Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2745916

White area is; Area sector with radius 6 cm and angle 120° + Area sector with radius 2 cm and angle 60° = (120*36π÷360) + (60*4π÷360) = 12π + ⅔(π) = ⅓(38π) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2745908

Green area is; Area circle with radius 5 units + 2(area sector with radius 20 units and angle 60°) - Area equilateral triangle with side 20 units. = π(5²) + 2(60÷360*20²π) - ½(20²)sin60 = 25π + ⅓(400π) - 100√(3) = ⅓(475π-300√(3)) square units. = 324.213756061 square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2745900

Let the radius of the inscribed circle be r. The radius of the semi circle is 5 units. Therefore; (2r)² + r² = 5² 4r² + r² = 25 r = √(5) units. Therefore shaded area/area blue is; Area semi circle with radius 5 units - Area circle with radius √(5) units. = ½(25π) - π√(5)² = ½(25π) - 5π = ½(25π-10π) = ½(15π) square units.

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2745892

Sir Mike Ambrose is the author of the question. AB is 2 cm. BC is 6 cm. The single side length of equilateral triangle BDE is; = ⅓(16√(3)) cm. Therefore; Area Orange is; Area triangle with two side length ⅓(10√(3)-6) cm and ⅓(36-16√(3)) cm, and angle 60° = ½(⅓(36-16√(3))*⅓(10√(3)-6)*½√(3) = ½(⅓(36-16√(3))*(5-√(3))) = ⅓(18-8√(3))(5-√(3)) = ⅓(90-40√(3)-18√(3)+24) = ⅓(114-58√(3)) cm²

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2745353

sin30 = a/12 a = 6 cm. Where a is the height of the green regular triangle. sin60 = 6/b ½√(3) = 6/b b = 12/√(3) b = 4√(3) cm. b is the side length of the green regular triangle. tan30 = c/(0.5*4√(3)) 1/√(3) = c/2√(3) c = 2 cm. cos60 = d/12 d = 6 cm. sin60 = e/12 e = 6√(3) cm. cos30 = f/12 f = 6√(3) cm. g = f-d+½(b) g = 6√(3)-6+2√(3) g = (8√(3)-6) cm. h = e-d-c h = (6√(3)-8) cm. Therefore, blue length is; Let it be j. j² = (8√(3)-6)²+(6√(3)-8)² j² =...

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/2744523

Let the side length of the regular pentagon be 2 units. a = ⅕*180(5-2) a = 108° a is the single interior angle of the regular pentagon. b² = 1²+2²-2*1*2cos108 b = 2.497212041 units. (2.497212041/sin108) = (1/sinc) c = 22.3861775592° d = 108-c d = 85.6138224408° e = 180-d e = 94.3861775592° f = 180-108-c f = 49.6138224408° Therefore, the required angle, x is; x = 180-e-f x = 180-94.3861775592-49.6138224408 x = 36°

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2744516

Let the side length of the ascribed regular hexagon be 1 unit. a = ⅙*180(6-2) a = 120° a is the single interior angle of the ascribed regular hexagon. b = ¼(a) b = 30° d = (1-c) c is the radius of the inscribed circle. Calculating c. tan30 = c/(1-c) √(3)c = 1-c (1+√(3))c = 1 c = 1/(1+√(3)) units. c = 0.3660254038 units. Recall. d = 1-c d = 0.6339745962 units. e² = 2-2cos120 e = √(3) units. f² = √(3)²+0.6339745962²-2√(3)*0.6339745962cos30 f = 1.2247448714...

Mathematics Question and Solution

https://onlineedumath.com/update/mathematics-question-and-solution/2744510

Calculating r, the radius of the blue inscribed circle. Therefore; Consider triangle OBC. (6-r)² = r² + x², and x = √(9+6r) - 3 It implies; (6-r)² = r² + (√(9+6r)-3)² 36-12r+r²=r²+9-6√(9+6r)+(9+6r) 36-12r=9-6√(9+6r)+9+6r 6√(9+6r) = 18r-18 √(9+6r) = 3r-3 9+6r = (3r-3)² 9+6r = 9r²-18r+9 24r = 9r² Therefore r, radius of the blue inscribed circle is; r ≠ 0 r = (8/3) units. Therefore area inscribed blue circle is; π(8/3)² = 64π/9 square units.

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.

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