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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3422425

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Sir Mike Ambrose is the author of the question.Let the two equal marked lengths of triangle ABC be 2 units.Therefore;Area triangle ABC is;Area triangle with height 5.75877048314 units and base (5.75877048314sin20) units.= 0.5*5.75877048314*5.75877048314sin20= 5.67128181961 space units.Area purple is;Area triangle with height 4 units and base (1.11985686148sin28.334490436) units.= 0.5*4*1.11985686148sin28.334490436= 1.06300877373 space units.It implies;Area Purple ÷ Area ABC to 2 d. p. is;1.0630...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3395146

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Sir Mike Ambrose is the author of the question.Calculating Area Triangle ABC ÷ Area Purple.Let the two equal side length of the ascribed isosceles triangle be 1 unit.Therefore;Area triangle ABC is;½(1*1sin36)= ½(sin36) space units.= 0.29389262615 square units.Area purple is;Area triangle with height 0.37567713613 units and base (0.15621533715sin138) units.= 0.5*0.37567713613*0.15621533715sin138= 0.01963447686 square units.It implies;Area Triangle ABC ÷ Area Purple to 2 d. p. is;0.29389262615...

Mathematics Questions and Solution

https://onlineedumath.com/update/mathematics-questions-and-solution/3032283

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Sir Mike Ambrose is the author of the question. Let the side of the inscribed square be 2 units. Therefore; Area S is; Area triangle with two side 1.03528 units and 0.53589838486 units, and angle 45° = 0.5*1.03528*0.53589838486sin45 = 0.1961531464 space units. Area R is; Area triangle with two side (2√(2)-2) units and 0.65961372961 units, and angle 72.367664609° = 0.5*(2√(2)-2)*0.65961372961sin72.36766460 = 0.26038499708 space units. Area T is; Area sector with radius 2√(2) units a...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2900060

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Sir Mike Ambrose is the author of the question. 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 units and 16 units, and height 1 unit - Area triangle with height 16 units...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2899870

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#Sir Mike Ambrose is the author of the question. Let the side length of the regular Pentagon be 1 unit. Therefore Area Pentagon is; (5*1)÷(4tan(180/5)) = 5/(4tan36) space units. = 1.72047740059 space units. Area star is; 5(area triangle with height 0.61803398875 units and base 0.61803398875sin36 units) + Area regular Pentagon with side 0.38196601125 units. = 5(0.5*0.61803398875*0.61803398875sin36) + ((5*0.38196601125²)/(4tan(180/5))) = 0.56128497072 + 0.25101426986 = 0.812...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2861830

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Sir Mike Ambrose is the author of the question. Let the side of the square be 2 units. Therefore; Area square is; 2² = 4 square units. Area S is; Area triangle with two side 0.98921899145 units and 1.02554900641 units, and angle 14.318078177° - Area sector with radius 2 units and angle 28.6361563539° + Area triangle with two side 2 units each, and angle 28.6361563539° = (0.5*0.98921899145*1.02554900641sin14.318078177) - (28.6361563539*2²π÷360) + (0.5*2²sin28.6361563539)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2858517

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Sir Mike Ambrose is the author of the question. Let the side of the inscribed square be 2 units. Therefore; Area S is; Area triangle with two side 1.03528 units and 0.53589838486 units, and angle 45° = 0.5*1.03528*0.53589838486sin45 = 0.1961531464 space units. Area R is; Area triangle with two side (2√(2)-2) units and 0.65961372961 units, and angle 72.367664609° = 0.5*(2√(2)-2)*0.65961372961sin72.367664609 = 0.26038499708 space units. Area T is; Area sector with...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2853602

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Sir Mike Ambrose is the author of the question. Let the two equal marked lengths of triangle ABC be 2 units. Therefore; Area triangle ABC is; Area triangle with height 5.75877048314 units and base (5.75877048314sin20) units. = 0.5*5.75877048314*5.75877048314sin20 = 5.67128181961 space units. Area purple is; Area triangle with height 4 units and base (1.11985686148sin28.334490436) units. = 0.5*4*1.11985686148sin28.334490436 = 1.06300877373 space units. It implies;...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2848589

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Sir Mike Ambrose is the author of the question. Shaded area is; Area under the curve at points (-1, 0) and (7, 0) - Area triangle with height 4 units and base 8 units. = ⅓(208) - ½(4*8) = ⅓(160) space units. Area S is; Area under the curve at points (-2, 0) and (7, 0) - Area triangle with height 4 units and base 8 units - Area triangle with height 1 unit and base (7/4) units - Shaded area. = 72 - ½(4*8) - ½(1(7/4)) - ⅓(160) = (43/24) space units. Area R is; Area tr...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2846881

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Sir Mike Ambrose is the author of the question. Let the side of the regular hexagon be 1 unit. Therefore; Area blue is; 2(area triangle with height 1 unit and base (⅓√(3)) units). = 2(½*1*(⅓√(3))) = ⅓√(3) space unit. Area orange is; Area triangle with height (3-√(3))/6 units and base (√(3)-1) units. = ½*(3-√(3))/6*(√(3)-1) = (2√(3)-3)/6 space units. It implies; Area Blue ÷ Area Orange Exactly is; ⅓√(3) ÷ (2√(3)-3)/6 = 2√(3) ÷ (2√(3)-3) = (2√(3)*(2√(3)+3))...
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