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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2289190

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Let the side length of the regular Pentagon be 1 unit. Therefore Area Pentagon is; (5*1)÷(4tan(180/5)) = 5/(4tan36) space unit. = 1.72047740059 space unit. Area star is; 5(area triangle with height 0.61803398875 unit and base 0.61803398875sin36 unit) + Area regular Pentagon with side 0.38196601125 unit. = 5(0.5*0.61803398875*0.61803398875sin36) + ((5*0.38196601125²)/(4tan(180/5))) = 0.56128497072 + 0.25101426986 = 0.81229924058 space unit. It implies; Area Star ÷ Area Pentagon to 3 d. p....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2289168

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At x = 0, y = 2. At y = 16, x = 1. Gradient of the curve at x = 1 or at the tangent is; dy/dx = 3In2*2^(3x+1) For x = 1 dy/dx = 48In2 Therefore coordinate Q is (x¹, 0). Calculating x¹. (0-16)/(x¹-1) = 48In2 x¹= (3In2-1)/3In2 It implies; Area Red exactly as a single fraction in its simplest form is; Area trapezoid with parallel sides 2 unit and 16 unit, and height 1 unit - Area triangle with height 16 unit and base (1-((3In2-1)/3In2)) = (½*1(2+16)) - (½(16)(1-((3In2-1)/3In2)) = 9 - (8(1/3...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2288018

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Let the side length of the be 2 unit. Therefore; BC = 2 unit. It implies; Area orange is; ½(1*2) = 1 square unit. Area green is; ½(1*2) + ½(2/√(5))(4/√(5)) = 1 + (4/5) = (9/5) Therefore; Area orange : Area green is; 1 : (9/5) =(5*1) : (5*(9/5)) = 5 : 9

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2288002

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Let the side length of the square be 1 unit. Therefore; Area square is; 1² = 1 square unit. It implies; Area R is; Area square - Area equilateral triangle with side 1 unit - 2(area sector with radius 1 unit and angle 30°) = 1 - ½(sin60) - 2(30π÷360) = 1 - ¼(√(3)) - (π/6) = (12-3√(3)-2π)/12 square unit. Therefore; Area R ÷ Area Square as a single fraction in terms of π is;  (12-3√(3)-2π)/12 ÷ 1 = (12-3√(3)-2π)/12

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2286772

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Notice. The ascribed triangle is isosceles with angles 70°, 70° and 40°. Let the two equal lengths be 1 unit. (1/sin70) = (a/sin40) a = (sin40/sin70) a = 0.6840402867 units. sin40 = b/a sin40 = b/0.6840402867 b = 0.4396926208 units. c² = 0.5²+0.4396926208² -2*0.5*0.4396926208cos20 c = 0.1736481777 units. It implies, the required angle x is; (0.1736481777/sin20) = (0.4396926208/sinx) x (required angle) = 60°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2285189

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a² = 11*11+10*10-2*11*10cos50 a = 8.92113926968 cm. (8.92113926968/sin50) = (10/sinb) b = 59.16920318624° c = 180-50-59.16920318624 c = 70.83079681376°  d = 0.5a d = 4.46056963484 cm. e² = 4.46056963484²+5.5²-2*4.46056963484*5.5cos59.16920318624 e = 5 cm. (5/sin59.16920318624) = (5.5/sinf) f = 70.83079681376° g² = 4.46056963484²+5²-2*4.46056963484*5cos70.83079681376 g = 5.5 cm. (5.5/sin70.83079681376) = (5/sinh) h = 59.16920318624° i = 180-59.16920318624-70.83079681376 i = 50° (j/tan...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2285187

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Let AB = 1 unit. Notice; AB = BE Therefore; Length AE is; √(2-2cos135) = √(2+√(2)) unit Length CG is; √(2-2cos45) = √(2-√(2)) unit. Notice; BD = BF = √(2) Therefore; Length DF is; √(4-4cos45) = √(4-2√(2)) unit. It implies; CG + DF = √(2-√(2))+√(4-2√(2)) = 1.84775906502 unit. And; AE = √(2+√(2)) = 1.84775906502 unit. Therefore; AE = CG + DF Which is; √(2+√(2)) = √(2-√(2))+√(4-2√(2)) Therefore; √(2+√(2)) = √(2+√(2)) Proved.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2285175

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Let BC be 1 unit. Therefore; AB = 2 unit. Radius of the circle, r is; r = √(5) unit. Therefore; Area R is; Area sector with radius √(5) unit and angle (2asin(√(2)/(2√(5))))° - 2(area triangle with height 1 unit and base (√(2)sin135) unit. (2asin(√(2)/(2√(5))))*(√(5))²*π÷360) - 2(½*1*√(2)*sin135) = (πasin(√(2)/(2√(5)))/36 - 1 = ((πasin(√(2)/(2√(5))) - 36)/36 = 0.60875277198 square unit. Area square is; 2² = 4 square unit. It implies; Area R ÷ Area Square as a single fraction is; ((...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2283798

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Let the side length of the regular pentagon be 1 unit. tan36 = a/0.5 a = 0.363271264 unit. b² = 0.363271264²+0.5² b = 0.61803398875 unit. tan72 = c/0.5 c = 1.53884176859 units. Area Green is; 0.5sin108+0.5*1.53884176859*0.5-0.5*0.5*0.363271264-0.5*0.61803398875*0.61803398875sin36 = 0.65716389015 square units. Calculating Area Blue. Let d be the radius of the inscribed blue circle. Calculating d. e = 0.5(90+36) e =63° (d/tan63) + (d/tan45) = a d = 0.363271264/((1/tan63)+(1/tan45)) d =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2283792

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At point P; x = (π/6)~30° y = √(3)-2 Gradient at point P is; dy/dx = 2/(1+2sinxcosx) At x = (π/6)~30° dy/dx = 8-4√(3) It implies; yQ exactly as a single fraction in terms of π is;  The coordinates are; ((π/6), (√(3)-2)) and (0, yQ) Gradient = 8-4√(3) Therefore; yQ = ⅓((2√(3)-4)π + 3(√(3)-2))
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