Can You Solve This Tricky Section of a Quarter Circle? | 2 Awesome Methods

Think you can solve this tricky section of a quarter circle? In this video, we explore two awesome methods to find the area of this fascinating shape.

We’ll start with a geometric approach, using visual reasoning to break down the problem step-by-step. Then, we’ll dive into an integral-based method, applying trigonometry and analytical geometry for a deeper mathematical perspective.

Which method will you prefer? Watch the video, try it yourself, and let us know in the comments!

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#MathChallenge #GeometryPuzzle #QuarterCircle #AnalyticalGeometry #Trigonometry #mathpuzzles

[FadkinsDiet]

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

[jimlocke]

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.

[bjorntorlarsson]

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.

[marioalb]

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 √)

[lasalleman]

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.

[bjorntorlarsson]

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.

[haroldlake]

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

[marioalb]

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

[LuisdeBritoCamacho]

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

[ManjeetRani-vn]

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

[babisstafulas]

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

[nandisaand]

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