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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2059473

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Sir Mike Ambrose is the author of the question. Aqua total in cm² is; tan30 = a/10 a = ⅓(10√(3)) cm. sin60 = 2/b b = ⅓(4√(3)) cm. c = a+b c = ⅓(14√(3)) cm. tan30 = d/2 d = ⅓(2√(3)) cm. e = c-d e = 4√(3) cm. Where e is the side length of the green inscribed regular hexagon. f = (120-atan(7√(3)/15))° g² = c²+10² g = ⅓(4√(93)) cm. h = ⅔(√(3)) cm. i = 6-htan60 i = 2 cm. j = 6-i j = 4 cm. k² = h²+2² k = ⅓(4√(3)) cm. It implies; Total Area Aqua exactly in cm² is; Area triangle with he...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2057021

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Let AC be 2 units. It implies; OA = AB = BC = DE = EF = 1 unit. Blue length = 1 unit. Calculating red length. a² = 5-4cos72 a = 1.94008557092 units. a is OD. (1.94008557092/sin72) = (1/sinb)  b = 29.35462809223° b is angle ADO. c = 108+29.35462809223 c = 137.35462809223° c is angle ODE. Calculating OE. (OE)² = 1.94008557092²+1-2*1.94008557092cos137.35462809223 OE = 2.76007862003 units. OE is the radius of the inscribed arc. (2.76007862003/sin137.35462809223) = (1/sind) d = 14.208372507...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2054859

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Sir Mike Ambrose is the author of the question. 12²=10²+10²-2*10*10cosa cosa = (200-144)/200 a = acos(56/200)° a = acos(7/25)° a = 73.73979529169° b = c = ½(180-73.73979529169) b = c = 53.13010235416° d² = 10²+10² d = 10√(2) cm e = ½d e = 5√(2) cm. Let PR = f. f² = 2(5√(2))²-2(5√(2))²cos(90+acos(7/25)) f = PR = 14 cm. g² = 6²+6² g = 6√(2) cm. Let PQ = h = QR h² = 50+72-2*60cos(90+53.13010235416) h = PQ = QR = 14.76482306023 cm. 14²=2(14.76482306023)²-2(14.76482306023)²cosi cosi = (2(1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2046134

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a² =10²-6² a = 8 cm. tanb = 3/4 b = atan(3/4)° (3/4) = c/10 c = ½(15) cm. c = 7.5 cm. sin(atan(3/4)) = d/8 d = 4.8 cm. e = 7.5-4.8 e = 2.7 cm. (3/4) = 2.7/f f = 3.6 cm. g² = 2.7²+3.6² g = 4.5 cm. h² = 10²-4.8² h = ⅖√(481) cm. h = 8.77268487978 cm. j = ⅖√(481)-3.6 j = 8.77268487978-3.6 j = 5.17268487978 cm. Or j = ⅖√(481)-(18/5) j = ⅕(2√(481)-18) cm. k = (108-atan(¾))° It implies; Area Red Exactly is; 0.5*⅕(2√(481)-18)*4.5sin(180-atan(3/4)) = 27(√(481)-9)/50 cm² = 6.98312458771 cm²...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2024003

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Sir Mike Ambrose is the author of the question. sin30 = a/12 a = 6 cm. sin60 = b/12 b = 6√(3) cm. Let the inscribed square side be c. Calculating c. (c/sin60)+(csin30) = 6 (⅓(2√(3))+½)c = 6 ⅙(4√(3)+3)c = 6 c = 36/(4√(3)+3) cm. (a) Area Blue exactly is; c² = (36/(4√(3)+3))² = 1296/(4√(3)+3)² = 1296/(48+24√(3)+9) = 1296/(57+24√(3)) = 432/(19+8√(3)) = 432(19-8√(3))/169 cm².   (b) Calculating Shaded area to 2 decimal places. Notice; Blue square side is; 3.62603375102 cm. tan60 = c/...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2019111

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Sir Mike Ambrose is the author of the question. Let r be the radius of the three congruent circles. Calculating r. (9+r)² = (9-r)²+(9-3r)² 81+18r+r² = 81-18r+r²+81-54r+9r² 36r = 81-54r+9r² 9r²-90r+81 = 0 r²-10r+9 = 0 r ≠ 9 units. r = 1 unit. a² = (2*9²) a = 9√(2) units. b² = 2(1)² b = √(2) units. c = a-b c = 8√(2) units. (8√(2))² = 10²+2²-2*2*10cosd 128 = 104-40cosd cosd = (104-128)/40 d = 126.86989764584° (8√(2)/sin126.86989764584) = (2/sine) e = 8.13010235416° f = 180-d-e f = 45° Th...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2017153

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Let AB be 1 unit. a² = 2-2cos108 a = 1.61803398875 units. a is BE. (1.61803398875/sin156) = (b/sin6) b = 0.41582338164 units. b is BP. (1.61803398875/sin156) = (c/sin18) c = 1.22929666779 units. c is EP. d = 72-18 d = 54° e² = 1+0.41582338164-2*0.41582338164cos54 e = 0.82709091528 units. e is CP. (0.82709091528/sin54) = (1/sinf) f = 78° Notice; f is the supplementary angle of angle BPC. f' = 180-f f' = 180-78 f' = 102° f' is angle BPC. g = 72-8 g = 66° g is angle DEP. h² = 1+1.2292966...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2016187

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Let the the side length of the inscribed hexagon side 2 units. a² = 2-2cos120 a = √(3) units. a = 1.73205080757 units. a is the side length of the inscribed square. Therefore; Area Square is; a² = √(3)² = 3 square units. Calculating Area Purple. sin30 = b/2 b = 1 unit. c = 2b+2 c = 2*1+2 c = 4 units. sin30 = d/1 d = ½ units. e = 4-½-√(3) e = ½(7-2√(3)) units. e = 1.76794919243 units. tan18 = f/1.76794919243  f = 0.57444151456 units. g = 2f g = 1.14888302912 units. g is the side lengt...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2009073

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Let AB = 1 unit. a² = 2-2cos108 a = 1.61803398875 units. a is AD. b = 108-36-18 b = 54° b is angle DAF. c = 108-36-30 c = 42° c is angle ADF. d = 180-54-42 d = 84° d is angle AFD. (1.61803398875/sin84) = (e/sin54) e = 1.31622742593 units. e is DF. (CF)² = 1.31622742593+1-2*1.31622742593cos30 CF = 0.6728163648 units. (0.6728163648/sin30) = (1.31622742593/sinf) f = 78° f is the supplementary angle of angle g. Calculating angle g. g+f = 180 g = 180-f g = 180-78 g = 102° g is angle DCF. T...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2008166

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Let the side length of the regular pentagon be 2 units. a² = 2²-1² a = √(3) units. b² = 5-4cos108 b = 2.49721204096 units. (2.49721204096/sin108) = (2/sinc) c = 49.61382244056° d = 90-c d = 40.38617755944° e² = 3+2.49721204096²-2√(3)*2.49721204096cos40.38617755944 e = 1.62694657231 units. (1.62694657231/sin40.38617755944) = (√(3)/sinf) f = 43.61382244075° g = 180-e-f g = 96° Let the radius of the inscribed circle be r. Calculating r. r(√(3)+1.62694657231+2.49721204096) = (1.6269465723...
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