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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3002001

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Let GD (inscribed square side) be a. Calculating a. 2(a/(tan60))+a = 12 a = 12/((2/√(3))+1) a = 5.56921938165 cm. b = ½(12-5.56921938165) b = 3.21539030918 cm. Where b is BD = CE. c = 3.21539030918+5.56921938165 c = 8.78460969083 cm. Where c is CD. d² = 3.21539030918²+12²-24*3.21539030918cos60 d = 10.75890566601 cm. Where d is AD. (10.75890566601/sin60) = (12/sine) e = 105° Where e is angle ADB. f = 105-90 f = 15° Where f is angle ADG. cos15 = 5.56921938165/g g = 5.7656801693 cm. Where...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3000310

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a² = 12²-6² a = 6√(3) cm. Where a is BD. b = ½(a) b = 3√(3) cm. Where b is BG = DG. c = atan(6/(3√(3)) c = 49.10660535087° Where c is angle AGD = angle BGE. d = 60-49.10660535087 d = 10.89339464913° Where d is angle BAE. e = 180-60-10.89339464913 e = 109.10660535087° Where e is angle BEA. (12/sin109.10660535087) = (f/sin60) f = 10.99818166789 cm. Where f is AE. (12/sin109.10660535087) = (g/sin10.89339464913) g = 2.4 cm. Where g is BE. h = 12-2.4 h = 9.6 cm....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2998547

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Let a be the radius of the inscribed circle. Calculating a. tan30 = a/6 a = 2√(3) cm. b = inscribed circle diameter. b = 4√(3) cm. 12² = c²+6² c = 144-36 c = √(108) c = 6√(3) cm. d = 6√(3)-4√(3) d = 2√(3) cm. Where d is AD. e² = 12²+(2√(3))²-2*12*2√(3)cos30 e = 2√(21) cm. Where e is BD. (2√(21)/sin30) = (12/sinf) f = 139.10660535087° Where f is angle BDA. g = 180-30-139.10660535087  g = 10.89339464913° Where g is angle ABD. h = 60-10.89339464913  h = 49.10660535087° Where h is angle FB...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2996921

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18² = 14²+16²-2*14*16cosa 448cosa = 14²+16²-18² a = acos((14²+16²-18²)/448) a = 73.39845040098° (18/sin73.39845040098) = (14/sinb) b = 48.18968510422° c = 180-48.18968510422-73.39845040098 c = 58.4118644948° cotd = cota+cotb+cotc cotd = 1.80748828181  d = acot(1.80748828181) d = 28.95373454979° e = 58.4118644948-d e = 29.45812994501° f = 180-28.95373454979-73.39845040098 f = 77.64781504923° (14/sin77.64781504923) = (g/sin73.39845040098)  g = 13.73434221224 units. It implies; Area Blue...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2995142

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a² = 11*11+10*10-2*11*10cos50 a = 8.92113926968 cm. (8.92113926968/sin50) = (10/sinb) b = 59.16920318624° c = 180-50-59.16920318624 c = 70.83079681376°  d = 0.5a d = 4.46056963484 cm. e² = 4.46056963484²+5.5²-2*4.46056963484*5.5cos59.16920318624 e = 5 cm. (5/sin59.16920318624) = (5.5/sinf) f = 70.83079681376° g² = 4.46056963484²+5²-2*4.46056963484*5cos70.83079681376 g = 5.5 cm. (5.5/sin70.83079681376) = (5/sinh) h = 59.16920318624° i = 180-59.16920318624-70.83079681376 i = 50° (j/tan...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2993552

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Sir Mike Ambrose is the author of the question. Radius of the semicircle is 2 units. Therefore; Shaded area exactly in decimal square units is; Area trapezoid with parallel sides 2 units and (6-3√(3)) units, and height 3 units - Area sector with radius 2 units and angle (120-atan(3-√(3)))°. = ½*3(2+(6-3√(3))) - ((120-atan(3-√(3)))*2*2π÷360) = ½(24-9√(3)) - 2.38279212814 = 4.20577136594 - 2.38279212814 = 1.8229792378 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2991948

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Radius, r1 of the inscribed semicircle is; sin60 = r1/6 r1 = 3√(3) cm. Radius, r2 of the inscribed circle is; tan60 = 3/r2 r2 = 3/√(3) r2 = √(3) cm. It implies; Area of the coloured inscribed circle is; π(r2)² = π(√(3)²) = 3π cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2990381

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Sir Mike Ambrose is the author of the question. Let the base of the parallelogram be 2 units. Therefore the area of the parallelogram is; 2(area triangle with height (sin70/sin35) unit and base 2sin75 unit). = 2*½*(sin70/sin35)*2sin75 = 3.16496046094 square units. Area shaded is; Area triangle with height 0.55803775975 unit and base 1.17917770336sin55 unit. = ½*0.55803775975*1.17917770336sin55 = 0.26951154209 square unit. It implies; The percentage of the shaded part of the parallelo...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2989231

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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 green is; Area triangle with height ½ units and base √(3) units + Area triangle with height 1.30930734142 units and base 1.13389341903 units. = ½*½*√(3) + ½*1.13389341903*1.30930734142 = ¼√(3) + 0.74230748896 = 1.17532019085 square units. Area orange is; Area triangle with height (2/5) units and base (½sin60) units. = ½*(2/5) )*½*sin60 = √(3)/20 square...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2989096

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Area yellow is; 2(area triangle with height (4.5/cos15) cm and base (4.5/cos15)sin60 cm + Area square with side (4.5/cos15) cm. = 2*½*(4.5/cos15)²(√(3)/2) + (4.5/cos15)² = (4.5/cos15)²(√(3)/2) + (4.5/cos15)² = 18.7961154131 + 21.7038845869 = 40.5 cm²
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