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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2084619

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Sir Mike Ambrose is the author of the question. Let the side length of the five inscribed congruent regular hexagon be 1 unit. It implies; The ascribed regular hexagon is; 2+(⅔) = ⅓(8) units. Area Green is; 2(area triangle with height ⅓ unit and base ⅓(sin60) units) + 2(area parallelogram with height ⅓√(3) unit and base ⅓(4) units) = 2(½*⅓*⅓*sin60) + 2(⅓*√(3)*⅓*4) = (√(3)/18) + (8√(3)/9) = (17√(3)/18) square units. Area Grey is; Area regular hexagon with side (8/3) unit...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2083545

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a² = 6²+2² a =√(40) a = 2√(10) units. b = ½(a) b = √(10) units. c² = 6²+5² c = √(61) units. d = (atan(5/6)-atan(1/3))° tan(atan(5/6)-atan(1/3)) = e/√(10) e = 1.23741299746 units. cos(atan(5/6)-atan(1/3))= √(10)/f f = 3.39576072866 units. Area Orange is; ½(6*5)-½(6*2)*½(√(10)*1.23741299746 = 15-6-1.95652173913  = 7.04347826087 square units. Calculating Purple Area. g = 90-atan(6/5) g = 39.80557109227° h = atan(√(10)/1.23741299746) h = 68.62937773061°  j = 180-39.80557109227-68.629377...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2083006

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Let the green square side be 1 unit. Area Green Square is; 1² = 1 square units. Calculating the total shaded area. a² = 1²*2 a = √(2) units. a is the radius of the quarter circle. tan15 = b/1 b = (2-√(3)) units. c = 1-b c = (√(3)-1) units. cos15 = 1/d d = (√(6)-√(2)) units. tan30 = e/√(2) e = ⅓(√(6)) units. Shaded Area is; Area sector with radius √(2) units and angle 30° - Area with height (√(6)-√(2)) units and base (√(3)-1)sin105 units + Area triangle with height ⅓(√(6)) units and ba...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2082617

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Sir Mike Ambrose is the author of the question. Let the side of the regular pentagon be 1 unit. It implies; Radius large circle, R is; = 0.580673 unit. Radius small circle, r is; = 0.168821 unit. (a) Radius Small Circle ÷ Radius Large Circle to 3 decimal places is; 0.168821 ÷ 0.580673 = 0.29073333873  ≈ 0.291 Calculating pentagon area. ½(5)*(1/(2tan(180/5))) = 1.72047740059 square units. Shaded Area is; Area triangle with height 1 unit and sin108 units - Area circle with radius 0.16882...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2081502

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tan30 = 6/a a = 6√(3) units. b = (4+6√(3)) units. b is the square side. Let the side length of the inscribed regular hexagon be x (Red Length). Calculating x. c = (120-atan(⅔))° d = 120-c d = atan(⅔)° It implies; (x/sin(120-atan(⅔))) = (e/sin(atan(⅔))) e = (4x/(2+3√(3))) units. Therefore; e+2x = 12 (4x/(2+3√(3)))+2x = 12 (4x+4x+6√(3)x)/(2+3√(3)) = 12 8x+6√(3)x = (24+36√(3)) x(4+3√(3)) = (12+18√(3)) x = (12+18√(3))/(4+3√(3)) x = (6(19-6√(3))/11) units. x = (114-36√(3))/11 units. x = 4...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2081237

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Let AB = 1 unit. Therefore; BC = 2AB = 2 units. It implies; AD = 4 units. Calculating Area Red. a² = 2²+1-2*2cos(180-(900/7)) a = 1.58304794386 units. (1.58304794386/sin(180-(900/7))) = (2/sinb) b = 180-81.02431846203 b = 98.97568153797°   c = 108-98.97568153797  c = 9.02431846203° d = 180-(180-(900/7))-98.97568153797 d = 29.59574703346° e = 180-51.42857142857-29.59574703346  e = 98.97568153797° f = 180-98.97568153797-9.02431846203 f = 72° (1.58304794386/sin72) = (g/sin98.97568153797)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2080370

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Sir Mike Ambrose is the author of the question. Let the side length of the large square be 1 unit. 2a² = 1 a = ½√(2) unit. b² = ½+1-√(2)cos(360-195) b = 1.69293396321 units. (1.69293396321/sin(360-195)) = (1/sinc) c = 8.79397688726° d = 180-165-8.79397688726 d = 6.20602311274° e² = 1.69293396321²+1-2*1.69293396321cos(45+8.79397688726) e = 1.36602540379 units. (1.36602540379/sin(45+8.79397688726)) = (1/sinf) f = 36.20602311308° g = f-d g = 30° It implies; Area Gold is; Area triangle w...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2078804

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Let the side length of the congruent yellow regular hexagon be 1 unit. a = 3sin30+1 a = 2.5 units. sin60 = b/1 b = 0.5√(3) units. c² = a²+b² c² = 2.5²+(0.5√(3))² c = √(7) units. c = 2.64575131106 units. c is the radius of the circle. d² = √(7)²-1 d = √(6) units. e = d-b e = √(6)-0.5√(3) e = 1.583464339 units. e is the height of the green inscribed triangle. Area Green Triangle is; Half the height, 1.583464339 units multiply the base, 1 unit of the triangle = 0.5*1.583464339*1 = 0.791732...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2078183

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Sir Mike Ambrose is the author of the question. The vertices coordinates of area purple are; (6, 12) ((24/5), (36/5)) ((180/29), (168/29)) (12, (21/2)) Area purple is; ½((6, 12) ((24/5), (36/5)) ((180/29), (168/29)) (12, (21/2)) (6, 12)) = ½((-72/5)+(-2448/145)+(-126/29)+81) = ½((-2088-2448-630+11745)/145) = ½(6579/145) = 6579/290 square units. = 22.68620689655 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2076443

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Calculating equation of the curve. y-6√(2) = a(x-6)² --- (1). Substituting x = 0, y = 0 to get a. It implies; -6√(2) = 36a a = -(1/(3√(2)) --- (2). Substituting (2) in (1). y-6√(2) = -(1/(3√(2))(x-6)² y = -(1/(3√(2))(x²-12x+36)+6√(2) y = -(x²/(3√(2))+2√(2)x-6√(2)+6√(2) y = -(x²/(3√(2))+2√(2)x --- (3). Where (3) is the equation of the curve. Calculating x, when y is ⅓(10√(2)). ⅓(10√(2)) = -(x²/(3√(2))+2√(2)x  (x²/(3√(2))-2√(2)x+⅓(10√(2)) = 0 x²-12x+20 = 0 Resolving the quadratic equation. (...
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