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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2789357

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Calculating Silver Area. a² = 6²+(0.5*16)² a = √(100) a = 10 cm. a is the radius of the inscribed sector. tanb = 8/6 b = atan(4/3)° c = 2b c = 2atan(4/3)° c is the angle of the inscribed sector. Area silver is; Area square with side length 16 cm - Area rectangle with length 8 cm and width 6 cm - Area sector with radius 10 cm and angle 2atan(4/3). = (16*16)-(8*6)-(2atan(4/3)π*10*10/360) = 256-48-(5atan(4/3)π/9) = 208-(5atan(4/3)π/9) = ⅑(1872-(5atan(4/3)π)) cm² = 115...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2789324

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Sir Mike Ambrose is the author of the question. Area green total exactly in decimal cm² is; Area triangle with two side (12-6√(3)) cm each and angle 60° + Area triangle with side 3.46410161514 cm and 2.5358983849 cm, and angle 60° + Area triangle with side 6 cm and 3.10582854123 cm, and angle 45° = 1.11920087684 + 3.80384757735 + 6.58845726812 = 11.5115057223 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2788622

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a² = 1²+4² a = √(17) units. a is the diameter of the circle. r, radius of the circle is; ½(a) r = ½√(17) units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2787686

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Notice! AB = BC Let AB = BC be a. b²+(0.5(8+10))² = a² b = √(a²-81) units. c = ½(10+8)-8 c = 1 unit. √(a²-81) - 1 9 - √(a²-81) Cross Multiply. a²-81 = 9 a² = 90 a = √(90) a = 3√(10) units. Calculating x. x²+(3√(10))² = 10² x² = 100-90 x² = 10 x = √(10) units. x = 3.1622776602 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2787678

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Sir Mike Ambrose is the author of the question. AC = 12 cm. AB = 10 cm. Therefore Area red exactly in cm² is; Area triangle with two side (165/90) cm and √(61) cm, and angle atan(5/6)° = ½*(165/90)*√(61)sin(atan(5/6)) = 4.58333333333 cm² = 4125/900 cm² = 55/12 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2786963

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Calculating blue area (area inscribed half circle). 15² = 13²+14²-2*13*14cosa 364cosa = 140 a = 67.380135052° (15/sin67.380135052) = (13/sinb) b = 53.1301023542° c = 180-a-b c = 59.4897625938° Let d be the radius of the inscribed blue half circle. Calculating d. sin59.4897625938 = d/e e = 1.1607142857d units. sin53.1301023542 = d/f f = 1.25d Notice. e+f = 15 1.1607142857d+1.25d = 15 2.4107142857d = 15 d = 6.2222222223 units. d = 56/9 units. It implies,...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2786035

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a = ½(360-90) a = ½(270) a = 135° b² = 6²+(4√(2))²-2*6*4√(3)cos135 b = 2√(29) units. b = 10.7703296143 units. 2c² = b² 2c² = (2√(29))² 2c² = 4*29 c² = 58 c = √(58) units. c is the radius of the quarter circle. Area yellow (quarter circle area) is; ¼(πc²) = ¼*π*√(58)² = ½(29π) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2785871

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Calculating Area Yellow. Let a be the radius of the ascribed semi circle. b²+6² = a² b = √(a²-36) units. b is the radius of the inscribed half circle. Therefore, area yellow is; ½(a²)π-½*π√(a²-36)² = ½(πa²)-½(πa²)+½*36π = 18π square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2785224

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Let the side length of the regular hexagon be 1 unit. a = ⅙*180(6-2) a = 120°  a is the single interior angle of the regular hexagon. sin30 = b/1 b = ½ units. cos30 = c/1 c = ½√(3) units. d = 1+b d = ½(3) units. e = 1+c e = ½(2+√(3)) units. tanf = d/e f = atan(1.5/(½(2+√(3)))) f = 38.793976887° sin38.793976887 = 1.5/g g = 2.394170171 units. h = ½(g) h = 1.1970850855 units. j = 90-f j = 51.206023113° k² = 1²+1.1970850855²-2*1.1970850855cos51.206023113...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2784424

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Sir Mike Ambrose is the author of the question. a² = 10²+16²-2*10*16cos120 a = 2√(129) units. Where a is length AC. (3√(129)/sin120) = (16/sinb) b = 37.58908946897° c = 90-b c = 52.41091053103°  d = 60-b d = 22.41091053103° e = 90-d e = 67.58908946897°  f = 90-60 f = 30° cos37.58908946897 = g/10 g = 7.92405815693 units. i = 30+37.58908946897 i = 67.58908946897° tan67.58908946897 = j/7.92405815693 j = 19.2148198619 units. Where j is the radius of the arc...
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