Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2730809

Area red is; Area semi circle with radius 5 cm - Area sector with radius 5 cm and angle (180-2atan(2))° - Area kite with diagonal lengths √(10²+5²) cm and √(200-200cos(2atan(½))) cm + Area sector with radius 10 cm and angle 2atan(½)° - 2(area quarter circle with radius 5 cm - area triangle with height and base 5 cm respectively). = 12.5π - 11.591190225 - 50 + 46.3647608999 - 14.2699081699 = 9.77357067487 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2730778

Area red is; Area square with length 5 cm - Area quarter circle with radius 5 cm - Area sector with radius 2.5 cm and angle (180-2atan(2))° - Area kite with diagonal lengths √(5²+(2.5)²) cm and √(50-50cos(2atan(½))) cm + 2(area sector with radius 5 cm and angle 2atan(½)°) - Area triangle with two side 5 cm each, and angle 2atan(½)° + Area sector with radius 2.5 cm and angle 2atan(2)° - Area triangle with two side 2.5 cm each, and angle 2atan(2)° = 5² - ¼(5²π) - (180-2atan(2))*2.5²π÷360...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2727569

Let a be the radius of the quarter circle. b = x° c = (45-x)° It implies; sinx = 1/a --- (1). sin(45-x) = 2/a --- (2). At (1). a = 1/sinx --- (3). Substituting (3) in (2). sin(45-x) = 2/(1/sinx) sin(45-x) = 2sinx sin45cosx-cos45sinx = 2sinx ½√(2)(cosx-sinx) = 2sinx cosx-sinx = 2√(2)sinx cosx = 2√(2)sinx+sinx cosx = (2√(2)+1)sinx cosx/sinx = (2√(2)+1) sinx/cosx = 1/(2√(2)+1) tanx = (2√(2)-1)/7 x = atan((2√(2)-1)/7) x = 14.6388065952° Calculating a, radi...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2726665

½*a*b = 10 a = (20/b) cm. a is CD. b is OD. b² - 30 (20/b)² - 10 Cross Multiply. 10b² = 12000/b² 10b⁴ = 12000 b⁴ = 1200 b⁴ = 1200 b² = √(1200) b² = 20√(3) b = √(20√(3)) b = 5.8856619128 cm. Again, b is OD. Recall. a = (20/b) And b = 5.8856619128 cm. a = 20/5.8856619128 a = 3.3980884897 cm. Again, a is CD. Therefore, the required red angle is; Let it be c. tanc = b/a c = atan(5.8856619128/3.3980884897) c = atan(1.7320508076) c = atan(√(3))° c = 60° A...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2720030

Notice. The complete plane shape is half a circle with radius 8 cm. Area green is; Area semi circle with radius 8cm - 2(area quarter circle with radius 8 cm - area triangle with height and base 8 cm respectively). = (½*8*8π)-2((¼*8*8π)-(½*8*8)) = 32π-2(16π-32) = 32π-32π+64 = 64 cm² Or (¼*8*8π)+(¼*8*8π)-(¼*8*8π)+(½*8*8)-(¼*8*8π)+(½*8*8) = 2(½*8*8) = 8*8 = 64 cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2714408

Pink area is; ½(area square with side 4 cm - area quarter circle with radius 4 cm) = ½(4*4 - ¼(16π)) = ½(16 - 4π) = 8 - 2π = 2(4-π) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2713691

Let the radius of the ascribed quarter circle be 1 unit. It implies; R = 1 unit. 2a² = 1² a² = ½ a = ½√(2) units. a is the side length of the inscribed square. b = 2R b = 2 units. b is the diameter of the complete circle. c² = 2²+(½√(2))²-2*2*½√(2)cos45 c² = 4+½-2 c = √(5/2) c = ½√(10) units. c = 1.5811388301 units. (1.5811388301/sin45) = (2/sind) d = 63.4349488229° sin63.4349488229 = e/(0.5√(2)) e = 0.632455532 units. cos63.4349488229 = f/(0.5√(2)) f = 0.316...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2711850

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/2709806

Let the ascribed quarter circle radius be 1 unit. 2a² = 1² a² = ½ a = ½√(2) units. a = 0.7071067812 units. a is the radius of the blue inscribed quarter circle. Area blue inscribed quarter circle (S2) is; ¼*πa² = ¼(½√(2))²π = ⅛π square units. c = (1-b) units. b is the radius of the yellow inscribed semi circle. It implies, calculate b. (1-b)² = b²+(½√(2))² 1-2b+b² = b²+½ 2b = ½ b = ¼ units. b = 0.25 units. Again, b is the radius of the yellow semi circle. Area...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2697637

Let the inscribed square side length be a. b² = 2a² b = √(2)a units. b is AC = BD, the diagonal of the inscribed square. a² = 2c² Where c is OA = OD. c² = ½(a²) c = √(a²/2) c = ½(√(2)a) units. d² = a²+(½(√(2)a))²-2a*½(√(2)a)cos135 d² = a²+½(a²)+a² d² = 2a²+½(a²) d² = ½(5a²) d = √(2.5)a unit. d is OP = OQ, the radius of the quarter circle. e = √(2.5)a-½(√(2)a) e = 0.8740320489a units. e is AP = DQ. Notice. PC = 2√(5) units. PC = 4.472135955 units. Calculati...

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