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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2755494

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Let the side length of the square be a. b²+(0.5a)² = 5² b = √(25-¼(a²)) units. c = a-b c = (a-√(25-¼(a²))) units. It implies, calculate a. 8² = (0.5a)²+(a-√(25-¼(a²)))² 64 = ¼(a²)+a²-2a√(25-¼(a²))+25-¼(a²) 39 = a²-2a√(25-¼(a²)) 2a√(25-¼(a²)) = a²-39 4a²(25-¼(a²)) = a⁴-78a²+1521 100a²-a⁴ = a⁴-78a²+1521 2a⁴-178a²+1521 = 0 Let p = a². 2p²-178p+1521 = 0 p²-89p = -(1521/2) (p-½(89))² = -(1521/2)+(-0.5*89)² (p-44.5)² = 1219.75 p = 44.5±√(1219.75) p ≠ 44.5-√(1219.75) p =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2751959

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a = 180-100-50 a = 30° (20/sin100) = (b/sin30) b = 10.1542661189 cm  c = ½(180-100) c = 40° Let d be the radius of the circle. tan40 = d/e e = 1.1917535926d . f = (10.1542661189-1.1917535926d) units. g = 180-50 g = 130° sin(0.5*130) = d/h h = 1.103377919d cm. Calculating d. cos(0.5*130) = (10.1542661189-1.1917535926d)/1.103377919d 10.1542661189-1.1917535926d = 0.4663076582d 1.6580612508d = 10.1542661189 d = 10.1542661189/1.6580612508 d = 6.1241803427 cm....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2750843

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Sir Mike Ambrose is the author of the question. Let the single side length of the inscribed regular pentagon be 2 units. Therefore; Area green is; Area rectangle with length 2 units and width 1.17557050458 units. = 2 * 1.17557050458 = 2.35114100916 square units. Area red is; Area regular pentagon with side length 2 units - Area triangle with two side 2 units each, and angle 108° - Area triangle with side 2 units and 0.7639320225 units, and angle 72° = ((10*2)/(4tan(180/...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2750835

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Shaded area is; Area sector with radius 10 units and angle acos(0.7) - Area triangle with height √(51) units and base 7 units - area sector with radius 3 units and angle 81.37307344° - area sector with radius 10 units and angle 17.2538531174° + area triangle with two side length 10 units, and angle 17.2538531174° = (acos(0.7)*1))π÷360) - ½(√(51)*7) - (9π*81.37307344)/360 -(100π*17.2538531174)/360 + ½(100)sin17.2538531174 = 39.7699415092 - 24.9949994999 - 6.39102624298 - 15.0568272777 +...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2750828

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Sir Mike Ambrose is the author of the question. Area Purple exactly in decimal square cm is; Area triangle with two side length 7.04301437734 cm and 7.01866670644 cm, and angle 100° = ½(7.04301437734*7.01866670644)sin100 = 24.3407893513 cm² ≈ 24.34 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2748466

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Let the side length of the inscribed square be a. b² = 2a² b = √(2)a cm. b is the diagonal of the inscribed square. c²+a² = (4+6)² c = √(100-a²) cm. d = c-a d = (√(100-a²)-a) cm. 14² = 6²+(4+6)²-2*6*10cose 196 = 136-120cose 120cose = -60 e = acos(-½) e = 120° f = 180-e f = 60° Calculating a. tan60 = a/(√(100-a²)-a) √(3)(√(100-a²)-a) = a √(3(100-a²))-√(3)a = a √(300-3a²) = a+√(3)a 300-3a² = a²+2√(3)a²+3a² 300 = 7a²+2√(3)a² (7+2√(3))a² = 300 a² = 28.66944636...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2747682

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Area green is; Area rectangle with length 18 units and width 16 units - Area triangle with height 12 units and base 16 units - Area semi circle with radius 10 units. = (18*16) - ½(16*12) - ½(100π) = 288 - 96 - 50π = (192-50π) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2745908

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Green area is; Area circle with radius 5 units + 2(area sector with radius 20 units and angle 60°) - Area equilateral triangle with side 20 units. = π(5²) + 2(60÷360*20²π) - ½(20²)sin60 = 25π + ⅓(400π) - 100√(3) = ⅓(475π-300√(3)) square units. = 324.213756061 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2743954

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Let a be the side length of the square. b² = 13²+a² b = √(169+a²) units. c² = 2a² c = √(2)a units. d²+5² = a² d = √(a²-25) units. e = b-d e = (√(169+a²)-√(a²-25)) units. Calculating a. 13² = (√(169+a²)-√(a²-25))²+5² 169-25 = (√(169+a²)-√(a²-25))² 144 = (√(169+a²)-√(a²-25))² 12 = (√(169+a²)-√(a²-25)) (12+√(a²-25))² = 169+a² 144+24√(a²-25)+a²-25 = 169+a² 144+24√(a²-25) = 194 24√(a²-25) = 50 12√(a²-25) = 25 144(a²-25) = 625 144a² = 625+(144*25) 144a² = 4225 a =...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2743132

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Calculating angle x. x = 180-½(360-100-160) x = 180-½(360-260) x = 180-½(100) x = 180-50 x = 130°
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