Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3133538

b = √(a) b = √(48) b = √(16*3) b = 4√(3) units. b is the radius of the ascribed quarter circle. c = ½(b) c = 2√(3) units. c is the radius of the inscribed half circle. tan30 = d/c ⅓√(3) = d/(2√(3)) d = 2 units. e = b-d e = (4√(3)-2) units. e = 4.92820323028 units. e is the side length of the inscribed regular triangle. f = ½(e) f = (2√(3)-1) units. tan60 = g/f g = √(3)(2√(3)-1) g = (6-√(3)) units. g = 4.26794919243 units. g shares the inscribed regular triangle to two equal right-angled t...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3129465

Let a be the radius of the circ. Calculating a. b = ½(12+2) b = 7 units. c = ½(4+6) c = 5 units. d = b-2 d = 7-2 d = 5 units. It implies; a² = c²+d² a = √(5²+5²) a = √(50) a = 5√(2) units. Again, a is the radius of the circle. Calculating blue area. e² = 6²+12² e = 6√(5) units. sinf = (0.5*6√(5))/(5√(2)) f = 71.5650511771° g = 2f g = 143.130102354° h² = 4²+2² h = 2√(5) units. sinj = (0.5*2√(5))/(5√(2)) j = 18.4349488229° k = 2j k = 36.86989...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3126411

Let the bigger inscribed semi circle radius be 2 units. Let the big inscribed semi circle radius be r. Let the inscribed circle radius be q. Calculating r. (2+r)² = 2²+(4-r)² 4+4r+r² = 4+16-8r+r² 4r = 16-8r 12r=16 3r=4 r=4/3 unit. Calculating q. (2+(4/3))²=(2+q)²+((4/3)+q)² (10/3)²=4+4q+q²+(16/9)+(8q/3)+q² (100/9)-(16/9)-4=2q²+4q+(8q/3) (84/9)-4=2q²+4q+(8q/3) 84-36=18q²+36q+24q 48=18q²+60q 6q²+20q-16=0 3q²+10q-8=0 Therefore; q≠-4...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3125099

Notice. The inscribed square area is 3 square units. Therefore the single side length of the inscribed square is √(3) units. The radius of the ascribed quarter circle will be; √((√(3))²+(√(3))²) = √(6) units. Notice that the green inscribed triangle is a right-angled triangle. The marked angle will be; 2(90-tan–¹(√(6)/√(3))) = 2(90-tan–¹(√(2))) = (180-2tan–(√(2))) The hypotenuse of the green inscribed right-angle triangle is; √((√(3))²+(√(6))²) = √(9) = 3 uni...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3123251

Let r be the radius of the quarter circle. Let x be the interior angle facing 1 cm. a = (x+45)° a is alternate to y, the angle formed by the meet point of r and 2 cm at the circumference of the circle. It implies; y = (x+45)° sinx = 1/r --- (1). cosy = 2/r cos(x+45) = 2/r cosxcos45-sinxsin45 = 2/r ½√(2)(cosx-sinx) = 2/r --- (2). Notice. At (1). Sinx = 1/r Where; 1 cm = opposite. r cm = hypotenuse. Therefore, let b be the adjacent. It implies; b²+1²...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3121528

Sir Mike Ambrose is the author of the question. Let the radius of the semi circle be 3 units. Therefore shaded area ÷ area square will be; (½(area quarter circle of radius 3 units - area triangle of height and width 3 units respectively) + area triangle of two sides ((3√(2)/2)-(3√(3)tan22.5)/2) units and ((3√(2)/2)-(3√(3)tan22.5)/2)xsin(tan–¹(3)-45)÷sin(180-tan–¹(3))) units, and the angle they form, angle 45°) ÷ (area square of single side length ½(3√(2)) units). {½((9π/4)-(9/2)) +...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3121510

Let the radius of the ascribed quarter circle be R. Let the radius of the inscribed circle be r. Therefore; 2²+(2r)²=R² 4+4r²=R² Therefore; R²-4r²=4 Multiplying through by ¼(π). It implies; ¼(πR²) - πr² = π Notice. ¼(πR²) = area of the ascribed quarter circle. πr² = area of the inscribed circle. Therefore area of the red shaded region in the given quarter circle is; π square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3120259

Notice. Calculating the required angle, alpha. 4 units is the radius of the quarter circle. It implies; 3² = 4²+4²-2*4*4cosa 9 = 32-32cosa 32cosa = 32-9 a = acos(23/32) a = 44.0486256741° b = ½(180-a) b = 67.975687163° c = 90-b c = 22.024312837° d² = 4²+2²-2*4*2cos22.024312837 d = 2.273236245 units. It implies, the required angle, alpha is; Let it be f. (2.273236245/sin22.024312837) = (4/sine) e = 41.2886037977 f = 180-e f = 138.7113962023° Again, f...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3115438

Notice. r, radius of the ascribed quarter circle is (25/√(2)) units. r = ½(25√(2)) units. (AD)² = 2(½(25√(2)))² (AD)² = 2*¼*625*2 AD = √(625) AD = 25 units. It implies; AB+BC+CD = AD 10+BC+4 = 25 BC = 25-14 BC = 11 units. a = 25-2(4) a = 17 units. b = ½(a) b = ½(17) units. sinc = ½(17)/(½(25√(2))) c = asin(17√(2)/50)° c = 28.7397952917° d = 2c d = 2asin(17√(2)/50)° e = ½(90-d) e = (45-asin(17√(2)/50))° f = ½(180-e) f = ½(180-45+asin(17√(2)/50)) f = (0.5(135+asin(17√(2)/50)))° g = f...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/3113330

Calculating Area Blue. a = 4+8 a = 12 cm. a is the radius of the ascribed quarter circle. b² = 12²+4² b = √(144+16) b = √(160) b = 4√(10) units. tanc = 12/4 c = atan(3)° (12/sin(60+atan(3))) = (4/sind) d = 14.4420678686° e = 180-60-atan(3)-d e = 33.9928809543° f = 90-e f = 56.0071190457° (12/sin(60+atan(3))) = (g/sin33.9928809543) g = 8.96692386239 cm. Therefore blue area is; (0.5*8.96692386239*4√(10)sin60)+(56.0071190457π*12²/360)-(0.5*12²sin56.0071190457) = 49.1138647086+70.38062...

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