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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2195528

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Let length AB be 3 unit. Therefore; AC = 20 unit. It implies; Area large square is; 20² = 40 square unit. Area orange is; Area triangle with two side 17 unit and 32.0319709782 unit, and angle (180-atan(10/7))° + Area triangle with two side 20 unit and 44.0408623507 unit, and angle 53.4269678346° - Area square with side 20 unit + Area triangle with height 3 unit and base 4.2857142857 unit. = (0.5*32.0319709782*17sin(180-atan(10/7))) + (0.5*44.0408623507*20sin53.4269678346) - 20² + (0.5*4.2...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2193352

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Let the side of the square be 2 unit. Therefore; Area square is; 2² = 4 square unit. Area S is; Area triangle with two side 0.98921899145 unit and 1.02554900641 unit, and angle 14.318078177° - Area sector with radius 2 unit and angle 28.6361563539° + Area triangle with two side 2 unit each, and angle 28.6361563539° (0.5*0.98921899145*1.02554900641sin14.318078177) - (28.6361563539*2²π÷360) + (0.5*2²sin28.6361563539) = 0.12544441261 - 0.99959042698 + 0.95849162297 = 0.0843456086 space unit....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2191627

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a² = 6²+3² a = √(45) a = 3√(5) units. a is the square side. tanb = (6/3) b = atan(2)° It implies; c = atan(½)° d² = (3√(5))²+(0.5(3√(5)))² d = (15/2) units. d is the radius of the arc. e = (atan(½)+atan(½))° e is the angle of the sector formed by the arc. tan(atan(½)) = f/(0.5*3√(5)) f = ¼(3√(5)) units. g = 3√(5)-¼(3√(5)) g = ¼(9√(5)) units. sin(atan(½)) = h/(¼(9√(5))) h = (9/4) units. cos(atan(½)) = j/(¼(9√(5))) j = (9/2) units. Therefore; Area Shaded exactly as a single fraction is...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2191257

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Let the side of the inscribed square be 2 unit. Therefore; Area S is; Area triangle with two side 1.03528 unit and 0.53589838486 unit, and angle 45° = 0.5*1.03528*0.53589838486sin45 = 0.1961531464 space unit. Area R is; Area triangle with two side (2√(2)-2) unit and 0.65961372961 unit, and angle 72.367664609° = 0.5*(2√(2)-2)*0.65961372961sin72.367664609 = 0.26038499708 space unit. Area T is; Area sector with radius 2√(2) unit and angle 30° - Area triangle with two side 2√(2) unit and 2 u...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2190434

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Sir Mike Ambrose is the author of the question. Area green total, in square unit decimal form is; Area triangle with height 6sin(atan(3/5)) unit and base 6cos(atan(3/5)) unit + Area triangle with height 4 unit and base 4tan(atan(2/5)) unit. = ½*6sin(atan(3/5))*6cos(atan(3/5)) + ½*4*4tan(atan(2/5)) = 7.94117647059 + 3.2 = 11.1411764706 square unit. = 11.14 square units to 2 decimal places.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2187818

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tana = 8/13 a = atan(8/13)° tan(atan(8/13)) = b/10 b = 80/13 units. It implies; Area Blue1 is; ½*10(80/13) = (400/13) square units. = 30.7692307692 square units. Calculating Area Blue2. c² = 10²+8² c = √(164) c = 2√(41) units. d² = 8²+13² d = √(233) units. 233 = 9+164-12√(41)cose 12√(41)cose = 173-233 e = acos((173-233)/(12√(41))) e = acos(-5/√(41))° e = 141.3401917459° f = e-90 f = (acos(-5/√(41))-90)° tan(acos(-5/√(41))-90) = g/3 g = 3.75 units. g = ¼(15) units. h = 12-g h = 12-¼(1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2187323

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Sir Mike Ambrose is the author of the question. Area D exactly in cm² is; Area trapezoid with parallel sides 4 cm and 8 cm, and height 4 cm - Area triangle with height 4 cm and base 3 cm - Area sector with radius 4 cm and angle atan(4/3). = (½*4(4+8)) - (½*4*3) - (atan(4/3)*π*4²÷360) = 24 - 6 - (2πatan(4/3))/45 = 18 - (2πatan(4/3))/45 = (810 - (2πatan(4/3)))/45 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2183313

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Let the side length of the regular hexagon be 1 unit. a² = 1²+1² a = √(2) units. b = 180-120 b = 60° c = 60+45 c = 105° d² = 2+1-2√(2)cos105 d = 1.9318516526 units. (1.9318516526/sin105) = (√(2)/sine) e = 45° f = 180-60-45 f = 75° (1/sin75) = (g/sin60) g = 0.8965754722 units. It implies; Area Green is; 0.5*0.8965754722sin45 = 0.3169872981 square units. = (sin60sin45)/2sin75 square units. h = ½(180-150) h = 15° j = 90-15 j = 75° k = 180-120 k = 60° (1/sin60) = (l/sin45) l = 0.81649...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2181662

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Let the pentagon side length be 1 unit. a = ½(180-108) a = 36° b² = 2-2cos36 b = 0.6180339887 units. Calculating red length (circumference of the inscribed red circle). Let the radius of the re inscribed circle be r. Calculating r. c = ½(180-36) c = 72° Therefore; ½(0.6180339887r+2r) = ½*sin36 2.6180339887r = sin36 r = 0.2245139883 units. Therefore, length red is; 2πr = 2*0.2245139883π = 1.4106629925 units. Calculating blue length. d² = 2-2cos108 d = 1.6180339887 units. 1.618033988...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2180850

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Let the side length of the regular pentagon be 2 units. Therefore; Arc radius is; = 2 units. Calculating Circle Radius. a = 180(3)/5 a =108° b = ½(180-108) b = 36° c = 108-36 c =72° cos72 = 1/d d = 3.2360679775 units. e = 90-72 e = 18° f = 27-18 f =9° sin27 = g/h h = g/sin27 h = 2.2026892646g units. Where g is the radius of the inscribed circle. It implies; (2+g)² = (2.2026892646g)²+3.2360679775²-2*3.2360679775*2.2026892646gcos9 4+4g+g² = 4.8518399964g²+10.472135955-14.0805880855g...
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