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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3141300

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Calculating Shaded Area. a = 4(4) a = 16 units. a is AB. b = a+4 b = 20 units. b is AC. It implies; 15²+20² = 7²+c² 225+400-49 = c² c = 24 units. c is CE. d = ½(c) d = 12 units. d is CD = DE. d' = atan(7/12) d' = 30.2564371635° d' is angle EDF. e² = 12²+7² e² = 144+49 e = √(193) units. e = 13.8924439894 units. e is DF. f = atan(15/20)+ atan(7/24) f = 53.1301023542° f is angle ACE. g² = 4²+12²-2*4*12cos53.1301023542 g = 10.1192885125 units. g is BD. (10.1192885125/sin53.1301023542) = (...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3140912

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Calculating alpha. Let it be a. Let the three equal lengths be 1 units each. b = 180-30-7.5 b = 142.5 ° b is angle AKC. Notice. Angle CAK = Angle KBC = 7.5° (1/sin30) = (c/sin7.5) c = 0.26105238444 units. c is CK. Calculating a. (0.26105238444/sin7.5) = (1/sind) d = 30° e = 180-30 e = 150° Therefore, a (alpha) is; a+b+e = 360 a = 360-142.5-150 a = 67.5°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3140911

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Calculating Shaded Area. a² = 1²-(4k)² a = √(1-16k²) units. 1 ~ √(1-16k²) 4k ~ b Cross Multiply. b = 4k√(1-16k²) units. Again. 1 ~ √(1-16k²)  √(1-16k²) ~ c Cross Multiply. c = (1-16k²) units. d = 1-c d = 1-(1-16k²) d = 16k² units. e = 1-b e = (1-4k√(1-16k²)) units. Therefore; (5k)² = (16k²)²+(1-4k√(1-16k²))² 25k² = 256k⁴+1-8k√(1-16k²)+16k²(1-16k²) 25k² = 256k⁴+1-8k√(1-16k²)+16k²-256k⁴ 9k² = 1-8k√(1-16k²) 8k√(1-16k²) = 1-9k² 64k²(1-16k²) = 1-18k²+81k⁴ 64k²-1024k⁴ = 1-18k²+81k⁴...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3139969

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a = x° b = (90-x)° R² = 6²+16²-2*6*16cosx R² = 292-192cosx  cosx = (292-R²)/192 --- (1). Notice. At (1). (292-R²) is adjacent. 192 is hypotenuse. Let c be opposite. c² = 192²-(292-R²)² c² = 36864-85264+584R²-R⁴ c = √(-R⁴+584R²-48400) units. It implies; sinx = √(-R⁴+584R²-48400)/192 --- (3). Again. R² = 10²+6²-2*10*6cos(90-x) R² = 136-120(cos90cosx+sin90sinx) R² = 136+120sinx --- (4). Substituting (3) in (4). R² = 136+120(√(-R⁴+584R²-48400)/192...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3139968

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Let the radius of the ascribed circle be 2 units. Therefore, radius of the bigger inscribed circle is 1 unit. a = (1+y) units  b = (2-y) units. c = (1-y) units. Calculating y, radius of the big inscribed circle. a² = c²+d² (1+y)² = (1-y)²+d² 1+2y+y² = 1-2y+y²+d² d² = 4y d = √(4y) d = 2√(y) units. It implies; d²+y² = b² (2√(y))²+y² = (2-y)² 4y+y² = 4-4y+y² 8y = 4 y = ½ units. Again, y is the radius of the big inscribed circle. Recall. d = 2√(y) And y = ½ units. d = 2√(½) d = √(2) unit...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3139965

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Notice. The ascribed quadrilateral is a square. Let it's side length be x. a²+x² = 3² a = √(9-x²) units.    2 ~ 3 b ~ x Cross Multiply. b = ⅓(2x) units. It implies; Calculating x. x²-½(x*√(9-x²)) = ½(x+⅓(2x))*x 2x²-x√(9-x²) = ⅓(5x)*x 2x-√(9-x²) = ⅓(5x) x = 3√(9-x²) x² = 81-9x² 10x² = 81 x = ⅒(9√(10)) units. x = 2.84604989415 units. Again, x is the side length of the square. cosc = 2.84604989415/3 c = 18.434948823° sin18.434948823 = d/2 d = 0.63245553204 units. e = x-d e = 2.8...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3139219

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Let OA be 1 unit. Area Triangle AOB is; ½*1*1  = ½ square units. Calculating Area Triangle ODG. a² =2(1)²-2(1)²cos30 a = √(2-√(3)) units. a = 0.51763809021 units. a is AC, the side length of the big square. b² = 2(0.51763809021)² b = 0.73205080758 units. b is AD, the diagonal of the big square. c² = 1²+0.73205080758²-2*0.73205080758cos120° c = 1.50597117916 units. c is OD. (1.50597117916/sin120) = (1/sind) d = 35.1039093611° d is angle ADO e = 180-35.1039093611-120 e = 24.8960906389° e...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3139214

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Let a = 1 unit. It implies; 1 ~ b b ~ e Cross Multiply. e = b² f = (1-b²) units. cos60 = g/(1-b) g = ½(1-b) units. h = 2g+1 h = (2-b) units. It implies; Calculating b. (2-b)² = (1-b²)²+1²-2(1-b²)cos120 4-4b+b² = 1-2b²+b⁴+1+1-b² 1-4b+4b² = b⁴ b⁴+0b³-4b²+4b-1 = 0 It implies; b ≠ 1 unit. b ≠ -(1+√(2)) units. b = (√(2)-1) units. b = 0.41421356237 units. Recall. e = b² e = (√(2)-1)² e = (3-2√(2)) units  e = 0.17157287525 units. It impl...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3139211

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Calculating length RT. a = ½(10) a = 5 units. a is half the diagonal of the rectangle. b = 8+a b = 13 units. It implies; (RT)²+5² = 13² RT = √(169-25) RT = √(144) RT = 12 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3139208

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Calculating x, length BE. a = 90-75 a = 15° a is angle CBE. sin15 = b/x b = (xsin15) units. b is CE. c = 5+b c = (5+(xsin15)) units. c is CD. cos15 = d/x d = (xcos15) units. d is BC. Notice. c = d (5+(xsin15)) = (xcos15) 5 = xcos15-xsin15 x(cos15-sin15) = 5 x = 5/(cos15-sin15) x = 5/(½√(2)) x = 5√(2) units.
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