Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2539465

Calculating blue area. It is; Area square with side length 3 cm + Area semi circle with radius 1 cm + Area semi circle with radius 2 cm. = (3*3)+(π*½*1²)+(π*½*2²) = 9+½(π)+2π = ½(18+π+4π) = ½(18+5π) cm² = 16.853981634 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2539396

cos30 = a/16 a = 8√(3) cm. cos30 = 10/b b = ⅓(20√(3)) cm. c = a-b c = 8√(3)-⅓(20√(3)) c = ⅓(4√(3)) units. d = 2c d = ⅓(2√(3)) cm. e = b+d e = ⅓(20√(3))+⅓(2√(3)) e = ⅓(22√(3)) cm e = 12.7017059222 cm. f² = 12.7017059222²+8²-16*12.7017059222cos30 f = 7.0237691686 cm. (7.0237691686/sin30) = (8/sing) g = 34.7150039538° h = 180-120-g h = 25.2849960462° tan25.2849960462 = 2/j j = 4.233901974 units. tan30 = k/10 k = ⅓(10√(3)) cm. l = k-j l = 1.5396007179...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2539293

Notice! Radius of the half circle is 9 units. Let the three equal lengths be b. 5² = b²+a²-2bacos45 25 = b²+a²-√(2)ab --- (1). 9² = b²+a²-2abcos135 81 = b²+a²+√(2)ab --- (2). Subtracting (1) from (2). 56 = 2√(2)ab 28 = √(2)ab a = 28/(√(2)b) --- (3). Substituting (3) in (2) to get b. 81 = b²+(28/(√(2)b))²+√(2)(28/(√(2)b))b 81 = b²+(392/b²)+28 53 = b²+(392/b²) 53b² = b⁴+392 Let b² = p. p²-53p +392 = 0 Calculating p. It implies; p = ½(53+√(1241)) p = 44.1139...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2538182

Sir Mike Ambrose is the author of the question. Calculating Yellow Area. It is; Area triangle with height 4sin(atan(⅔)) units and base 8√(13)/9 units - Area triangle with height 1 unit and base ⅔ unit - Area triangle with height ¼ unit and base ⅔ unit = ½*4sin(atan(⅔))*(8√(13)/9) - ½*1*⅔ - ½*¼*⅔ = (32/9)-⅓-(1/12) = (113/36) square units. Area Orange is; Area triangle with height ⅓(10sin(atan(¾))) units and base (5√(13)/9) units. = ½*⅓(10sin(atan(3/2)))*(5√(13)/9) = (25/9) sq...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2538034

a²+2² = 8² a² = 64-4 a = √(60) a= 2√(15) units. a is BD b = ½(a) b = √(15) units. c = ½(AB) c = 4 units. c is the radius of the half circle. cosd = √(15)/4 d = acos(¼√(15))° d is angle ABD. e = 2d e = 2acos(¼√(15))° e is angle ABC. Therefore, the required length, ? (BC) is; Observing CAH, Trigonometry ratio. cos(2acos(¼√(15))) = (BC)/8 BC = 8cos(2acos(¼√(15))) BC = 8*⅞ BC = 7 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2537987

Let the two equal lengths be a units each. a² = 2b² b² = a²/2 b = ½√(2)a unit. It implies; 14+½√(2)a* ½√(2)a = 2a*2a 14+½(a)² = 4a² 14 = 4a²-½(a)² 14 = ½(7)a² 2 = ½(a)² a² = 4 a = 2 units. b = ½√(2)a And a = 2 units. b = ½√(2)*2 b = √(2) units. Therefore; 2a is; = 2*2 = 4 units. c² = 2²+2² c² = 8 c = 2√(2) units. d = b+c d = (√(2)+2√(2)) d = 3√(2) units. Therefore, area blue is; Area square with side 3√(2) units - Blue Area. = (3√(2))²-7 = 18-7...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2537952

a² = 4²+2² a = √(20) a = 2√(5) units. a is the diameter of the circle. b = ½(a) b = √(5) units. b is the radius of the circle. tanc = 4/2 c = atan(2)° d = 180-c d = (180-atan(2))° It implies, the required angle, ? is; Let it be e. Observing Sine Rule. (4/sin(180-atan(2))) = (e/sin45) e = 3.1622776602 units. Again, e is the required length.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2537563

Let the small inscribed regular triangle side length be x. It implies; Calculating x. 2² = x²+(3x)²-2*x*3xcos60 4 = 10x²-3x² 4 = 7x² x = ⅐(2√(7)) units. Therefore; 3x is; = 3*⅐(2√(7)) = ⅐(6√(7)) units. It implies, area blue is; ½*x*3xsin60 = ½*⅐(2√(7))*⅐(6√(7))*½√(3) = ⅐(3√(3)) square units. = 0.742307489 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2537524

tan60 = √(3)/a a = 1 unit. b = 8-a b = 7 units. tanc = √(3)/7 c = atan(√(3)/7)° d = 2c d = 2atan(√(3)/7)° e = 180-60-d e = (120-2atan(√(3)/7))° f = ½(e) f = (60-atan(√(3)/7))° Calculating R, radius of the bigger circle. It implies; tan(60-tan(√(3)/7)) = R/g g = R/tan(60-atan(√(3)/7)) units. tan60 = R/h h = R/tan60 units. It implies; g+h = 8 (R/tan(60-atan(√(3)/7)))+(R/tan60) = 8 1.5396007178R = 8 R = 8/1.5396007178 R = 5.1961524227 units. Again, R...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2537314

a² = 2²+b² a = √(4+b²) a is AB. b is the height of the yellow inscribed triangle. 0.5*b*c = 6 bc = 12 c = 12/b units. c is the base of the inscribed yellow triangle. d² = b²+(12/b)² d² = (b⁴+12)/b² d = √(b⁴+12)/b units. 2 - √(4+b²) b - e Cross Multiply. e = ½√(4+b²)b units. e is BC. f = ½(e) f = ¼√(4+b²)b units. ¼√(4+b²)b - b g - 2 Cross Multiply. g = ½√(4+b²) units. It implies; Calculating b. 12+(2*b)+(½√(4+b²)*½√(4+b²)b) = ½√(4+b²)b*√(4+b²) 12+2b+¼(4+b...

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