A Very Nice Geometry Problem | You should be able to solve this!

A Very Nice Geometry Problem | You should be able to solve this!
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[@danilopapa3853]

No need for complicated trigonometry to (trivially) calculate R: ∆ ACB is a triangle inscribed in a semicircle. This means that ∆ ACB is a right triangle in C; therefore, its angle at C is 90°. Since we know its angle at A (30°), then we deduce its angle at B, equal to 60°. So ∆ ACB is half of an equilateral triangle of side 2*R Hence, CB = R Applying the Pythagorean theorem: 5^2 = (2R)^2 - R^2 = 3R^2 So, we have R = 5/sqrt(3)

[@jimlocke9320]

As in the video, let O be the center of the circle. O must be the midpoint of diameter AB. Construct OC and drop a perpendicular from O to AC, labelling the intersection as point D. Note that D is the midpoint and that ΔADO and ΔCDO are congruent 30°-60°-90° right triangles, with long sides of length 2.5. Therefore, OD has length (2.5)/(√3) and OA, which is also the radius, has length (5)/(√3).

[@santiagoarosam430]

Si "O" es el centro de la circunferencia→ Ángulos en 0: 60º+60º+60º=180º ; ángulo ABC=60º ; ángulo ACB=30º+60º → Distancia vertical de C a AB =h=5/2 → r√3/2=5/2→ r=5√3/3 → Área rosa =(Triángulo AOC)+(Sector circular radio=r y ángulo central 60º) =(rh/2)+(πr²/6) =(25√3/12)+(25π/18) =7,9717..... Gracias y saludos.

[@PrithwirajSen-nj6qq]

Join BC ABC will be a 30-60-90 triangle AB=2R BC =R Then 4R^2=R^2+25 R=5/√3 Area of ABC=1/2*R*5 =25/2√3 AOC =1/2ABC=25/4√3--(1) Sector BOC=22/7*25/3*1/6--(2) Adding 1 and 2 we may get the required area

[@murdock5537]

φ = 30°; ∆ ABC → CAB = φ; ABC = 2φ; BCA = 3φ; AC = 5; BC = r; AB = A0 + B = r + r → r = 5√3/3 = AO = BO = CO; area ∆ AOC = (1/2)sin⁡(φ)5r = 25√3/12 area (1/6) circle = 25π/18 → shaded area = (25/36)(2π + 3√3)

[@prime423]

Draw CB.We have a 30-60-90 right triangle. The problem is then trivial.,

[@skwest]

1. Diameter = 10√3/3 (30-60-90 right triangle) 2. BC = radius = 5√3/3 (same) 3. [△ACB] = (5•5√3/3)/2 = 25√3/6 4. Altitude of △ACB, segment CD (perp. to diameter, w/D lying on diameter) = h = 5/2 (another 30-60-90 right triangle). 5. Purple area = area of sector OBC + area of △AOC (where O is the center of the circle). 6. Area of sector OBC = (1/6) area of circle = (1/6)π(5√3/3)² = 25π/18 7. Area of △AOC = ½bh = ½(5√3/3)(5/2) = 25√3/12. 8. Sum = 25π/18 + 25√3/12 = (50π + 75√3)/18 Seems too complicated.

[@AmirgabYT2185]

S=25(3√3+2π)/36≈7,97

[@user-sw9lb2zs6e]

Thaleskreis und Kreisbogen addieren geht auch!

[@quigonkenny]

Let O be the center of the circle, at the midpoint of AB. Draw CB. As A and B are ends of the diameter and C is a point on the circumference, ∠BCA = 90°. The purple area will be equal to the sum of the areas of triangle ∆BCA and the circular segment subtended by minor arc CB. A xircular segment is the area enclosed by a chord and the arc subtending that chord, and is the difference between the area of the triangle formed by the chord and the center of the circle, and the area of the circular sector corresponding to the chord. As ∆BCA is a right triangle and ∠CAB = 30°, ∆BCA is a 30-60-90 special right triangle, AB = 2BC, and CA = √3BC. Triangle ∆BCA: CA = √3BC 5 = √3BC BC = 5/√3 AB = 2BC = 2(5/√3) = 10/√3 = 2r A = bh/2 = 5(5/√3)/2 = 25/2√3 As ∠A is 30° and A is on the circumference, the angle of the circle subtended by arc CB from the center O is (2)30° = 60°. Draw radius OC. As ∠OBC = ∠COB = 60° and CB = OB = OC = 5/√3, then ∠BCO = 60° as well and ∆BCO is an equilateral triangle. As noted previously, the area of the circular segment CB is equal to the area of the triangle formed by chord BC and center O (the equilateral triangle ∆BCO) subtracted from the area of the sector CB. Circular segment CB: A = (60°/360°)πr² - √3s²/4 A = π(5/√3)²/6 - √3(5/√3)²/4 A = (25π/3)/6 - √3(25/3)/4 A = 25π/18 - 25√3/12 Purple area: A = 25/2√3 + 25π/18 - 25√3/12 A = 50√3/12 + 25π/18 - 25√3/12 A = 75√3/36 + 50π/36 A = (25/36)(3√3+2π) ≈ 7.972

[@prossvay8744]

Area=(75√3+50π)/36

[@michaeldoerr5810]

I found this really easy to comprehend. Does that mean that I should able to do this?

[@user-pe4bu1qp9k]

Bro it depends on your age among people who have the same age

[@klatis84]

too easy