Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3253069

Sir Mike Ambrose is the author of the question. Calculating Red Area ÷ Area ABCD Exactly. Let the side length of the regular dodecagon be 1 unit. Therefore; The side length square ABCD is; 1 + 2cos30 = (1+√(3)) units. Area square ABCD is; (1+√(3))² = 2(2+√(3)) square units. Area red is; Area triangle with two side 1 unit and ½(1+√(3)), and angle 60° = ½(1*½(1+√(3)))sin60° = (3+√(3))/8 square units. Therefore; Area red ÷ Area square ABCD is; (3+...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3250380

Calculating red area, area regular hexagon. Let x be the side length of the red regular hexagon. a² = 2x²-2x²cos120 a² = 3x² a = √(3)x units. b²+3² = x² b = √(x²-9) units. sinc = √(x²-9)/x --- (1). sinc = 5/√(3)x --- (2). Calculating x. Equating (1) and (2). √(x²-9)/x = 5/√(3)x Cross Multiply. 5x = x√(3x²-27x) 5 = √(3x²-27) 25+27 = 3x² 52 = 3x² x² = 52/3 x = √(52/3) x = 2√(13)/√(3) x = ⅓(2√(39)) units. x = 4.16333199893 units. Again, x i...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3250371

Calculating the radius of the ascribed half circle. Let r be the radius of the semi circle. Therefore; A right-angled triangle with hypotenuse 6 units, and adjacent height and base, (r-7) units and √(r²-49) units respectively is derived. It implies; 6²=(r-7)²+√(r²-49)² 36=r²-14r+49+r²-49 2r²-14r-36=0 r²-7r-18=0 (r-(7/2))²=18+(7/2)² (r-(7/2))²=(121/4) r = (7/2)±(11/2) r = ½(7±11) Therefore; r ≠ ½(-4) ≠ -2 units. r = ½(18) = 9 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3249646

Calculating Area Blue. It implies; A1 = 64π cm², therefore radius A1, x = 8 cm. A2 = 32π cm², therefore radius A1, y = 4√(2) cm. A3 = (16/3)π cm², therefore radius A1, z = 4/√(3) cm. Therefore; Area blue is; Area triangle with two length 10.3094010768 cm and 7.96625532625 cm, and angle 95.85811003° - Area sector with radius (4/√(3)) cm and angle 95.85811003° - Area sector with radius 4√(2) cm and angle 48.6724043189° -Area sector with radius 8 cm and angle 35.4694856511°....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3249575

Sir Mike Ambrose is the author of the question. Calculating Area yellow ÷ Area green. Let the side length of the regular hexagon be 1 unit. Therefore; Area green is; Area triangle with height 1 unit and base ⅓√(3) units. = ½(⅓√(3)*1) = √(3)/6 square units. Area yellow is; Area triangle with two length 0.40929915222 units and 1 units, and angle (90-atan(⅓(2√(3))))° = ½(0.40929915222*1)sin(90-atan(⅓(2√(3)))) = 0.13397459621 square units. It implies; Area yel...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3249572

Let P be the required area. Calculating P. a² = 6²+3² a² = 45 a = √(45) a = 3√(5) cm. b = a-x b = (3√(5)-x) cm. tanc = 6/3 c = atan(2)° d = atan(½) f = 90+d f = (90+atan(½))° Calculating x ½*3(3√(5)-x)sin(90+atan(½)) = ½*(3+3)xsin(atan(2)) 9-1.3416407865x = 2.683281573x 4.0249223595x = 9 x = (9/4.0249223595) x = 2.2360679775 cm. Calculating area S. It implies; Area S = 0.5*2.2360679775*6sin(atan(2)) Area S = 6 cm² The required area, P is;...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3249570

Calculating the inscribed half circle's diameter. Let a be the radius of the inscribed half circle. b = (1+2a) units. c = (1+a) units. d²+a² = (1+a)² d² = 1+2a+a²-a² d = √(1+2a) units. Therefore; Calculating a. 2(12*a)+a*√(1+2a) = 12(1+2a) 24a+√(1+2a)a = 12+24a √(1+2a)a = 12 a²(1+2a) = 12² a²+2a³ = 144 2a³+a² = 144 It implies; a = 4 units. Therefore, diameter of the inscribed half circle is; 2a = 2*4 = 8 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3247297

Calculating Angle x. Let the two equal lengths be 1 unit each. (a/sin15) = (1/sin30) a = 0.51763809021 units. b = 180-(180-15-30) b = 45° c² = 1²+0.51763809021²-2*0.51763809021cos45 c = 0.73205080757 units. Therefore, the required angle x is; (0.73205080757sin45) = (0.51763809021/sinx) sinx = 0.5 x = asin(0.5) x = 30°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3247292

Calculating Semicircle's Area. Let r be the radius of the half circle. a² = 6²+(½(4))² a² = 36+4 a = √(40) a = 2√(10) units. It implies; 2r² = a² 2r² = (2√(10))² 2r² = 40 r² = 20 r = √(20) units. Therefore, area half circle is; ½(πr²) = ½*π√(20)² = ½*π*20 = 10π square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3246178

Sir Mike Ambrose is the author of the question. Area Blue ÷ Area ABC to 2 decimal places. a = 360-140-80 a = 140° cos20 = 1/b b = 1.06417777248 units. (1.06417777248/sin30) = (c/sin70) c = 2 units. d² = 4+1.064177772482²-4*1.06417777248cos140 d = 2.89711998506 units. (2.89711998506/sin140) = (2/sine) e = 26.34291130662° f = 40-e f = 13.65708869338° g = 20+26.34291130662 g = 46.34291130662° tan46.34291130662 = h/2 h = 2.09602104183 units. Area ABC is; 0...

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Mathematics Question and Solution

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