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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3017238

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Sir Mike Ambrose is the author of the question  Let the side of the regular Pentagon be 2 units. Therefore; Area orange is; Area triangle with height 1.7013016167 units and base 1.8345763534sin(18) units = ½*1.7013016167*1.8345763534sin(18) = 0.48224693327 square units. Area blue is; Area trapezoid with parallel sides 0.21654236465 units and 1 unit, and height 1.9021130326 units. ½*1.9021130326(0.21654236465+1) = 1.15700054326 square units. It implies; Area Orang...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3015837

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a² = 2*12²-2*12²cos120 a = 12√(3) cm. a = 20.78460969083 cm. a = AB. b = 2a b = 24√(3) cm. b = 41.56921938165 cm. b = AC. c² = 12²+41.56921938165²-24*41.56921938165cos30 c = 31.74901573277 cm. c = CO1 (31.74901573277/sin30) = (12/sind) d = 10.89339464913° d = Angle O1CA. e = 180-10.89339464913-30 e = 139.10660535087° e = Angle AO1C. f = asin(12/31.74901573277)  f = 22.2076542986° f = Angle O1CD. cos22.2076542986 = g/31.74901573277 g = 29.39387691339 cm. g = C...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3014345

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a² =2*12² a = 12√(2) cm. a = AC. 12b+12b+12√(2)b = 12*12 b = 144/(24+12√(2)) b = 12-6√(2) cm. b = 3.51471862576 cm. b = radius of the inscribed circle. BF = ½(12√(2)) BF = 6√(2) cm. c² = (6√(2))²+3.51471862576²-12√(2)*3.51471862576cos45 c = 6.49435320175 cm. c = EF. (6.49435320175/sin45) = (6√(2)/sind) d = 112.5° d = Angle BEF. e = 2(180-112.5-45) e = 45° f = 112.5-90 f = 22.5° cos22.5 = 12/g g = 12.98870640351 cm. g = EG. FG = 12.98870640351-6.494353201...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3012696

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a² = 15²+12²-2*12*15cos120 a = 23.43074902772 cm. Where a is the side length of the equilateral triangle ABC. (23.43074902772/sin120) = (12/sinb) b = 26.32950349168° c = 60-b c = 33.67049650832° d² = 15²+15²-2*15*15cos120 d = 25.98076211353 cm. e = atan(7.5/25.98076211353) e = 16.10211375199° f = 90-e f = 73.89788624801° g = 180-f-c g = 72.43161724367° (15/sin72.43161724367) = (h/sin73.89788624801) h = 15.11661227595 cm. i = a-h  i = 8.31413675177 cm. Let...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3011353

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Sir Mike Ambrose is the author of the question. Area Green exactly is;  Area trapezoid with parallel sides 3 cm and (6-2√(3)) cm, and height (6-3√(3)) cm. = ½(6-3√(3))(3+(6-2√(3))) = ½(6-3√(3))(9-2√(3)) = ½(72-39√(3)) cm² = 3(24-13√(3))/2 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3009704

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a² = 12²-6² a = 6√(3) cm. a = 10.39230484541 cm. a = circle's radius = AD = AG. b² = 2*10.39230484541²-2*10.39230484541²cos120 b = 18 cm. b = DG. c = 12-10.39230484541 c = 1.60769515459 cm. c = CD. d² = 12²+1.60769515459²-24*1.60769515459cos60 d = 11.28239078631 cm. d = BD. (11.28239078631/sin60) = (1.60769515459/sine) e = 7.088663081° e = Angle CBD. f = 60-7.088663081 f = 52.911336919° f = Angle DBG. (12/sing) = (10.39230484541/sin52.911336919) g = 112.9113369...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3007952

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Let AB be 5 units. Radius, r of the inscribed circle is; 5*(2/5) r = 2 units. a = ⅛(180*6) a = 135° 2b² = 25 b = ½(5√(2)) units. c = 180-0.5(135)-45 c = 67.5° sin22.5 = ½(5√(2))/d d = 9.23879532511 units. Where d is the side length of the regular triangle (purple area). Area purple is; 0.5*9.23879532511*9.23879532511sin60 = 36.95994598698 square units. Calculating Area Shaded. e = 360-135-45-67.5-60 e = 52.5° f² = 50-50cos135 f = 9.23879532511 u...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3006311

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AB = BC = AC = 6 cm. cos60 = 2/BE BE = 4 cm. (DE)² = 4²-2² DE = 2√(3) cm. (CE)² = (2√(3))²+4² CE = 12+16 CE = 2√(7) cm EF = (2√(7)-4) cm. EF = 1.29150262213 cm. AE = 6-4 AE = 2 cm. Angle AEF = 180-(atan(2/2√(2√(3)))+atan(4/2√(3))) Angle AEF = 100.89339464913° Therefore; Area Shaded, Triangle AEF is; 0.5*2*1.29150262213sin100.89339464913 = 1.26823039846 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3004810

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a = ½(180-150) a = 15° b = 90-15 b = 75° (c/sin45) = (12/sin60) c = 9.79795897113 cm. (12/sin60) = (d/sin45) d = 9.79795897113 cm. e² = 9.79795897113²+12²-24*9.79795897113cos135  e = 20.15631110909 cm. (12/sinf) = (20.15631110909/sin135) f = 24.89609063899° g = 120-24.89609063899 g = 95.10390936101° h = 180-24.89609063899-75 h = 80.10390936101° (9.79795897113/sin80.10390936101) = (i/sin75) i = 9.60704430088 cm. Therefore; Area Shaded Region is; 0.5*9...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3003488

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The radius of the circle is 6 cm. a = atan(2)° Where a is angle AOB = angle COE. Angle BOE = 180-atan(2) = 116.56505117708° Angle OBE = ½(180-116.56505117708) = 31.71747441146° b = 90+31.71747441146 b = 121.71747441146° Where b is angle ABE. (12/sin31.71747441146) = (c/sin121.71747441146) c = 19.416407865 cm. Where c is AE. sin(atan2) = d/19.416407865 d = 17.366563146 cm.  cos(atan2) = e/19.416407865 e = 8.683281573 cm. f = 12-8.683281573   f = 3.316718427 c...
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