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Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2450034

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a = (x+17) cm. a is the length of the the ascribed rectangle. b = a-9 b = (x+17)-9 b = (x+8) cm. b is the width of the ascribed rectangle and also, the diameter of the inscribed semi circle and the radius of the inscribed quarter circle. c² = 2(8+x)² c² = 2(64+16x+x²) c = √(128+32x+2x²) cm. d = ½(b) d = ½(x+8) cm. d is the radius of the inscribed semi circle. Therefore, calculating x. Observing the side length ratios of similar plane shapes (right-angled triangle). x - ½...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2449884

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a² = 2²+2²-2*2*2cos120 a = √(12) a = 2√(3) units. a is the side length of each of the congruent 4 inscribed regular hexagon. Calculating the side length of the ascribed regular hexagon. sin30 = c/a ½ = c/(2√(3)) c = √(2) units. b = a+2(2c+a) b = 2√(3)+2(2√(3)+2√(3)) b = 2√(3)+2(4√(3)) b = 10√(3) units. Again, b is the side length of the ascribed regular hexagon. (10√(3))² = 2d²-2d²cos120 300 = 3d² d² = 100 d = 10 units. d is the side length of the ascribed regular hexagon. sin60 = e/(2√(3...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2444075

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a = 10 units. a is the radius of the circle. b = (20-2x) units. c = ½(b) c = ½(20-2x) c = (10-x) units. It implies; Calculating x. 10² = 2c² 100 = 2(10-x)² 100 = 200-40x+2x² 2x²-40x+100 = 0 x²-20x+50 = 0 Resolving the above quadratic equation via completing the square approach to get x. ( x-10)² = -50+(-10)² ( x-10)² = 50 x-10 = ±√(50) x = 10±5√(2) It implies; x ≠ 10+5√(2) x = 10-5√(2) units. x = 2.9289321881 units. The length, y of the side of the square is;...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2443447

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Let the radius of the inscribed circle be x. a = (3-x) units. b = 7-3 b = 4 units. c = (7-x) units. d = a+c d = (3-x)+(7-x) d = (10-2x) units. Therefore, x (radius of the inscribed circle) is; (2x)²+b² = d² (2x)²+4² = (10-2x)² 4x²+16= 100-40x+4x² 16 = 100-40x 40x = 100-16 40x = 84 10x = 21 x = 2.1 units. Again, x is the radius of the inscribed circle.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2440596

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Let the radius of the ascribed quarter circle be a. It implies; a² = 5²+b² b = √(a²-25) units. 6² = b²+c² c² = 36-b² c = √(36-√(a²-25)²) c = √(36-(a²-25)) c = √(61-a²) units. d = 5+c d = (5+√(61-a²)) units. Calculating a. (5+√(61-a²))² = a²+6² 25+10√(61-a²)+61-a² = a²+36 10√(61-a²)-a² = a²+36-86 100(61-a²) = (2a²-50)² 100(61-a²) = 4a⁴-200a²+2500 6100-100a² = 4a⁴-200a²+2500 4a⁴-100a²-3600 = 0 a⁴-25a²-900 = 0 Let a² = p. p²-25p-900 = 0 Resolving the above quadratic equation via completing...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2439064

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Calculating angle x. Let AB be 1 unit. a = 9-20  a = 70° a is angle ABH. b = 180-10-70 b = 100° (c/sin70) = (1/sin100) c = 0.9541888941 units. (d/sin10) = (1/sin100) d = 0.1763269807 units. sin20 = e/1 e = 0.3420201433 units e is BH. cos20 =f/1 f = 0.9396926208 units. f is AH. g = e-d g = 0.3420201433-0.1763269807 g = 0.1656931626 units. h = 180-20-30 h = 130° (1/sin130) = (j/sin30) j = 0.6527036447 units. k = f-j k = 0.9396926208-0.6527036447 k = 0.2869889761 units. Therefore, the...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2434402

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Calculating x. Notice! AD = 100 cm. Therefore; Area triangle ABD + Area triangle ACD = Area triangle ABC 0.5*2x*100sin60+0.5*x*100sin60 = 0.5*2x*xsin120 100x+50x = x² x = 100+50 x = 150 cm. It implies; AB = 2*x AB = 2*150 AB = 300 cm. AC = x AC = 150 cm. Therefore, calculating BC. Let it be y. y² = 300²+150²-2*300*150cos120 y² = 157500 y = √(157500) y = 150√(7) cm. y = 396.8626966597 cm.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2431654

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Calculating x, side length of the square. a² = 2x² a = √(2)x units. a is BD, diagonal of the square. tanb = x/(0.5x) b = atan(2)° c = 2b c = 2atan(2)° c is angle DEF. d² = 2(0.5x)²-2(0.5x)²cos(2atan(2)) d = 0.894427191x units. d is DF. e = ½(180-2atan(2)) e = 26.5650511771° e is angle EDF. f = 90-45-26.5650511771 f = 18.4349488229° f is angle FDB. Therefore; x, the square side is; 0.5*0.894427191x*√(2)xsin18.4349488229 = 20 0.2x² = 20 (x²/5) = 20 x² = 100 x = √(100) x = 10 units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2428076

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Let the side length of the regular hexagon be 2 units. a = ⅙(180(6-2)) a = 120° a is the single interior angle of the regular hexagon. b² = 2²+1²-2*2*1cos120 b = √(7) units. b = 2.6457513111 units. (2.6457513111sin120) = (1/sinc) c = 19.1066053509° d = 120-c d = 120-19.1066053509 d = 100.8933946491° Therefore, e is; e = 0.5*√(7)*2cos100.8933946491 e = √(7)cos100.8933946491 e = 2.5980762114 square units. e is the area of green plus orange inscribed plane shape (inscribed scalene triangle)...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2413845

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x = 12.416666 - (1). 10x = 124.16666 - (2). 100x = 1241.6666 - (3). 1000x = 12416.6666 - (4). Subtracting (3) from (4). 900x = 11175 It implies x (p/q) is; x = 11175/900 x = 2235/180 x = 447/36 x = 149/12
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