Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2753087

Let a be the radius of the quarter circle. b = ½(a) units. tanc = b/a c = atan(½)° d = atan(2)° e = d-c e = (atan(2)-atan(½))° Therefore, the required length, x is; tan(atan(2)-atan(½)) = x/8 x = 8tan(atan(2)-atan(½)) x = 6 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2750886

Total red area is; 3(area square with side 5 units - area quarter circle with radius 5 units) + Area triangle with height 5 units and base 10 units - Area sector with radius 5 units and angle 2atan(½)° - Area triangle with two side 5 units each, and angle (180-2atan(½))°. = 3(25-¼(25π)+½(50)-(2atan½÷360*25π)-½(25)sin(180-2atan½) = 75-18.75π+25-11.591190225-10 = 19.5039475202 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2750851

Calculating x. a = 4+1 a = 4 units. a is the radius of the quarter circle. b²+4² = 5² b = √(25-16) b = 3 units. tanc = 3/4 c = atan(3/4)° d = 90-c d = atan(4/3)° It implies; x² = 5²+5²-2*5*5cos(atan(4/3)) x = 4.472135955 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2747671

Area Blue is; Area square with side length 4 cm - Area quarter circle with radius 4 cm + 2(area square with side length 2 cm). = (4*4)-(¼*4*4π)+2(2*2) = 16-4π+8 = 24-4π = 4(6-π) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2745071

Let the side length of the square be a. b = ½(a) cm. b is the radius of the inscribed half circle. c = (a-2) cm. d²+2² = (a-2)² d² = a²-4a+4-4 d = √(a²-4a) cm. d is the radius of the bigger inscribed quarter circle. e = b+d e = (½(a)+√(a²-4a)) cm. Calculating a. a²+(½(a))² = (½(a)+√(a²-4a))² ¼(5a²) = ¼(a²)+a√(a²-4a)+a²-4a 4a = a√(a²-4a) 4 = √(a²-4a) a²-4a-16 = 0 (a-2)² = 16+(-2)² (a-2)² = 20 a = 2±√(20) a = 2±2√(5) It implies; a ≠ 2-2√(5) a = 2+2√(5) cm....

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2744501

Let the radius of the quarter circle be 2 units. Therefore; OD = BD = ½(2) OD = BD = 1 units. a² = 1²+2²-2*1*2cos45 a = 1.4736257582 units. a is CD. (1.4736257582/sin45) = (2/sinb) b = 73.6750500631° Therefore, the regular angle, theta is; Let it be c. c = 90-b c = 90-73.6750500631 c = 16.3249499369° Again, c is the required angle, theta.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2738443

Purple area is; Area triangle with height 4 units and base 8 units - (area square with side 4 units - area quarter circle with radius 4 units) - area sector with radius 4 units and angle (180-2atan(½))° + area triangle with two side 4 units each, and angle (180-2atan(½))°. = ½(4*8) - (4²-¼(4²π)) - (180-2atan(½))*4²π÷360 + ½(4²)sin(180-2atan(½)) = 16 -16 + 4π - 17.7143794847 + 6.4 = 4π - 11.314379485 = 1.25199112936 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2736984

Let the radius of the quarter circle be a. b = (a-4) units. c = (a-2) units. It implies; a² = (a-4)²+(a-2)² a² = a²-8a+16+a²-4a+4 a²-12a+20 = 0 a²-2a-10a+20 = 0 a(a-2)-10(a-2) = 0 (a-2)(a-10) = 0 It implies; a = 10 units. Again, a is the radius of the quarter circle. Therefore, area quarter circle is; ¼*10*10π = 25π square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2735481

Blue area is; 8(area quarter circle with radius 4 cm - area triangle with height and base 4 cm each) - 8(area quarter circle with radius 2 cm - area triangle with height and base 2 cm each) = 8(4π-8) - 8(π-2) = 32π-64-8π+16 = 24π-48 = 24(π-2) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2733657

a = (2+x) units. a is the radius of the ascribed sector. b = 2x units. x is the radius of the inscribed quarter circle. Calculating x. (2+x)² = x²+(2x)² 4+4x+x² = 5x² x²-x-1 = 0 (x-½)² = 1+(-½)² (x-½)² = (5/4) x = ½±√(5/4) x = ½±½√(5) It implies; x = ½(1+√(5)) units. x = 1.6180339887 units. Therefore, area inscribed quarter circle is; ¼(x²)π = ¼*½(1+√(5))*½(1+√(5))π = π(1+√(5))²/16 = 2.0561990865 square units.

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