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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2017107

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sin60 = a/20 a = 10√(3) cm. a+b = 20 b = (20-10√(3)) cm. c = 20-2b c = 20-2(20-10√(3)) c = 20-40+20√(3) c = (20√(3)-20) cm.  c = 14.64101615138 cm. Therefore; Shaded Area Red is; ½(14.64101615138)²+4((30π*20*20/360)-(½*20*20sin30)) = 107.17967697245+4(104.71975511966-100) = 107.17967697245+4*4.71975511966 = 126.05869745109 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2017071

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Notice; Radius r of the circle is 6 units. Area Red is; 6(area sector with radius 6 units and angle 6°-area triangle with base 6 units and height 6sin60 units) = 6(60π*36/360 - ½*36sin60) = 6(6π-9√(3)) = 18(2π-3√(3)) square units. = 19.56659192051 square units.

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/2014569

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Let AB be 2 units. a = atan(½)° b² = 2²+1² b = √(5) units. Let r be the radius of the inscribed circle. Calculating r. sin(atan½) = r/c c = r/sin(atan½) units. It implies; 1+r+(r/sin(atan½)) = √(5) r+(r/sin(atan½)) = √(5)-1 r+√(5)r = √(5)-1 r = (√(5)-1)/(1+√(5)) r = ½(3-√(5)) units. (a) Circle Radius ÷ AB exactly is; ½(3-√(5)) ÷ 2 = ¼(3-√(5)) (b) C Area ÷ Area Shaded is; C area is; 2² - ½*4sin(90-2atan½) - ½*4sin(2atan½) - (180-2atan2)π/360 - ½*sin(2atan2) = 4 - (6/5) - (8/5) - 0.46...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2014567

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Let r be the radius of the inscribed semi circle. Calculating r. r² = (r-4)²+(r-8)² r² = r²-8r+16+r²-16r+64 r²-24r+80 = 0 It implies; r ≠ 4 cm. r = 20 cm. a² = 20²+16² a = 4√(41) cm. (4/sinb) = (4√(41)/sin45) b = 6.34019174591° c² = 2(20)² c = 20√(2) cm. Area A is; ½*4*20√(2)sin45 - ½(4²) - (6.34019174591*π*20²/360) = 40 - 8 - 22.13144423478  = 9.86855576522 cm² d = atan(20/12) d = atan(5/3)° Therefore; e = atan(⅗)° f² = 20²+12² f = 4√(34) cm. (4/sing) = (20/sin(atan⅗)) g = 5.9061...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2012023

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Area Shaded exactly in square units as a single fraction is; Area triangle with height 6√(3) units and base 6 units - Area triangle with height 2√(3) units and base 2 units - Area sector with radius 2 units and angle 120° - Area sector with radius 6 units and angle 60° = ½*6√(3)*6 - ½*2√(3)*2 - (120π*4/360) - (60π*36/360) = 18√(3) - 2√(3) - ⅓(4π) - 6π = 16√(3) - ⅓(4π) - 6π = ⅓(48√(3)-4π-18π) = ⅓(48√(3)-22π) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2012019

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Let the side of the equilateral triangle be 2 units. Let the side of the square be a. Calculating a. tan60 = a/1 a = √(3) units. Area square is; a² = √(3)² = 3 square units. Calculating area green. b² = 2√(3)² b = √(6) units. c = 180-60-45 c = 75° (√(3)/sin75) = (d/sin60) d = 1.55291427062 units. e = b-d e = 0.89657547217 units. (f/sin45) = (√(3)/sin75) f = 1.26794919243 units. (1.26794919243/sin45) = (0.89657547217/sing) g = 30° sin30 = h/(√(3)+1) h = ½(√(3)+1) units. Theref...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2011345

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Let the shortest length of the composite plane shape be 1 unit. (1/sin12) = (a/sin153) a = 2.18357369878 units. (2.18357369878/sin27) = (b/sin99) b = 4.75051853365 units. c² = 4.75051853365+1-2*4.75051853365cos39 c = 4.02290116111 units. Calculating the required angle x. (4.02290116111/sin39) = (1/sinx) x = 9°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2010957

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Let the a side length of the yellow regular triangle be 2 units. Therefore, a side length of the regular pentagon is 2 units. Area Yellow is; 0.5*2*2sin60 = 2sin60 = √(3) square units. = 1.73205080757 square units. Calculating r, radius of the circle. a = 108+0.5*60 a = 108+30 a = 138° cos30 = 1/b b = 1.15470053838 units. r² = 2²+1.15470053838²-2*2*1.15470053838cos138 r = 2.96070468869 units. (2.96070468869/sin138) = (1.15470053838/sinc) c = 15.12743321513° d = 180-108-15.12743321513...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2010716

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Notice; GH = 8 units. Calculating the side length of the regular pentagon. a = 108-90 a = 18° a is angle EFG. b = 180-108-18 b = 54° b is angle EGF. (8/sin108) = (c/sin18) c = 2.59935756986 units. c is EG. d = 90-54 d = 36° d is angle DGH. cos36 = 8/e e = 9.88854382 units. e is DG. Therefore the regular pentagon side length (DE) is; c+e = 2.59935756986+9.88854382 = 12.48790138986 units. f = ½(12.48790138986) f = 6.24395069493 units. f is AJ. tan36 = g/8 g = 5.81234022404 units. g is...
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