Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2975982

Sir Mike Ambrose is the author of the question. Area Blue Exactly in decimal cm² is; Area triangle with height 3.84187454245 cm and base 5sin(180-atan(12/5)-atan(4/5)) = ½*3.84187454245*5sin(180-atan(12/5)-atan(4/5)) = 0.5*3.84187454245*5sin(73.960056694) = 9.23076923075 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2974204

Let the inscribed square side be a. tan60 = a/b a = √(3)b cm. Where b is BS = CR It implies; b+√(3)b+b = 10 (2+√(3))b = 10 b = 10(2-√(3)) cm. Therefore; a = √(3)*10(2-√(3)) a = 10(2√(3)-3) cm. a = 4.64101615138 cm. sin60 = 10(2√(3)-3)/c c = 20(2-√(3)) cm. Where c is CQ. d² = 10²+(10(2-√(3)))²-20*10(2-√(3))cos60 d = 8.96575472168 cm. Where d is AS. (8.96575472168/sin60) = (10(2-√(3))/sine) e = 15° Where e is angle PAS. f = 180-60-15 f = 105° Where f is angle BSA. g = 105-90 g = 15° Whe...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2972559

Sir Mike Ambrose is the author of the question. (a) Area yellow exactly in decimal cm² is; Area triangle with height 25.2274259042 cm and base 14.5sin(½atan(20/21)) cm = ½*25.2274259042*14.5sin(½atan(20/21)) = 67.9269230771 cm² (b) Let beta be y. Therefore; y = 90 - (atan(20/21) + ½(atan(20/21))) y = ½(180 - (3atan(20/21)))° y = 24.595771541° It implies; Tan(y) exactly in decimal is; Tan(½(180 - (3atan(20/21)))) = Tan(24.595771541) = 0.45774647887

Mathematics teaching and learning

https://onlineedumath.com/update/mathematics-teaching-and-learning/2971120

Area ABED is; Area trapezium with parallel sides (4√(2)sin15) cm and (4+4√(2)sin15) cm, and height 4√(3) cm. = ½*4√(3)((4√(2)sin15)+(4+4√(2)sin15)) = 2√(3)((4√(2)sin15)+(4+4√(2)sin15)) = 24 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2969557

Blue area is; Area sector with radius 12 units and angle 30° - Area sector with radius 6 units and angle 60° - Area triangle with height 6 units and base 6sin120 units. = (30π*12²/360) - (60π*6²/360) - (½*6²*sin120) = 12π - 6π - 9√(3) = 6π-9√(3) = 3(2π-3√(3)) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2967876

Let the length of the ascribed regular hexagon be 2 units. Therefore it area will be; (2*2*6)/(4tan(180/6)) = 6/tan30 = 6√(3) square units. The green areas is; Area rectangle with length 2√(3) units and width 1 unit + Area regular hexagon with side ⅔√(3) units + Area regular hexagon with side ⅓√(3) unit. = (2√(3)*1) + (6*(⅔√(3))²/(4tan(180/6)) + (6*(⅓√(3))²/(4tan(180/6)) = 2√(3) + 2√(3) + ½(√(3)) = 4√(3) + ½(√(3)) = ½(9√(3)) square units. It implies; Shaded...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2967867

Let the side of the ascribed regular hexagon be 2 units. Area regular hexagon is; (2*2*6)/(4tan(180/6)) = 6/tan30 = 6√(3) square units. Area green is; Area kite with perpendicular lengths 4 units and ⅓(4√(3)) units respectively + Area kite with perpendicular lengths 2 units and ⅓(2√(3)) units respectively. = ½(4*⅓(4√(3))) + ½(2*⅓(2√(3))) = ⅓(8√(3)) + ⅓(2√(3)) = ⅓(10√(3)) square units. It implies; Shaded fraction is; Area Green ÷ Area Regular Hexagon. = ⅓(10√(3)) ÷ 6√(3) = (10√(3)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2966130

a² = 4²+4²-2*4*4cos150 a = 7.72740661031 cm. Where a is BE. b = angle ABE = ½(180-150) b = 15° c = angle AFE = 180-45-15 c = 120° (4/sin120) = (d/sin45) d = 3.26598632371 cm. Where d is EF. Notice; EF = CF Therefore; Length CF = 3.26598632371 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2966102

tan15 = a/4 a = 1.07179676972 cm. Where a is AD. b² = 1.07179676972²+4² b = 4.14110472164 cm. Where b is DF. c = 4-1.07179676972 c = 2.92820323028 cm. Where c is CD. sin60 = d/2.92820323028 d = 2.53589838487 cm. Where d is DE = BE. e² = 16+16-2*16cos150 e = 7.72740661031 cm. Where e is BF. f² = 7.72740661031²+2.53589838487²-2*2.53589838487*7.72740661031cos45 f = 6.19925645689 cm. Where f is EF. (6.19925645689/sin45) = (2.53589838487/sing) g = 16.81321456806° Where g is angle BFE. h = 1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2964663

a = acos(3/5) Where a is angle GCA. tan(acos(3/5)) = b/3 b = 4 units. Where b is AG. c = atan(⅔) Where c is angle GBE. d = 180-acos(3/5)-atan(⅔) d = 93.17983011986° Where d is angle BEC. (6/sin93.17983011986 ) = (e/sin(acos(⅗))) e = 4.80740170062 units. Where e is BE. Area Parallelogram BCDE is; 2(area triangle with height 4.80740170062 units and base 6sin(atan(⅔)) units. = 2(0.5*4.80740170062*6sin(atan(⅔))) = 16 square units.

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Mathematics Question and Solution

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