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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2625003

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Purple area will be; (⅓) of (1/6) the hexagon area. = (1/18) of 1 square unit. = (1/18) square unit.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2624669

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a² = 3+1-2√(3)cos120 a = 2.394170171 units. a is AC. (2.394170171/sin120) = (1/sinb) b = 21.206023113° sinc = √(2)/2.394170171 c = 36.2060231125° c is angle BAC. Therefore, the required angle, beta is; Let it be d. d = c-b d = 36.2060231125-21.206023113 d = 15°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2623547

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Area Blue will be; Area sector with radius 3 unit and angle 60° - Area equilateral triangle with two lengths 2 unit each and angle 60°, the angle they form - 2(Area sector with radius 1 unit and angle 120°) = (60*9π/360) - ½(4sin60) -2(120π/360) = ½(3π) - √(3) - ⅓(2π) = (9π - 4π - 6√(3))/6 = (5π-6√(3))/6 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2623536

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AD is; √(8²+3²-2*8*3cos60) = √(73-24) = √(49) = 7 unit. Therefore radius of area S1 is; 3r+7r+8r=12√(3) r = (12√(3))/18 r = ⅔(√(3)) unit. Area S2 = π(⅔(√(3)))² = ⅓(4π) square unit. Therefore radius of area S2 is; 5r+7r+8r=20√(3) r = (20√(3))/20 r = √(3) unit. Area S1 = π(√(3))² = 3π square unit. It implies; S1 ÷ S2 is; ⅓(4π) ÷ 3π = 4π ÷ 9π = 4/9

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2622247

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Let the side of the ascribed regular hexagon be 2 units. Therefore; Area Blue is; Area triangle with height 1.5 units and base ½(√(3)) units. = ½*1.5*½(√(3)) = ⅛(3√(3)) square units. Area Red is; Area triangle with height (5√(3)/11)sin30 units and base ⅛(5) unit. = ½*(5√(3)/11)sin30*⅛(5) = (25√(3)/352) square units. Area Green is; Area triangle with height 1.5 units and base ½(3√(3)) units - Area Red. = ½*1.5*½(3√(3)) - (25√(3)/352) = ⅛(9√(3)) - (25√(3)/352) = (371√(3)/352) square units...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2621800

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Calculating the area of the inscribed circle. a = (9+r) cm. b = (9-r) cm. It implies; (9+r)² = c²+(9-r)² 81+18r+r²-(81-18r+r²) = c² c² = 36r c = 6√(r) cm. d = (6√(r)-9) cm. e = (9-r) cm. Therefore, Calculating r. (9-r)² = r²+(6√(r)-9)² 81-18r+r² = r²+36r-108√(r)+81 108√(r) = 54r 108²r = (54r)² 11664r = 2916r² r = 11664/2916 r = 4 cm. Therefore, area inscribed green circle is; πr² = π(4²) = 16π cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2621785

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Let Phi be a. let Beta be b. Notice. b = 2a And b = 45° It implies; 2a = 45 a = 22.5° Let CF = c. tan22.5 = c/d d = c/tan22.5 units. d is CE. Therefore; c+d = 12 c+(c/tan22.5) = 12 3.4142135624c = 12 c = 3.5147186258 units. d = c/tan22.5 units. And c = 3.5147186258 units. d = 3.5147186258/tan22.5 d = 8.4852813743 units. Calculating BC = CD = DE, the side length of the square. Let it be e. 2e² = 8.4852813743² e² = 36 square units. Where e² is...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2621767

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Calculating yellow inscribed triangle area. It implies; Calculating a, the inscribed triangle opposite the short length of the ascribed triangle. 36+0.5*8*3sina = 0.5*(8+3)*(4+8)Sina a is thr 36+12sina = 66sina 36 = 66sina-12sina 54sina = 36 sina = 36/54 sina = 2/3 a = asin(2/3) a = 41.8103148958° Therefore, the required area, yellow inscribed triangle area is; 0.5*3*8sin41.8103148958 = 8 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2620260

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tana = 1/3 a = atan(⅓)° b = 90-a b = atan(3)° cos(atan(3)) = c/6 c = 1.8973665961 units. d² = 1.8973665961²+3²-2*3*1.8973665961*(1/3) d = 1.3416407865 units. (1.3416407865/sin(atan(⅓))) = (3/sine) e = 45° It implies, the required angle, alpha is; Alpha = e = 45°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2619989

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Let the single side length of the regular hexagon be 2 unit. Therefore; Area Red is; 4 - ½*½(2/(√(3)+1)) = 4 - ((√(3)-1)/2) = ½(9-√(3)) square unit. Area Yellow is; ½(√(3)+2)(√(3)+1) - ½(√(3)) = ½(5+2√(3)) square unit. It implies; Area Red ÷ Area Yellow exactly as a single fraction is; ½(9-√(3)) ÷ ½(5+2√(3)) = (9-√(3))/(5+2√(3))
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