - 3r +3r/4 BL = 12 - 9r/4 LN = 5r/4 + r = 9r/4 Triangle ∆LNB: BL/LN = BC/AC (12-9r/4)/(9r/4) = 15/9 = 5/3 3(12-9r/4) = 5(9r/4) 36 - 27r/4 = 45r/4 72r/4 = 36 18r = 36 r = 36/18 = 2 Combined circle area: A = πr² + πr² = 2πr² A = 2π2² = 8π sq units

Actually, this problem is most easily solved using the formula for the radius of the in-circle of a right triangle: R = (1/2)( a + b - c ) In this case we have b = 12 - 2R, a = (3/4)b , c = (5/4)b, and we get a linear equation for R. No need to use Pythagorean Theorem.

I bisected the angle at B to get the two congruent right triangles BPQ and BPE that have the radius as one side. And from there on. I won't type it out here because I then went about it more roundabout than what is probably necessary. The crux is to find a relationship between the triangle and a circle center.

Really appreciate your reminders of pertinent theorems as you solved the problem. I saw a 3-4-5 triangle and used trig to get r but I really like your purely geometric approach.