Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3137614

Let the side length of the inscribed regular triangle be a. 2b² = a² b = √(a²/2) b = ½√(2)a units. c = 2b c = √(2)a units. c is the side length of the ascribed square. d² = b²+c² d² = (√(2)a)²+(½√(2)a)² d² = 2a²+¼(2a²) d² = ¼(10a²) d = ½√(10)a units. tane = √(2a)/(½√(2)a) e = atan(2)° f = 180-45-60-atan(2) f = 11.5650511771° Calculating a. 0.5*a*0.5√(10)asin11.5650511771 = 24 0.15849364905a² = 24 a² = 151.425625846 a = 12.3055120107 units. Again, a is the is the side length of the inscr...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3137609

Let AB be 1 unit. Let BC be a. Therefore; CD = (1-a) units. It implies; tan60 = b/(0.5a) 2b = √(3)a b = ½√(3)a units. b is the height of the bigger inscribed equilateral triangle. c = 1-b c = ½(2-√(3)a) units. d = ½(a) units. tan60 = 2e/(1-a) 2e = √(3)-√(3)a e = ½(√(3)-√(3)a) units. e is the height of the big inscribed equilateral triangle. f = b-e f = ½√(3)a-½(√(3)-√(3)a) f = ½(2√(3)a-√(3)) units. g = ½(a)+½(1-a) g = ½(1) g = ½ units. It implies;...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3136722

Let angle AOB be a. Let b = AD = OB = OD. b is the radius of the quarter circle. Calculating b. 2² = 2b²-2b²cosa 2b²cosa = 2b²-4 b²cosa = b²-2 cosa = (b²-2)/b² --- (1). 4² = b²+(2b)²-4b²cosa 16 = 5b²-4b²cosa 4b²cosa = 5b²-16 cosa = (5b²-16)/(4b²) --- (2). Equating (1) and (2). (b²-2)/b² = (5b²-16)/(4b²) 4(b²-2) = 5b²-16 4b²-8 = 5b²-16 b² = 8 b = √(8) b = 2√(2) units. Again, b is the radius of the quarter circle.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3136720

Let r = 1 unit. a = (R+1) units. b = (R-1) units. c²+b² = a² c²+(R-1)² = (R+1)² c² = R²+2R+1-(R²-2R+1) c = √(4R) c = 2√(R) units. d = R+2-R d = 2 units. e = R+R e = 2R units. Calculating R. (2R)² = 2²+(2√(R))² 4R² = 4+4R R²-R-1 = 0 (R-½)² = (-½)²+1 (R-½)² = (5/4) R = ½±½√(5) It implies; R ≠ ½(1-√(5)) units. R = ½(1+√(5)) units. Therefore; R÷r is; ½(1+√(5))÷1 = ½(1+√(5))

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3135861

Notice. The ascribed quadrilateral is regular (square). Let a be it's side length. ½*ab = 3 ab = 6 b = (6/a) units. b is the base of the red area. ½*ac = 4 ac = 8 c = (8/a) units. c is the base of the green area d = a-(6/a) d = (a²-6)/a units. d is the height of the yellow area. e = a-(8/a) e = (a²-8)/a units. e is the base of the yellow area. Therefore; Calculating a. ½*d*e = 5 ½*(a²-6)/a*(a²-8)/a = 5 ((a²-6)(a²-8))/a² = 10 (a⁴-14a²+48) = 10a² a⁴-24a²...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3135859

tan60 = a/(0.5*8) a = 4tan60 a = 4√(3) cm. a is the diameter of the circle. b = ½(a) b = 2√(3) cm. b is the radius of the circle. c = ¼(8) c = 2 cm. c is the radius of the two equal half circles. Calculating shaded area. It is; 2(area sector with radius 2√(3) cm and angle 120° - area triangle with height 2√(3) cm and base 2√(3)sin120 cm)+2(area sector with radius 2√(3) cm and angle 60° - area triangle with height 2√(3) cm and base 2√(3)sin60)+2(area sector with radius 2 cm and angle 60° -...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3135856

a = 4+4 a = 8 units. a is the radius of the quarter circle. cosb = ⅛(0.5*4) b = acos(¼)° cosb = c/8 c = 8cos(acos(¼)) c = ¼(8) c = 2 units. c is the width of the inscribed rectangle. sinb = d/8 d = 8sin(acos(¼)) d = 7.74596669241 units. Therefore, area inscribed rectangle is; cd = 2*7.74596669241 = 15.4919333848 square units.r

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3135177

Let AD = BC = 1 unit. a = 180-100-30 a = 50° a is angle BAC. (1/sin50) = (b/sin30) b = 0.65270364467 units. b is AC. Therefore, the required angle theta is, let it be c. (0.65270364467/sinc) = (1/sin100) sinc = 0.65270364467sin100 sinc = 0.64278760969 c = asin(0.64278760969) c = 40°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3135172

Let BD = CD = 1 unit. a = 180-45-30 a = 105° a is angle CAD. (1/sin105) = (b/sin45) b = 0.73205080757 units. b is AC. c² = 0.73205080757²+2²-2*2*0.73205080757cos30 c = 1.41421356237 units. c is AB. ( 1.41421356237/sin30) = (2/sind) d = 45° e = 180-d e = 180-45 e = 135° e is angle BAC. Therefore, the required angle alpha is, let it be f. f = e-a f = 135-105 f = 30°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3135154

a = 180-60 a = 120° b = 180-120-45 b = 15° (1/sin15) = (c/sin45) c = 2.73205080757 units. sin60 = d/c d = 2.73205080757sin60 d = 2.36602540378 units. cos60 = e/c e = 2.73205080757cos60 e = 1.36602540378 units.ñ f = 2-e f = 2-1.36602540378 f = 0.63397459622 units. Therefore, the required angle is, let it be g. tang = d/f g = atan(2.36602540378/0.63397459622) g = 75°

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