Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2573455

Let a be the side length of the inscribed square. b² = 2a² b = √(2)a units. b is the radius of the half circle. tanc = a/√(2)a c = atan(1/√(2))° d² = a²+(√(2)a)² d² = a²+2a² d = √(3)a units. e = (2+√(3)a) units. f = 2b f = 2√(2)a f = √(2)a unit. Calculating a. cos(atan(1/√(2))) = (2+√(3)a)/(2√(2)a) 0.8164965809 = (2+√(3)a)/(2√(2)a) 2.3094010767a-√(3)a = 2 0.5773502691a = 2 a = 3.4641016156 units. a = 2√(3) units. Again, a is the side length of the inscribed green square. Area green squ...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2573357

Let a be the radius of the inscribed semi circle. Therefore; Calculating a. (2a)²+a² = 25² 5a² = 25*25 a = 5√(5) units. 2a is; 2*5√(5) = 10√(5) units. Therefore, red shaded region is; Area triangle with height 10√(5) units and base 5√(5) units - Area square with side 5√(5) units + Area quarter circle with radius 5√(5) units. = ½*10√(5)*5√(5)-(5√(5))²+¼(5√(5))²π = 125-125+¼(125π) = ¼(125π) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2572556

a² = 3²+4² a = 5 units. a is the diagonal of the ascribed rectangle. 3b+5b+4b = 12 b = 1 unit. b is the radius of the inscribed circle. c = 3-1 c = 2 units. tand = 2/1 d = atan(2)° e = 360-90-2d e = (270-2atan(2))° e = 143.1301023542° f = ½(180-e) f = ½(36.8698976458) f = 18.4349488229° Therefore, the required angle, Beta, let it be g. g = 90-45-f g = 45-18.4349488229 g = 26.5650511771° Therefore tang (tangent beta) is; tang = tan26.5650511771 = ½

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2572533

Calculating Area Blue. a = ⅙*180(6-2) a = 120° a is the single interior angle of the regular hexagon. sin30 = b/5 b = 2.5 c. c = 2b+5 c = 10 cm. d = ½(60) d = 30° sin60 = e/5 2e = 5√(3) e = ½(5√(3)) cm. Therefore, area blue is; (0.5*5*5sin120)+(0.5*10*½(5√(3))sin30) = ¼(25√(3))+¼(25√(3)) = ½(25√(3)) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2572501

tana = 3/4 a = atan(3/4)° b = ½(180-a) b = ½(180-atan(3/4)) b = 71.5650511771° Therefore, r, radius of the quarter circle is; sin71.5650511771 = r/5 r = 4.7434164903 units. Or c = 4+5 c = 9 units. (2r)² = 3²+9² 4r² = 9+81 4r² = 90 2r = √(90) r = ½√(90) r = ½(3√(10)) units. r = 4.7434164903 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2572288

Sir Mike Ambrose is the author of the question. Equation of the curve is; y = 2x-¼(x²) The vertex coordinate (maximum turning point) is at; (4, 4) Therefore; Shaded Area in square units is; Area square with side 4 units - Area rectangle with length (39/16) units and width (3/2) units - (area under the curve at x = (3/2) and x = 4) = 16-((39/16)*(3/2))-(835/96) = 16-(117/32)-(835/96) = (175/48) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2572264

Calculating Area of the inscribed black circle. Let a be the radius of the inscribed black circle. b = (2+a) units. c = (4-a) units. It implies; Calculating a. (4-a)² = (2+a)²+2² 16-8a+a² = 4+4a+a²+4 16-8a = 8+4a 8 = 12a a = ⅔ units. Again, a is the radius of the inscribed black circle. Area inscribed black circle is; πa² = π*⅔*⅔ = ⅑(4π) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2572210

a = 180-120+50 a = 110° a is angle ABC. b² = 320²+190²-2*320*190cos110 b = 424.3697084246 km. (424.3697084246/sin110) = (320/sinc) c = 45.1198323317° d = 360-60-c d = 254.8801676683° d ≈ 254.880° d is the bearing of A from C. Therefore, considering the cardinal point e = d-180 e = 74.880° It implies; e = S74.880°W Again, e is the bearing of A from C.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2571701

Let a be the area of the ascribed circle. b = ½(a) units. c = (a+2) units. d = ½(c) d = ½(a+2) units. d is the radius of the big inscribed circle. e = (a-2) units. f = ½(e) f = ½(a-2) units. f is the radius of the small inscribed circle. Therefore; Calculating a, radius of the ascribed circle. ½(a)*½(a) = 2*(½(a+2)+½(a+2)-2) ¼(a²) = 2(a) It implies; a² = 8a a = 8 units. Again, a is the radius of the ascribed square. d = ½(a+2) And a = 8 units. d = ½(8+2) d = 5 units. Again, d is the r...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2571651

Calculating area green inscribed quarter circle. Let it's radius be a. It implies; 2a² = 4² a² = 8 a = 2√(2) units. Therefore, area green inscribed quarter circle is; ¼*π*a² = ¼*π*2√(2)*2√(2) = ¼*π*8 = 2π square units.

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