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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1944040

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Let the side of the regular decagon which is equal the side of the regular octagon be 1 unit. Calculating Area Gold. a = ⅛(180*6) a = 135° b = ½(360-2(135)) b = 45° sin45 = c/1 c = ½(√(2)) units. d = 1+2c d = (1+√(2)) units. Area Gold is; Area regular octagon with side 1 unit - Area trapezoid with parallel sides 1 unit and (1+√(2)) units and height ½(√(2)) units. = ½(8*1)*(1/(2tan(180/8))-0.5(1+1+√(2))*(√(2)/2) = 4.82842712475-1.20710678119  = 3.62132034356 square units. Calculating...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1943729

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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 = 112.91133691899° g = Angle BEG. Angle...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1942412

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Let the side of the regular pentagon be 1 unit. a = 360-2(108)-90 a = 54° tan72 = b/0.5 b = 1.53884176859 units. tan54 = 1.53884176859/c c = 1.11803398875 units. d = c-0.5 d = 0.61803398875 unit. (e/tan36)+(e/tan27) = d Where e is the radius of the inscribed circle. e = 0.61803398875/((1/tan36)+(1/tan27)) e = 0.18509595408 unit. f = 180-72-54 f = 54° Therefore; g = d = 0.61803398875 unit. Shaded Area Red is; Area triangle with height 0.61803398875 units and base 0.61803398875sin72...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1942103

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a² = 6²+8²-2*6*8cos30 a = 4.10628314132 cm. a = DE. (4.10628314132/sin30) = (8/sinb) b = 103.06431342951° b = Angle ADE. Angle AED = 180-30-103.06431342951  = 46.93568657049° tan30 = c/6 c = 3.46410161514 cm. c = CH. d = 8-2*3.46410161514  d = 1.07179676972 cm. e = 1.07179676972+3.46410161514 e = 4.53589838486 cm. f = atan(6/4.53589838486)  f = 52.91133691903° g = 90-52.91133691903  g = 37.08866308097° h = 37.08866308097-30 h = 7.08866308097° i = 60+46.93568657049 i = 106.93568657049°...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1941077

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Calculating r, radius of the circle. a = 2+½(1) a = (2½) a = (5/2) units. r² = (½)²+b² b = √(r²-(¼)) units. c = 3-b c = (3-√(r²-(¼))) units. It implies; r² = (5/2)²+(3-√(r²-¼))² r² = (25/4)+9-6√(r²-¼)+r²-¼ 6√(r²-¼) = 6+9 6√(r²-¼) = 15 2√(r²-¼) = 5 4r²-1 = 25 4r² = 26 √(4)r = √(26) 2r =√(26) r = ½√(26) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1940597

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Let a side length of the three congruent regular pentagon be 1 unit. Single interior angle of the three congruent regular pentagon is; ⅕(180*3) = 108° a² = 1²+0.5²-2*1*0.5cos108 a = 1.24860602048 units. (1.24860602048/sin108) = (1/sinb)  b = 49.6138224408° c² = 2-2cos108 c = 1.61803398875 units. d = c+1+1 d = 3.61803398875 units. e² = 1+3.61803398875²-2*3.61803398875cos108 e = 4.04057395938 units. (4.04057395938/sin108) = (1/sinf) f = 13.6138224408° g = 180-108-13.6138224408 g = 58.386...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1940540

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a² = 8²-2² a = √(60) a = 2√(15) units. Let the radius of the inscribed circle be r. Considering similar triangle ratios, calculating r. r = 2√(15) 1 = 2 Cross Multiply. r = √(15) units. b² = 1+r² b² = 1+√(15)² b = √(16) b = 4 units. c = 2+4+√(15)+x c = (6+√(15)+x) units. d² = (x+√(15))²-√(15)² d² = x²+2√(15)x+15-15 d = √(x²+2√(15)x) Again, considering similar triangle ratios to get x. √(15) = 2√(15) √(x²+2√(15)x) = (6+√(15)+x) Cross Multiply √(15)(6+√(15)+x) = 2√(15)*√(x²+2√(15)x) (6√(1...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1938983

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Notice; Angle DCE = -(Angle BCG) Let Angle DCE be x. Calculating K. 4k² = 4k²+k²-4k*kcosx cosx = (¼) --- (1). 144 = 4k²+k²-2k²(-cosx) 144 = 5k²-2k²(-cosx) Substituting (1) in (2). 144 = 5k²-2k²(-¼) 144 = 6k² k² = 24 k = 2√(6) units. It implies; 2k is 4√(6) units. Total square areas is; (4√(6))²+(2√(6))² = 16*6+4*6 = 6(16+4) = 6(20) = 120 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1938157

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Let a be the side of the inscribed square. 24² = a²+(2a)² 576 = a²+4a² 576 = 5a² a = √(576/5) a = 10.733126292 cm. b = ½(24-10.733126292)  b = 6.633436854 cm. b = AM = DR. c = atan(6.633436854/(6+10.733126292)) c = 21.62463392714° c = Angle BPQ. tan21.62463392714 = d/10.733126292 d = 4.25488424946 cm. d = MQ. e = 0.5(10.733126292)+4.25488424946 e = 9.62144739546 cm. e = MO. f = 9.62144739546+12 f = 21.62144739546 cm. f = DM. g = atan(12/21.62144739546) g = 29.03047057929° g = Angle CMD...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/1937325

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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
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